Introduction to highdimensional data
Overview
Teaching: 30 min
Exercises: 20 minQuestions
What are highdimensional data and what do these data look like in the biosciences?
What are the challenges when analysing highdimensional data?
What statistical methods are suitable for analysing these data?
How can Bioconductor be used to access highdimensional data in the biosciences?
Objectives
Explore examples of highdimensional data in the biosciences.
Appreciate challenges involved in analysing highdimensional data.
Explore different statistical methods used for analysing highdimensional data.
Work with example data created from biological studies.
What are highdimensional data?
Highdimensional data are defined as data with many features (variables observed). In recent years, advances in information technology have allowed large amounts of data to be collected and stored with relative ease. As such, highdimensional data have become more common in many scientific fields, including the biological sciences, where datasets in subjects like genomics and medical sciences often have a large numbers of features. For example, hospital data may record many variables, including symptoms, blood test results, behaviours, and general health. An example of what highdimensional data might look like in a biomedical study is shown in the figure below.
Researchers often want to relate such features to specific patient outcomes (e.g. survival, length of time spent in hospital). However, analysing highdimensional data can be extremely challenging since standard methods of analysis, such as individual plots of features and linear regression, are no longer appropriate when we have many features. In this lesson, we will learn alternative methods for dealing with highdimensional data and discover how these can be applied for practical highdimensional data analysis in the biological sciences.
Challenge 1
Descriptions of four research questions and their datasets are given below. Which of these scenarios use highdimensional data?
 Predicting patient blood pressure using: cholesterol level in blood, age, and BMI measurements, collected from 100 patients.
 Predicting patient blood pressure using: cholesterol level in blood, age, and BMI, as well as information on 200,000 single nucleotide polymorphisms from 100 patients.
 Predicting the length of time patients spend in hospital with pneumonia infection using: measurements on age, BMI, length of time with symptoms, number of symptoms, and percentage of neutrophils in blood, using data from 200 patients.
 Predicting probability of a patient’s cancer progressing using gene expression data from 20,000 genes, as well as data associated with general patient health (age, weight, BMI, blood pressure) and cancer growth (tumour size, localised spread, blood test results).
Solution
 No. The number of features is relatively small (4 including the response variable since this is an observed variable).
 Yes, this is an example of highdimensional data. There are 200,004 features.
 No. The number of features is relatively small (6).
 Yes. There are 20,008 features.
Now that we have an idea of what highdimensional data look like we can think about the challenges we face in analysing them.
Why is dealing with highdimensional data challenging?
Most classical statistical methods are set up for use on lowdimensional data (i.e. with a small number of features, $p$). This is because lowdimensional data were much more common in the past when data collection was more difficult and time consuming.
One challenge when analysing highdimensional data is visualising the many variables. When exploring lowdimensional datasets, it is possible to plot the response variable against each of features to get an idea which of these are important predictors of the response. With highdimensional data, the large number of features makes doing this difficult. In addition, in some highdimensional datasets it can also be difficult to identify a single response variable, making standard data exploration and analysis techniques less useful.
Let’s have a look at a simple dataset with lots of features to understand some of the challenges we are facing when working with highdimensional data. For reference, all data used throughout the lesson are described in the data page.
Challenge 2
For illustrative purposes, we start with a simple dataset that is not technically highdimensional but contains many features. This will illustrate the general problems encountered when working with many features in a highdimensional data set.
First, make sure you have completed the setup instructions here. Next, let’s load the
prostate
dataset as follows:library("here") prostate < readRDS(here("data/prostate.rds"))
Examine the dataset (in which each row represents a single patient) to:
 Determine how many observations ($n$) and features ($p$) are available (hint: see the
dim()
function). Examine what variables were measured (hint: see the
names()
andhead()
functions). Plot the relationship between the variables (hint: see the
pairs()
function). What problem(s) with highdimensional data analysis does this illustrate?Solution
dim(prostate) # print the number of rows and columns
names(prostate) # examine the variable names head(prostate) # print the first 6 rows
names(prostate) # examine column names
[1] "lcavol" "lweight" "age" "lbph" "svi" "lcp" "gleason" [8] "pgg45" "lpsa"
pairs(prostate) # plot each pair of variables against each other
The
pairs()
function plots relationships between each of the variables in theprostate
dataset. This is possible for datasets with smaller numbers of variables, but for datasets in which $p$ is larger it becomes difficult (and time consuming) to visualise relationships between all variables in the dataset. Even where visualisation is possible, fitting models to datasets with many variables is difficult due to the potential for overfitting and difficulties in identifying a response variable.
Note that function documentation and information on function arguments will be useful throughout
this lesson. We can access these easily in R by running ?
followed by the package name.
For example, the documentation for the dim
function can be accessed by running ?dim
.
Locating data with R  the
here
packageIt is often desirable to access external datasets from inside R and to write code that does this reliably on different computers. While R has an inbulit function
setwd()
that can be used to denote where external datasets are stored, this usually requires the user to adjust the code to their specific system and folder structure. Thehere
package is meant to be used in R projects. It allows users to specify the data location relative to the R project directory. This makes R code more portable and can contribute to improve the reproducibility of an analysis.
As well as many variables causing problems when working with highdimensional data, having relatively few observations ($n$) compared to the number of features ($p$) causes additional challenges. To illustrate these challenges, imagine we are carrying out least squares regression on a dataset with 25 observations. Fitting a best fit line through these data produces a plot shown in the lefthand panel of the figure below.
However, imagine a situation in which the number of observations and features in a dataset are almost equal. In that situation the effective number of observations per feature is low. The result of fitting a best fit line through few observations can be seen in the righthand panel below.
In the first situation, the least squares regression line does not fit the data perfectly and there is some error around the regression line. But, when there are only two observations the regression line will fit through the points exactly, resulting in overfitting of the data. This suggests that carrying out least squares regression on a dataset with few data points per feature would result in difficulties in applying the resulting model to further datsets. This is a common problem when using regression on highdimensional datasets.
Another problem in carrying out regression on highdimensional data is dealing with correlations between explanatory variables. The large numbers of features in these datasets makes high correlations between variables more likely. Let’s explore why high correlations might be an issue in a Challenge.
Challenge 3
Use the
cor()
function to examine correlations between all variables in theprostate
dataset. Are some pairs of variables highly correlated using a threshold of 0.75 for the correlation coefficients?Use the
lm()
function to fit univariate regression models to predict patient age using two variables that are highly correlated as predictors. Which of these variables are statistically significant predictors of age? Hint: thesummary()
function can help here.Fit a multiple linear regression model predicting patient age using both variables. What happened?
Solution
Create a correlation matrix of all variables in the
prostate
datasetcor(prostate)
lcavol lweight age lbph svi lcp lcavol 1.0000000 0.194128286 0.2249999 0.027349703 0.53884500 0.675310484 lweight 0.1941283 1.000000000 0.3075286 0.434934636 0.10877851 0.100237795 age 0.2249999 0.307528614 1.0000000 0.350185896 0.11765804 0.127667752 lbph 0.0273497 0.434934636 0.3501859 1.000000000 0.08584324 0.006999431 svi 0.5388450 0.108778505 0.1176580 0.085843238 1.00000000 0.673111185 lcp 0.6753105 0.100237795 0.1276678 0.006999431 0.67311118 1.000000000 gleason 0.4324171 0.001275658 0.2688916 0.077820447 0.32041222 0.514830063 pgg45 0.4336522 0.050846821 0.2761124 0.078460018 0.45764762 0.631528245 lpsa 0.7344603 0.354120390 0.1695928 0.179809410 0.56621822 0.548813169 gleason pgg45 lpsa lcavol 0.432417056 0.43365225 0.7344603 lweight 0.001275658 0.05084682 0.3541204 age 0.268891599 0.27611245 0.1695928 lbph 0.077820447 0.07846002 0.1798094 svi 0.320412221 0.45764762 0.5662182 lcp 0.514830063 0.63152825 0.5488132 gleason 1.000000000 0.75190451 0.3689868 pgg45 0.751904512 1.00000000 0.4223159 lpsa 0.368986803 0.42231586 1.0000000
round(cor(prostate), 2) # rounding helps to visualise the correlations
lcavol lweight age lbph svi lcp gleason pgg45 lpsa lcavol 1.00 0.19 0.22 0.03 0.54 0.68 0.43 0.43 0.73 lweight 0.19 1.00 0.31 0.43 0.11 0.10 0.00 0.05 0.35 age 0.22 0.31 1.00 0.35 0.12 0.13 0.27 0.28 0.17 lbph 0.03 0.43 0.35 1.00 0.09 0.01 0.08 0.08 0.18 svi 0.54 0.11 0.12 0.09 1.00 0.67 0.32 0.46 0.57 lcp 0.68 0.10 0.13 0.01 0.67 1.00 0.51 0.63 0.55 gleason 0.43 0.00 0.27 0.08 0.32 0.51 1.00 0.75 0.37 pgg45 0.43 0.05 0.28 0.08 0.46 0.63 0.75 1.00 0.42 lpsa 0.73 0.35 0.17 0.18 0.57 0.55 0.37 0.42 1.00
As seen above, some variables are highly correlated. In particular, the correlation between
gleason
andpgg45
is equal to 0.75.Fitting univariate regression models to predict age using gleason and pgg45 as predictors.
model_gleason < lm(age ~ gleason, data = prostate) model_pgg45 < lm(age ~ pgg45, data = prostate)
Check which covariates have a significant efffect
summary(model_gleason)
Call: lm(formula = age ~ gleason, data = prostate) Residuals: Min 1Q Median 3Q Max 20.780 3.552 1.448 4.220 13.448 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 45.146 6.918 6.525 3.29e09 *** gleason 2.772 1.019 2.721 0.00774 **  Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 7.209 on 95 degrees of freedom Multiple Rsquared: 0.0723, Adjusted Rsquared: 0.06254 Fstatistic: 7.404 on 1 and 95 DF, pvalue: 0.007741
summary(model_pgg45)
Call: lm(formula = age ~ pgg45, data = prostate) Residuals: Min 1Q Median 3Q Max 21.0889 3.4533 0.9111 4.4534 15.1822 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 62.08890 0.96758 64.17 < 2e16 *** pgg45 0.07289 0.02603 2.80 0.00619 **  Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 7.193 on 95 degrees of freedom Multiple Rsquared: 0.07624, Adjusted Rsquared: 0.06651 Fstatistic: 7.84 on 1 and 95 DF, pvalue: 0.006189
Based on these results we conclude that both
gleason
andpgg45
have a statistically significant univariate effect (also referred to as a marginal effect) as predictors of age (5% significance level).Fitting a multivariate regression model using both both
gleason
andpgg45
as predictorsmodel_multivar < lm(age ~ gleason + pgg45, data = prostate) summary(model_multivar)
Call: lm(formula = age ~ gleason + pgg45, data = prostate) Residuals: Min 1Q Median 3Q Max 20.927 3.677 1.323 4.323 14.420 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 52.95548 9.74316 5.435 4.3e07 *** gleason 1.45363 1.54299 0.942 0.349 pgg45 0.04490 0.03951 1.137 0.259  Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 7.198 on 94 degrees of freedom Multiple Rsquared: 0.08488, Adjusted Rsquared: 0.06541 Fstatistic: 4.359 on 2 and 94 DF, pvalue: 0.01547
Although
gleason
andpgg45
have statistically significant univariate effects, this is no longer the case when both variables are simultaneously included as covariates in a multivariate regression model.
Including highly correlated variables such as gleason
and pgg45
simultaneously the same regression model can lead to problems
in fitting a regression model and interpreting its output. Although each variable
appears to be associated with the response individually, the model cannot distinguish
the contribution of each variable to the model. This can also increase the risk
of overfitting since the model may fit redundant variables to noise rather
than true relationships.
To allow the information from variables to be included in the same model despite high levels of correlation, we can use dimensionality reduction methods to collapse multiple variables into a single new variable (we will explore this dataset further in the dimensionality reduction lesson). We can also use modifications to linear regression like regularisation, which we will discuss in the lesson on highdimensional regression.
What statistical methods are used to analyse highdimensional data?
We have discussed so far that highdimensional data analysis can be challenging since variables are difficult to visualise, leading to challenges identifying relationships between variables and suitable response variables; we may have relatively few observations compared to features, leading to overfitting; and features may be highly correlated, leading to challenges interpreting models. We therefore require alternative approaches to examine whether, for example, groups of observations show similar characteristics and whether these groups may relate to other features in the data (e.g. phenotype in genetics data).
In this course, we will cover four methods that help in dealing with highdimensional data: (1) regression with numerous outcome variables, (2) regularised regression, (3) dimensionality reduction, and (4) clustering. Here are some examples of when each of these approaches may be used:

Regression with numerous outcomes refers to situations in which there are many variables of a similar kind (expression values for many genes, methylation levels for many sites in the genome) and when one is interested in assessing whether these variables are associated with a specific covariate of interest, such as experimental condition or age. In this case, multiple univariate regression models (one per each outcome, using the covariate of interest as predictor) could be fitted independently. In the context of highdimensional molecular data, a typical example are differential gene expression analyses. We will explore this type of analysis in the Regression with many outcomes episode.

Regularisation (also known as regularised regression or penalised regression) is typically used to fit regression models when there is a single outcome variable or interest but the number of potential predictors is large, e.g. there are more predictors than observations. Regularisation can help to prevent overfitting and may be used to identify a small subset of predictors that are associated with the outcome of interest. For example, regularised regression has been often used when building epigenetic clocks, where methylation values across several thousands of genomic sites are used to predict chronological age. We will explore this in more detail in the Regularised regression episode.

Dimensionality reduction is commonly used on highdimensional datasets for data exploration or as a preprocessing step prior to other downstream analyses. For instance, a lowdimensional visualisation of a gene expression dataset may be used to inform quality control steps (e.g. are there any anomalous samples?). This course contains two episodes that explore dimensionality reduction techniques: Principal component analysis and Factor analysis.

Clustering methods can be used to identify potential grouping patterns within a dataset. A popular example is the identification of distinct cell types through clustering cells with similar gene expression patterns. The Kmeans episode will explore a specific method to perform clustering analysis.
Using Bioconductor to access highdimensional data in the biosciences
In this workshop, we will look at statistical methods that can be used to visualise and analyse highdimensional biological data using packages available from Bioconductor, open source software for analysing high throughput genomic data. Bioconductor contains useful packages and example datasets as shown on the website https://www.bioconductor.org/.
Bioconductor packages can be installed and used in
R
using theBiocManager
package. Let’s load theminfi
package from Bioconductor (a package for analysing Illumina Infinium DNA methylation arrays).library("minfi")
browseVignettes("minfi")
We can explore these packages by browsing the vignettes provided in Bioconductor. Bioconductor has various packages that can be used to load and examine datasets in
R
that have been made available in Bioconductor, usually along with an associated paper or package.Next, we load the
methylation
dataset which represents data collected using Illumina Infinium methylation arrays which are used to examine methylation across the human genome. These data include information collected from the assay as well as associated metadata from individuals from whom samples were taken.library("here") library("ComplexHeatmap") methylation < readRDS(here("data/methylation.rds")) head(colData(methylation))
DataFrame with 6 rows and 14 columns Sample_Well Sample_Name purity Sex Age <character> <character> <integer> <character> <integer> 201868500150_R01C01 A07 PCA0612 94 M 39 201868500150_R03C01 C07 NKpan2510 95 M 49 201868500150_R05C01 E07 WB1148 95 M 20 201868500150_R07C01 G07 B0044 97 M 49 201868500150_R08C01 H07 NKpan1869 95 F 33 201868590193_R02C01 B03 NKpan1850 93 F 21 weight_kg height_m bmi bmi_clas Ethnicity_wide <numeric> <numeric> <numeric> <character> <character> 201868500150_R01C01 88.4505 1.8542 25.7269 Overweight Mixed 201868500150_R03C01 81.1930 1.6764 28.8911 Overweight IndoEuropean 201868500150_R05C01 80.2858 1.7526 26.1381 Overweight IndoEuropean 201868500150_R07C01 82.5538 1.7272 27.6727 Overweight IndoEuropean 201868500150_R08C01 87.5433 1.7272 29.3452 Overweight IndoEuropean 201868590193_R02C01 87.5433 1.6764 31.1507 Obese Mixed Ethnic_self smoker Array Slide <character> <character> <character> <numeric> 201868500150_R01C01 Hispanic No R01C01 2.01869e+11 201868500150_R03C01 Caucasian No R03C01 2.01869e+11 201868500150_R05C01 Persian No R05C01 2.01869e+11 201868500150_R07C01 Caucasian No R07C01 2.01869e+11 201868500150_R08C01 Caucasian No R08C01 2.01869e+11 201868590193_R02C01 Finnish/Creole No R02C01 2.01869e+11
methyl_mat < t(assay(methylation)) ## calculate correlations between cells in matrix cor_mat < cor(methyl_mat)
cor_mat[1:10, 1:10] # print the topleft corner of the correlation matrix
The
assay()
function creates a matrixlike object where rows represent probes for genes and columns represent samples. We calculate correlations between features in themethylation
dataset and examine the first 100 cells of this matrix. The size of the dataset makes it difficult to examine in full, a common challenge in analysing highdimensional genomics data.
Further reading
 Buhlman, P. & van de Geer, S. (2011) Statistics for HighDimensional Data. Springer, London.
 Buhlman, P., Kalisch, M. & Meier, L. (2014) Highdimensional statistics with a view toward applications in biology. Annual Review of Statistics and Its Application.
 Johnstone, I.M. & Titterington, D.M. (2009) Statistical challenges of highdimensional data. Philosophical Transactions of the Royal Society A 367:42374253.
 Bioconductor ethylation array analysis vignette.
 The Introduction to Machine Learning with Python course covers additional methods that could be used to analyse highdimensional data. See Introduction to machine learning, Tree models and Neural networks. Some related (an important!) content is also available in Responsible machine learning.
Other resources suggested by former students
 Josh Starmer’s youtube channel.
Key Points
Highdimensional data are data in which the number of features, $p$, are close to or larger than the number of observations, $n$.
These data are becoming more common in the biological sciences due to increases in data storage capabilities and computing power.
Standard statistical methods, such as linear regression, run into difficulties when analysing highdimensional data.
In this workshop, we will explore statistical methods used for analysing highdimensional data using datasets available on Bioconductor.
Regression with many outcomes
Overview
Teaching: 70 min
Exercises: 50 minQuestions
How can we apply linear regression in a highdimensional setting?
How can we benefit from the fact that we have many outcomes?
How can we control for the fact that we do many tests?
Objectives
Perform and critically analyse highdimensional regression.
Understand methods for shrinkage of noise parameters in highdimensional regression.
Perform multiple testing adjustment.
DNA methylation data
For the following few episodes, we will be working with human DNA methylation data from flowsorted blood samples, described in data. DNA methylation assays measure, for each of many sites in the genome, the proportion of DNA that carries a methyl mark (a chemical modification that does not alter the DNA sequence). In this case, the methylation data come in the form of a matrix of normalised methylation levels (Mvalues), where negative values correspond to unmethylated DNA and positive values correspond to methylated DNA. Along with this, we have a number of sample phenotypes (eg, age in years, BMI).
Let’s read in the data for this episode:
library("here")
library("minfi")
methylation < readRDS(here("data/methylation.rds"))
Note: the code that we used to download these data from its source is available here
methylation
is actually a special Bioconductor SummarizedExperiment
object that summarises lots of different information about the data.
These objects are very useful for storing all of the information
about a dataset in a highthroughput context. The structure of SummarizedExperiment
objects is described in the vignettes on
Bioconductor.
Here, we show how to extract the information for analysis.
We can extract
 the dimensions of the dataset using
dim()
. Importantly, in these objects and data structures for computational biology in R generally, observations are stored as columns and features (in this case, sites in the genome) are stored as rows. This is in contrast to usual tabular data, where features or variables are stored as columns and observations are stored as rows;  assays, (normalised methylation levels here), using
assay()
;  samplelevel information using
colData()
.
dim(methylation)
[1] 5000 37
You can see in this output that this object has a dim()
of
$5000 \times 37$, meaning it has
5000 features and 37 observations. To
extract the matrix of methylation Mvalues, we can use the
assay()
function.
methyl_mat < assay(methylation)
The distribution of these Mvalues looks like this:
hist(methyl_mat, xlab = "Mvalue")
You can see that there are two peaks in this distribution, corresponding to features which are largely unmethylated and methylated, respectively.
Similarly, we can examine the colData()
, which represents the
samplelevel metadata we have relating to these data. In this case, the
metadata, phenotypes, and groupings in the colData
look like this for
the first 6 samples:
head(colData(methylation))
Sample_Well  Sample_Name  purity  Sex  Age  weight_kg  height_m  bmi  bmi_clas  Ethnicity_wide  Ethnic_self  smoker  Array  Slide 

A07  PCA0612  94  M  39  88.45051  1.8542  25.72688  Overweight  Mixed  Hispanic  No  R01C01  201868500150 
C07  NKpan2510  95  M  49  81.19303  1.6764  28.89106  Overweight  IndoEuropean  Caucasian  No  R03C01  201868500150 
E07  WB1148  95  M  20  80.28585  1.7526  26.13806  Overweight  IndoEuropean  Persian  No  R05C01  201868500150 
G07  B0044  97  M  49  82.55381  1.7272  27.67272  Overweight  IndoEuropean  Caucasian  No  R07C01  201868500150 
H07  NKpan1869  95  F  33  87.54333  1.7272  29.34525  Overweight  IndoEuropean  Caucasian  No  R08C01  201868500150 
B03  NKpan1850  93  F  21  87.54333  1.6764  31.15070  Obese  Mixed  Finnish/Creole  No  R02C01  201868590193 
In this episode, we will focus on the association between age and methylation. The following heatmap summarises age and methylation levels available in the methylation dataset:
library("ComplexHeatmap")
age < methylation$Age
# sort methylation values by age
order < order(age)
age_ord < age[order]
methyl_mat_ord < methyl_mat[, order]
# plot heatmap
Heatmap(methyl_mat_ord,
cluster_columns = FALSE,
show_row_names = FALSE,
show_column_names = FALSE,
name = "Mvalue",
row_title = "Feature",
column_title = "Sample",
top_annotation = columnAnnotation(age = age_ord))
Challenge 1
Why can we not just fit many linear regression models relating every combination of feature (
colData
and assays) and draw conclusions by associating all variables with significant model pvalues?Solution
There are a number of problems that this kind of approach presents. For example:
 If we perform 5000 tests for each of 14 variables, even if there were no true associations in the data, we’d be likely to observe some strong spurious associations that arise just from random noise.
 We may not have a representative sample for each of these covariates. For example, we may have very small sample sizes for some ethnicities, leading to spurious findings.
 Without a research question in mind when creating a model, it’s not clear how we can interpret each model, and rationalising the results after the fact can be dangerous; it’s easy to make up a “story” that isn’t grounded in anything but the fact that we have significant findings.
In general, it is scientifically interesting to explore two modelling problems using the three types of data:

Predicting methylation levels using age as a predictor. In this case, we would have 5000 outcomes (methylation levels across the genome) and a single covariate (age).

Predicting age using methylation levels as predictors. In this case, we would have a single outcome (age) which will be predicted using 5000 covariates (methylation levels across the genome).
The examples in this episode will focus on the first type of problem, whilst the next episode will focus on the second.
Measuring DNA Methylation
DNA methylation is an epigenetic modification of DNA. Generally, we are interested in the proportion of methylation at many sites or regions in the genome. DNA methylation microarrays, as we are using here, measure DNA methylation using twochannel microarrays, where one channel captures signal from methylated DNA and the other captures unmethylated signal. These data can be summarised as “Beta values” ($\beta$ values), which is the ratio of the methylated signal to the total signal (methylated plus unmethylated). The $\beta$ value for site $i$ is calculated as
\[\beta_i = \frac{ m_i } { u_{i} + m_{i} }\]where $m_i$ is the methylated signal for site $i$ and $u_i$ is the unmethylated signal for site $i$. $\beta$ values take on a value in the range $[0, 1]$, with 0 representing a completely unmethylated site and 1 representing a completely methylated site.
The Mvalues we use here are the $\log_2$ ratio of methylated versus unmethylated signal:
\[M_i = \log_2\left(\frac{m_i}{u_i}\right)\]Mvalues are not bounded to an interval as Beta values are, and therefore can be easier to work with in statistical models.
Regression with many outcomes
In highthroughput studies, it is common to have one or more phenotypes or groupings that we want to relate to features of interest (eg, gene expression, DNA methylation levels). In general, we want to identify differences in the features of interest that are related to a phenotype or grouping of our samples. Identifying features of interest that vary along with phenotypes or groupings can allow us to understand how phenotypes arise or manifest. Analysis of this type is sometimes referred to using the term differential analysis.
For example, we might want to identify genes that are expressed at a higher level in mutant mice relative to wildtype mice to understand the effect of a mutation on cellular phenotypes. Alternatively, we might have samples from a set of patients, and wish to identify epigenetic features that are different in young patients relative to old patients, to help us understand how ageing manifests.
Using linear regression, it is possible to identify differences like
these. However, highdimensional data like the ones we’re working with
require some special considerations. A first consideration, as we saw
above, is that there are far too many features to fit each onebyone as
we might do when analysing lowdimensional datasets (for example using
lm()
on each feature and checking the linear model assumptions). A
second consideration is that statistical approaches may behave
slightly differently when applied to very highdimensional data, compared to
lowdimensional data. A third consideration is the speed at which we can
actually compute statistics for data this large – methods optimised for
lowdimensional data may be very slow when applied to highdimensional
data.
Ideally when performing regression, we want to identify cases like this, where there is a clear association, and we probably “don’t need” statistics:
or equivalently for a discrete covariate:
However, often due to small differences and small sample sizes, the problem is more difficult:
And, of course, we often have an awful lot of features and need to prioritise a subset of them! We need a rigorous way to prioritise genes for further analysis.
Fitting a linear model
So, in the data we have read in, we have a matrix of methylation values $X$ and a vector of ages, $y$. One way to model this is to see if we can use age to predict the expected (average) methylation value for sample $j$ at a given locus $i$, which we can write as $X_{ij}$. We can write that model as:
[\mathbf{E}(X_{ij}) = \beta_0 + \beta_1 \text{Age}_j]
where $\text{Age}_j$ is the age of sample $j$. In this model, $\beta_1$
represents the unit change in mean methylation level for each unit
(year) change in age. For a specific CpG, we can fit this model and get more
information from the model object. For illustration purposes, here we
arbitrarily select the first CpG in the methyl_mat
matrix (the one on its first row).
age < methylation$Age
# methyl_mat[1, ] indicates that the 1st CpG will be used as outcome variable
lm_age_methyl1 < lm(methyl_mat[1, ] ~ age)
lm_age_methyl1
Call:
lm(formula = methyl_mat[1, ] ~ age)
Coefficients:
(Intercept) age
0.902334 0.008911
We now have estimates for the expected methylation level when age equals 0 (the intercept) and the change in methylation level for a unit change in age (the slope). We could plot this linear model:
plot(age, methyl_mat[1, ], xlab = "Age", ylab = "Methylation level", pch = 16)
abline(lm_age_methyl1)
For this feature, we can see that there is no strong relationship between methylation and age. We could try to repeat this for every feature in our dataset; however, we have a lot of features! We need an approach that allows us to assess associations between all of these features and our outcome while addressing the three considerations we outlined previously. Before we introduce this approach, let’s go into detail about how we generally check whether the results of a linear model are statistically significant.
Hypothesis testing in linear regression
Using the linear model we defined above, we can ask questions based on the estimated value for the regression coefficients. For example, do individuals with different age have different methylation values for a given CpG? We usually do this via hypothesis testing. This framework compares the results that we observed (here, estimated linear model coefficients) to the results you would expect under a null hypothesis associated to our question. In the example above, a suitable null hypothesis would test whether the regression coefficient associated to age ($\beta_1$) is equal to zero or not. If $\beta_1$ is equal to zero, the linear model indicates that there is no linear relationship between age and the methylation level for the CpG (remember: as its name suggests, linear regression can only be used to model linear relationships between predictors and outcomes!). In other words, the answer to our question would be: no!
The output of a linear model typically returns the results associated with the null hypothesis described above (this may not always be the most realistic or useful null hypothesis, but it is the one we have by default!). To be more specific, the test compares our observed results with a set of hypothetical counterexamples of what we would expect to observe if we repeated the same experiment and analysis over and over again under the null hypothesis.
For this linear model, we can use tidy()
from the broom
package to
extract detailed information about the coefficients and the associated
hypothesis tests in this model:
library("broom")
tidy(lm_age_methyl1)
# A tibble: 2 × 5
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) 0.902 0.344 2.62 0.0129
2 age 0.00891 0.0100 0.888 0.381
The standard errors (std.error
) represent the statistical uncertainty in our
regression coefficient estimates (often referred to as effect size). The test
statistics and pvalues (statistic
and p.value
) represent measures of how (un)likely it would be to observe
results like this under the “null hypothesis”. For coefficient $k$ in a
linear model (in our case, it would be the slope), the test statistic calculated in statistic
above is
a tstatistic given by:
[t_{k} = \frac{\hat{\beta}{k}}{SE\left(\hat{\beta}{k}\right)}]
$SE\left(\hat{\beta}_{k}\right)$ measures the uncertainty we have in our effect size estimate. Knowing what distribution these tstatistics follow under the null hypothesis allows us to determine how unlikely it would be for us to observe what we have under those circumstances, if we repeated the experiment and analysis over and over again. To demonstrate how the tstatistics are calculated, we can compute them “by hand”:
table_age_methyl1 < tidy(lm_age_methyl1)
tvals < table_age_methyl1$estimate / table_age_methyl1$std.error
all.equal(tvals, table_age_methyl1$statistic)
[1] TRUE
We can see that the tstatistic is just the ratio between the coefficient estimate and the standard error.
Calculating the pvalues is a bit more tricky. Specifically, it is the proportion of the distribution of the test statistic under the null hypothesis that is as extreme or more extreme than the observed value of the test statistic. This is easy to observe visually, by plotting the theoretical distribution of the test statistic under the null hypothesis (see next callout box for more details about it):
The redish shaded region represents the portion of the distribution of the test statistic under the null hypothesis that is equal or greater to the value we observe for the intercept term. As our null hypothesis relates to a 2tailed test (as the null hypothesis states that the regression coefficient is equal to zero, we would reject it if the regression coefficient is substantially larger or smaller than zero), the pvalue for the test is twice the value of the shaded region. In this case, the shaded region is small relative to the total area of the null distribution; therefore, the pvalue is small ($p=0.013$). The blueish shaded region represents the same measure for the slope term; this is larger, relative to the total area of the distribution, therefore the pvalue is larger than the one for the intercept term ($p=0.381$). The pvalue is a function of the test statistic: the ratio between the effect size we’re estimating and the uncertainty we have in that effect. A large effect with large uncertainty may not lead to a small pvalue, and a small effect with small uncertainty may lead to a small pvalue.
Calculating pvalues from a linear model
Manually calculating the pvalue for a linear model is a little bit more complex than calculating the tstatistic. The intuition posted above is definitely sufficient for most cases, but for completeness, here is how we do it:
Since the statistic in a linear model is a tstatistic, it follows a student t distribution under the null hypothesis, with degrees of freedom (a parameter of the student tdistribution) given by the number of observations minus the number of coefficients fitted, in this case $37  2 = 35$. We want to know what portion of the distribution function of the test statistic is as extreme as, or more extreme than, the value we observed. The function
pt()
(similar topnorm()
, etc) can give us this information.Since we’re not sure if the coefficient will be larger or smaller than zero, we want to do a 2tailed test. Therefore we take the absolute value of the tstatistic, and look at the upper rather than lower tail. In the figure above the shaded areas are only looking at “half” of the tdistribution (which is symmetric around zero), therefore we multiply the shaded area by 2 in order to calculate the pvalue.
Combining all of this gives us:
pvals < 2 * pt(abs(tvals), df = lm_age_methyl1$df, lower.tail = FALSE) all.equal(table_age_methyl1$p.value, pvals)
[1] TRUE
Challenge 2
In the model we fitted, the estimate for the intercept is 0.902 and its associated pvalue is 0.0129. What does this mean?
Solution
The first coefficient in a linear model like this is the intercept, which measures the mean of the outcome (in this case, the methylation value for the first CpG) when age is zero. In this case, the intercept estimate is 0.902. However, this is not a particularly noteworthy finding as we do not have any observations with age zero (nor even any with age < 20!).
The reported pvalue is associated to the following null hypothesis: the intercept ($\beta_0$ above) is equal to zero. Using the usual significance threshold of 0.05, we reject the null hypothesis as the pvalue is smaller than 0.05. However, it is not really interesting if this intercept is zero or not, since we probably do not care what the methylation level is when age is zero. In fact, this question does not even make much sense! In this example, we are more interested in the regression coefficient associated to age, as that can tell us whether there is a linear relationship between age and methylation for the CpG.
Fitting a lot of linear models
In the linear model above, we are generally interested in the second regression coefficient (often referred to as slope) which measures the linear relationship between age and methylation levels. For the first CpG, here is its estimate:
coef_age_methyl1 < tidy(lm_age_methyl1)[2, ]
coef_age_methyl1
# A tibble: 1 × 5
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 age 0.00891 0.0100 0.888 0.381
In this case, the pvalue is equal to 0.381 and therefore we cannot reject the null hypothesis: there is no statistical evidence to suggest that the regression coefficient associated to age is not equal to zero.
Now, we could do this for every feature (CpG) in the dataset and rank the results based on their test statistic or associated pvalue. However, fitting models in this way to 5000 features is not very computationally efficient, and it would also be laborious to do programmatically. There are ways to get around this, but first let us talk about what exactly we are doing when we look at significance tests in this context.
Sharing information across outcome variables
We are going to introduce an idea that allows us to take advantage of the fact that we carry out many tests at once on structured data. We can leverage this fact to share information between model parameters. The insight that we use to perform information pooling or sharing is derived from our knowledge about the structure of the data. For example, in a highthroughput experiment like a DNA methylation assay, we know that all of the features were measured simultaneously, using the same technique. This means that generally, we expect the baselevel variability for each feature to be broadly similar.
This can enable us to get a better estimate of the uncertainty of model parameters than we could get if we consider each feature in isolation. So, to share information between features allows us to get more robust estimators. Remember that the tstatistic for coefficient $\beta_k$ in a linear model is the ratio between the coefficient estimate and its standard error:
[t_{k} = \frac{\hat{\beta}{k}}{SE\left(\hat{\beta}{k}\right)}]
It is clear that large effect sizes will likely lead to small pvalues, as long as the standard error for the coefficent is not large. However, the standard error is affected by the amount of noise, as we saw earlier. If we have a small number of observations, it is common for the noise for some features to be extremely small simply by chance. This, in turn, causes small pvalues for these features, which may give us unwarranted confidence in the level of certainty we have in the results (false positives).
There are many statistical methods in genomics that use this type of
approach to get better estimates by pooling information between features
that were measured simultaneously using the same techniques. Here we
will focus on the package limma
, which is an established software
package used to fit linear models, originally for the gene expression
microarrays that were common in the 2000s, but which is still in use in
RNAseq experiments, among others. The authors of limma
made some
assumptions about the distributions that these follow, and pool
information across genes to get a better estimate of the uncertainty in
effect size estimates. It uses the idea that noise levels should be
similar between features to moderate the estimates of the test
statistic by shrinking the estimates of standard errors towards a common
value. This results in a moderated tstatistic.
The process of running a model in limma
is somewhat different to what you
may have seen when running linear models. Here, we define a model matrix or
design matrix, which is a way of representing the
coefficients that should be fit in each linear model. These are used in
similar ways in many different modelling libraries.
What is a model matrix?
R fits a regression model by choosing the vector of regression coefficients that minimises the differences between outcome values and predicted values using the covariates (or predictor variables). To get predicted values, we multiply the matrix of predictors by the coefficients. The latter matrix is called the model matrix (or design matrix). It has one row for each observation and one column for each predictor, plus one additional column of ones (the intercept column). Many R libraries contruct the model matrix behind the scenes, but
limma
does not. Usually, the model matrix can be extracted from a model fit using the functionmodel.matrix()
. Here is an example of a model matrix for the methylation model:design_age < model.matrix(lm_age_methyl1) # model matrix head(design_age)
(Intercept) age 201868500150_R01C01 1 39 201868500150_R03C01 1 49 201868500150_R05C01 1 20 201868500150_R07C01 1 49 201868500150_R08C01 1 33 201868590193_R02C01 1 21
dim(design_age)
[1] 37 2
As you can see, the model matrix has the same number of rows as our methylation data has samples. It also has two columns  one for the intercept (similar to the linear model we fit above) and one for age. This happens “under the hood” when fitting a linear model with
lm()
, but here we have to specify it directly. The limma user manual has more detail on how to make model matrices for different types of experimental design, but here we are going to stick with this simple twovariable case.
We pass our methylation data to lmFit()
, specifying
the model matrix. Internally, this function runs lm()
on each row of
the data in an efficient way. The function eBayes()
, when applied to the
output of lmFit()
, performs the pooled estimation of standard errors
that results in the moderated tstatistics and resulting pvalues.
library("limma")
design_age < model.matrix(lm_age_methyl1) # model matrix
fit_age < lmFit(methyl_mat, design = design_age)
fit_age < eBayes(fit_age)
To obtain the results of the linear models, we can use the topTable()
function. By default, this returns results for the first coefficient in
the model. As we saw above when using lm()
, and when we defined
design_age
above, the first coefficient relates to the intercept term,
which we are not particularly interested in here; therefore we specify
coef = 2
. Further, topTable()
by default only returns the top 10
results. To see all of the results in the data, we specify
number = nrow(fit_age)
to ensure that it returns a row for every row
of the input matrix.
toptab_age < topTable(fit_age, coef = 2, number = nrow(fit_age))
head(toptab_age)
logFC AveExpr t P.Value adj.P.Val B
cg08446924 0.02571353 0.4185868 6.039068 5.595675e07 0.002797837 5.131574
cg06493994 0.01550941 2.1057877 5.593988 2.239813e06 0.005599533 3.747986
cg17661642 0.02266668 2.0527722 5.358739 4.658336e06 0.006048733 3.019698
cg05168977 0.02276336 2.2918472 5.346500 4.838987e06 0.006048733 2.981904
cg24549277 0.01975577 1.7466088 4.939242 1.708355e05 0.011508818 1.731821
cg04436528 0.01943612 0.7033503 4.917179 1.828563e05 0.011508818 1.664608
The output of topTable
includes the coefficient, here termed a log
fold change logFC
, the average level (aveExpr
), the tstatistic t
,
the pvalue (P.Value
), and the adjusted pvalue (adj.P.Val
). We’ll
cover what an adjusted pvalue is very shortly. The table also includes
B
, which represents the logodds that a feature is signficantly
different, which we won’t cover here, but which will generally be a 11
transformation of the pvalue. The coefficient estimates here are termed
logFC
for legacy reasons relating to how microarray experiments were
traditionally performed. There are more details on this topic in many
places, for example this tutorial by Kasper D.
Hansen
Now we have estimates of effect sizes and pvalues for the association between methylation level at each locus and age for our 37 samples. It’s useful to create a plot of effect size estimates (model coefficients) against pvalues for each of these linear models, to visualise the magnitude of effects and the statistical significance of each. These plots are often called “volcano plots”, because they resemble an eruption.
plot(toptab_age$logFC, log10(toptab_age$P.Value),
xlab = "Effect size", ylab = bquote(log[10](pvalue)),
pch = 19
)
In this figure, every point represents a feature of interest. The xaxis represents the effect size observed for that feature in a linear model, while the yaxis is the $\log_{10}(\text{pvalue})$, where larger values indicate increasing statistical evidence of a nonzero effect size. A positive effect size represents increasing methylation with increasing age, and a negative effect size represents decreasing methylation with increasing age. Points higher on the yaxis represent features for which we think the results we observed would be very unlikely under the null hypothesis.
Since we want to identify features that have different methylation levels in different age groups, in an ideal case there would be clear separation between “null” and “nonnull” features. However, usually we observe results as we do here: there is a continuum of effect sizes and pvalues, with no clear separation between these two classes of features. While statistical methods exist to derive insights from continuous measures like these, it is often convenient to obtain a list of features which we are confident have nonzero effect sizes. This is made more difficult by the number of tests we perform.
Challenge 3
The effect size estimates are very small, and yet many of the pvalues are well below a usual significance level of p < 0.05. Why is this?
Solution
Because age has a much larger range than methylation levels, the unit change in methylation level even for a strong relationship is very small!
As we mentioned, the pvalue is a function of both the effect size estimate and the uncertainty (standard error) of that estimate. Because the uncertainty in our estimates is much smaller than the estimates themselves, the pvalues are also small.
If we predicted age using methylation level, it is likely we would see much larger coefficients, though broadly similar pvalues!
It is worthwhile considering what exactly the effect of the moderation
or information sharing that limma
performs has on our results. To do
this, let us compare the effect sizes estimates and pvalues from the two
approaches.
These are exactly identical! This is because limma
does not perform
any sharing of information when estimating effect sizes. This is in
contrast to similar packages that apply shrinkage to the effect size
estimates, like DESeq2
. These often use information sharing to shrink
or moderate the effect size estimates, in the case of DESeq2
by again
sharing information between features about sampletosample variability.
In contrast, let us look at the pvalues from limma
and R’s builtin lm()
function:
We can see that for the vast majority of features, the results are
broadly similar. There seems to be a minor general tendency for limma
to produce smaller pvalues, but for several features, the pvalues from
limma are considerably larger than the pvalues from lm()
. This is
because the information sharing tends to shrink large standard error
estimates downwards and small estimates upwards. When the degree of
statistical significance is due to an abnormally small standard error
rather than a large effect, this effect results in this prominent
reduction in statistical significance, which has been shown to perform
well in case studies. The degree of shrinkage generally depends on the
amount of pooled information and the strength of the evidence
independent of pooling. For example, with very few samples and many
features, information sharing has a larger effect, because there are a
lot of genes that can be used to provide pooled estimates, and the
evidence from the data that this is weighed against is relatively
sparse. In contrast, when there are many samples and few features, there
is not much opportunity to generate pooled estimates, and the evidence
of the data can easily outweigh the pooling.
Shrinkage methods like these ones can be complex to implement and understand, but it is useful to develop an intuition about why these approaches may be more precise and sensitive than the naive approach of fitting a model to each feature separately.
Challenge 4
 Try to run the same kind of linear model with smoking status as covariate instead of age, and making a volcano plot. Note: smoking status is stored as
methylation$smoker
. We saw in the example in the lesson that this information sharing can lead to larger pvalues. Why might this be preferable?
Solution
The following code runs the same type of model with smoking status:
design_smoke < model.matrix(~methylation$smoker) fit_smoke < lmFit(methyl_mat, design = design_smoke) fit_smoke < eBayes(fit_smoke) toptab_smoke < topTable(fit_smoke, coef = 2, number = nrow(fit_smoke)) plot(toptab_smoke$logFC, log10(toptab_smoke$P.Value), xlab = "Effect size", ylab = bquote(log[10](p)), pch = 19 )
Being a bit more conservative when identifying features can help to avoid false discoveries. Furthermore, when rejecting the null hypothesis is based more on a small standard error resulting from abnormally low levels of variability for a given feature, we might want to be a bit more conservative in our expectations.
Shrinkage
Shrinkage is an intuitive term for an effect of information sharing, and is something observed in a broad range of statistical models. Often, shrinkage is induced by a multilevel modelling approach or by Bayesian methods.
The general idea is that these models incorporate information about the structure of the data into account when fitting the parameters. We can share information between features because of our knowledge about the data structure; this generally requires careful consideration about how the data were generated and the relationships within.
An example people often use is estimating the effect of attendance on grades in several schools. We can assume that this effect is similar in different schools (but maybe not identical), so we can share information about the effect size between schools and shrink our estimates towards a common value.
For example in
DESeq2
, the authors used the observation that genes with similar expression counts in RNAseq data have similar dispersion, and a better estimate of these dispersion parameters makes estimates of fold changes much more stable. Similarly, inlimma
the authors made the assumption that in the absence of biological effects, we can often expect the technical variation in the measurement of the expression of each of the genes to be broadly similar. Again, better estimates of variability allow us to prioritise genes in a more reliable way.There are many good resources to learn about this type of approach, including:
The problem of multiple tests
With such a large number of features, it would be useful to decide which features are “interesting” or “significant” for further study. However, if we were to apply a normal significance threshold of 0.05, it would be likely we end up with a lot of false positives. This is because a pvalue threshold like this represents a $\frac{1}{20}$ chance that we observe results as extreme or more extreme under the null hypothesis (that there is no assocation between age and methylation level). If we carry out many more than 20 such tests, we can expect to see situations where, despite the null hypothesis being true, we observe observe signifiant pvalues due to random chance. To demonstrate this, it is useful to see what happens if we permute (scramble) the age values and run the same test again:
set.seed(123)
age_perm < age[sample(ncol(methyl_mat), ncol(methyl_mat))]
design_age_perm < model.matrix(~age_perm)
fit_age_perm < lmFit(methyl_mat, design = design_age_perm)
fit_age_perm < eBayes(fit_age_perm)
toptab_age_perm < topTable(fit_age_perm, coef = 2, number = nrow(fit_age_perm))
plot(toptab_age_perm$logFC, log10(toptab_age_perm$P.Value),
xlab = "Effect size", ylab = bquote(log[10](p)),
pch = 19
)
abline(h = log10(0.05), lty = "dashed", col = "red")
Since we have generated a random sequence of ages, we have no reason to suspect that there is a true association between methylation levels and this sequence of random numbers. However, you can see that the pvalue for many features is still lower than a traditional significance level of $p=0.05$. In fact, here 226 features are significant at p < 0.05. If we were to use this fixed threshold in a real experiment, it is likely that we would identify many features as associated with age, when the results we are observing are simply due to chance.
Challenge 5
 If we run 5000 tests, even if there are no true differences, how many of them (on average) will be statistically significant at a threshold of $p < 0.05$?
 Why would we want to be conservative in labelling features as significantly different? By conservative, we mean to err towards labelling true differences as “not significant” rather than vice versa.
 How could we account for a varying number of tests to ensure “significant” changes are truly different?
Solution
 By default we expect $5000 \times 0.05 = 250$ features to be statistically significant under the null hypothesis, because pvalues should always be uniformly distributed under the null hypothesis.
 Features that we label as “significantly different” will often be reported in manuscripts. We may also spend time and money investigating them further, computationally or in the lab. Therefore, spurious results have a real cost for ourselves and for others.
 One approach to controlling for the number of tests is to divide our significance threshold by the number of tests performed. This is termed “Bonferroni correction” and we’ll discuss this further now.
Adjusting for multiple tests
When performing many statistical tests to categorise features, we are effectively classifying features as “nonsignificant” or “significant”, that latter meaning those for which we reject the null hypothesis. We also generally hope that there is a subset of features for which the null hypothesis is truly false, as well as many for which the null truly does hold. We hope that for all features for which the null hypothesis is true, we accept it, and for all features for which the null hypothesis is not true, we reject it. As we showed in the example with permuted age, with a large number of tests it is inevitable that we will get some of these wrong.
We can think of these features as being “truly different” or “not truly different”^{1}. Using this idea, we can see that each categorisation we make falls into four categories:
Label as different  Label as not different  

Truly different  True positive  False negative 
Truly not different  False positive  True negative 
If the null hypothesis was true for every feature, then as we perform more and more tests we’d tend to correctly categorise most results as negative. However, since pvalues are uniformly distributed under the null, at a significance level of 5%, 5% of all results will be “significant” even though we would expect to see these results, given the null hypothesis is true, simply by chance. These would fall under the label “false positives” in the table above, and are also termed “false discoveries.”
There are two common ways of controlling these false discoveries. The first is to say, when we’re doing $n$ tests, that we want to have the same certainty of making one false discovery with $n$ tests as we have if we’re only doing one test. This is “Bonferroni” correction,^{2} which divides the significance level by the number of tests performed, $n$. Equivalently, we can use the nontransformed pvalue threshold but multiply our pvalues by the number of tests. This is often very conservative, especially with a lot of features!
p_raw < toptab_age$P.Value
p_fwer < p.adjust(p_raw, method = "bonferroni")
plot(p_raw, p_fwer, pch = 16, log="xy")
abline(0:1, lty = "dashed")
abline(v = 0.05, lty = "dashed", col = "red")
abline(h = 0.05, lty = "dashed", col = "red")
You can see that the pvalues are exactly one for the vast majority of tests we performed! This is not ideal sometimes, because unfortunately we usually don’t have very large sample sizes in health sciences.
The second main way of controlling for multiple tests is to control the false discovery rate (FDR).^{3} This is the proportion of false positives, or false discoveries, we’d expect to get each time if we repeated the experiment over and over.
 Rank the pvalues
 Assign each a rank (1 is smallest)
 Calculate the critical value \(q = \left(\frac{i}{m}\right)Q\), where $i$ is rank, $m$ is the number of tests, and $Q$ is the false discovery rate we want to target.^{4}
 Find the largest pvalue less than the critical value. All smaller than this are significant.
FWER  FDR 

+ Controls probability of identifying a false positive  + Controls rate of false discoveries 
+ Strict error rate control  + Allows error control with less stringency 
 Very conservative   Does not control probability of making errors 
 Requires larger statistical power   May result in false discoveries 
Challenge 6
 At a significance level of 0.05, with 100 tests performed, what is the Bonferroni significance threshold?
 In a gene expression experiment, after FDR correction with an FDRadjusted pvalue threshold of 0.05, we observe 500 significant genes. What proportion of these genes are truly different?
 Try running FDR correction on the
p_raw
vector. Hint: checkhelp("p.adjust")
to see what the method is called.
Compare these values to the raw pvalues and the Bonferroni pvalues.Solution
The Bonferroni threshold for this significance threshold is \(\frac{0.05}{100} = 0.0005\)
We can’t say what proportion of these genes are truly different. However, if we repeated this experiment and statistical test over and over, on average 5% of the results from each run would be false discoveries.
The following code runs FDR correction and compares it to noncorrected values and to Bonferroni:
p_fdr < p.adjust(p_raw, method = "BH") plot(p_raw, p_fdr, pch = 16, log="xy") abline(0:1, lty = "dashed") abline(v = 0.05, lty = "dashed", col = "red") abline(h = 0.05, lty = "dashed", col = "red")
plot(p_fwer, p_fdr, pch = 16, log="xy") abline(0:1, lty = "dashed") abline(v = 0.05, lty = "dashed", col = "red") abline(h = 0.05, lty = "dashed", col = "red")
Feature selection
In this episode, we have focussed on regression in a setting where there are more features than observations. This approach is relevant if we are interested in the association of each feature with some outcome or if we want to screen for features that have a strong association with an outcome. If, however, we are interested in predicting an outcome or if we want to know which features explain the variation in the outcome, we may want to restrict ourselves to a subset of relevant features. One way of doing this is called regularisation, and this is the topic of the next episode. An alternative is called feature selection. This is covered in the subsequent (optional) episode.
Further reading
limma
tutorial by Kasper D. Hansenlimma
user manual. The
VariancePartition
package has similar functionality tolimma
but allows the inclusion of random effects.
Footnotes

“True difference” is a hard category to rigidly define. As we’ve seen, with a lot of data, we can detect tiny differences, and with little data, we can’t detect large differences. However, both can be argued to be “true”. ↩

Bonferroni correction is also termed “familywise” error rate control. ↩

This is often called “BenjaminiHochberg” adjustment. ↩

People often perform extra controls on FDRadjusted pvalues, ensuring that ranks don’t change and the critical value is never smaller than the original pvalue. ↩
Key Points
Performing linear regression in a highdimensional setting requires us to perform hypothesis testing in a way that lowdimensional regression may not.
Sharing information between features can increase power and reduce false positives.
When running a lot of null hypothesis tests for highdimensional data, multiple testing correction allows retain power and avoid making costly false discoveries.
Multiple testing methods can be more conservative or more liberal, depending on our goals.
Regularised regression
Overview
Teaching: 110 min
Exercises: 60 minQuestions
What is regularisation?
How does regularisation work?
How can we select the level of regularisation for a model?
Objectives
Understand the benefits of regularised models.
Understand how different types of regularisation work.
Apply and critically analyse regularised regression models.
Introduction
This episode is about regularisation, also called regularised regression or penalised regression. This approach can be used for prediction and for feature selection and it is particularly useful when dealing with highdimensional data.
One reason that we need special statistical tools for highdimensional data is that standard linear models cannot handle highdimensional data sets – one cannot fit a linear model where there are more features (predictor variables) than there are observations (data points). In the previous lesson, we dealt with this problem by fitting individual models for each feature and sharing information among these models. Now we will take a look at an alternative approach that can be used to fit models with more features than observations by stabilising coefficient estimates. This approach is called regularisation. Compared to many other methods, regularisation is also often very fast and can therefore be extremely useful in practice.
First, let us check out what happens if we try to fit a linear model to highdimensional
data! We start by reading in the methylation
data from the last lesson:
library("here")
library("minfi")
methylation < readRDS(here("data/methylation.rds"))
## here, we transpose the matrix to have features as rows and samples as columns
methyl_mat < t(assay(methylation))
age < methylation$Age
Then, we try to fit a model with outcome age and all 5,000 features in this dataset as predictors (average methylation levels, Mvalues, across different sites in the genome).
# by using methyl_mat in the formula below, R will run a multivariate regression
# model in which each of the columns in methyl_mat is used as a predictor.
fit < lm(age ~ methyl_mat)
summary(fit)
Call:
lm(formula = age ~ methyl_mat)
Residuals:
ALL 37 residuals are 0: no residual degrees of freedom!
Coefficients: (4964 not defined because of singularities)
Estimate Std. Error t value Pr(>t)
(Intercept) 2640.474 NaN NaN NaN
methyl_matcg00075967 108.216 NaN NaN NaN
methyl_matcg00374717 139.637 NaN NaN NaN
methyl_matcg00864867 33.102 NaN NaN NaN
methyl_matcg00945507 72.250 NaN NaN NaN
[ reached getOption("max.print")  omitted 4996 rows ]
Residual standard error: NaN on 0 degrees of freedom
Multiple Rsquared: 1, Adjusted Rsquared: NaN
Fstatistic: NaN on 36 and 0 DF, pvalue: NA
You can see that we’re able to get some effect size estimates, but they seem very high! The summary also says that we were unable to estimate effect sizes for 4,964 features because of “singularities”. We clarify what singularities are in the note below but this means that R couldn’t find a way to perform the calculations necessary to fit the model. Large effect sizes and singularities are common when naively fitting linear regression models with a large number of features (i.e., to highdimensional data), often since the model cannot distinguish between the effects of many, correlated features or when we have more features than observations.
Singularities
The message that
\[(X^TX)^{1}X^Ty,\]lm()
produced is not necessarily the most intuitive. What are “singularities” and why are they an issue? A singular matrix is one that cannot be inverted. R uses inverse operations to fit linear models (find the coefficients) using:where $X$ is a matrix of predictor features and $y$ is the outcome vector. Thus, if the matrix $X^TX$ cannot be inverted to give $(X^TX)^{1}$, R cannot fit the model and returns the error that there are singularities.
Why might R be unable to calculate $(X^TX)^{1}$ and return the error that there are singularities? Well, when the determinant of the matrix is zero, we are unable to find its inverse. The determinant of the matrix is zero when there are more features than observations or often when the features are highly correlated.
xtx < t(methyl_mat) %*% methyl_mat det(xtx)
[1] 0
Correlated features – common in highdimensional data
In highdimensional datasets, there are often multiple features that contain redundant information (correlated features). If we visualise the level of correlation between sites in the
methylation
dataset, we can see that many of the features represent the same information  there are many offdiagonal cells, which are deep red or blue. For example, the following heatmap visualises the correlations for the first 500 features in themethylation
dataset (we selected 500 features only as it can be hard to visualise patterns when there are too many features!).library("ComplexHeatmap") small < methyl_mat[, 1:500] cor_mat < cor(small) Heatmap(cor_mat, column_title = "Featurefeature correlation in methylation data", name = "Pearson correlation", show_row_dend = FALSE, show_column_dend = FALSE, show_row_names = FALSE, show_column_names = FALSE )
Challenge 1
Consider or discuss in groups:
 Why would we observe correlated features in highdimensional biological data?
 Why might correlated features be a problem when fitting linear models?
 What issue might correlated features present when selecting features to include in a model one at a time?
Solution
 Many of the features in biological data represent very similar information biologically. For example, sets of genes that form complexes are often expressed in very similar quantities. Similarly, methylation levels at nearby sites are often very highly correlated.
 Correlated features can make inference unstable or even impossible mathematically.
 When we are selecting features one at a time we want to pick the most predictive feature each time. When a lot of features are very similar but encode slightly different information, which of the correlated features we select to include can have a huge impact on the later stages of model selection!
Regularisation can help us to deal with correlated features, as well as effectively reduce the number of features in our model. Before we describe regularisation, let’s recap what’s going on when we fit a linear model.
Coefficient estimates of a linear model
When we fit a linear model, we’re finding the line through our data that minimises the sum of the squared residuals. We can think of that as finding the slope and intercept that minimises the square of the length of the dashed lines. In this case, the red line in the left panel is the line that accomplishes this objective, and the red dot in the right panel is the point that represents this line in terms of its slope and intercept among many different possible models, where the background colour represents how well different combinations of slope and intercept accomplish this objective.
Mathematically, we can write the sum of squared residuals as
[\sum_{i=1}^N ( y_ix’_i\beta)^2]
where $\beta$ is a vector of (unknown) covariate effects which we want to learn by fitting a regression model: the $j$th element of $\beta$, which we denote as $\beta_j$ quantifies the effect of the $j$th covariate. For each individual $i$, $x_i$ is a vector of $j$ covariate values and $y_i$ is the true observed value for the outcome. The notation $x’_i\beta$ indicates matrix multiplication. In this case, the result is equivalent to multiplying each element of $x_i$ by its corresponding element in $\beta$ and then calculating the sum across all of those values. The result of this product (often denoted by $\hat{y}_i$) is the predicted value of the outcome generated by the model. As such, $y_ix’_i\beta$ can be interpreted as the prediction error, also referred to as model residual. To quantify the total error across all individuals, we sum the square residuals $( y_ix’_i\beta)^2$ across all the individuals in our data.
Finding the value of $\beta$ that minimises the sum above is the line of best fit through our data when considering this goal of minimising the sum of squared error. However, it is not the only possible line we could use! For example, we might want to err on the side of caution when estimating effect sizes (coefficients). That is, we might want to avoid estimating very large effect sizes.
Challenge 2
Discuss in groups:
 What are we minimising when we fit a linear model?
 Why are we minimising this objective? What assumptions are we making about our data when we do so?
Solution
 When we fit a linear model we are minimising the squared error. In fact, the standard linear model estimator is often known as “ordinary least squares”. The “ordinary” really means “original” here, to distinguish between this method, which dates back to ~1800, and some more “recent” (think 1940s…) methods.
 Squared error is useful because it ignores the sign of the residuals (whether they are positive or negative). It also penalises large errors much more than small errors. On top of all this, it also makes the solution very easy to compute mathematically. Least squares assumes that, when we account for the change in the mean of the outcome based on changes in the income, the data are normally distributed. That is, the residuals of the model, or the error left over after we account for any linear relationships in the data, are normally distributed, and have a fixed variance.
Model selection using training and test sets
Sets of models are often compared using statistics such as adjusted $R^2$, AIC or BIC. These show us how well the model is learning the data used in fitting that same model ^{1}. However, these statistics do not really tell us how well the model will generalise to new data. This is an important thing to consider – if our model doesn’t generalise to new data, then there’s a chance that it’s just picking up on a technical or batch effect in our data, or simply some noise that happens to fit the outcome we’re modelling. This is especially important when our goal is prediction – it’s not much good if we can only predict well for samples where the outcome is already known, after all!
To get an idea of how well our model generalises, we can split the data into two  a “training” and a “test” set. We use the “training” data to fit the model, and then see its performance on the “test” data.
One thing that often happens in this context is that large coefficient values minimise the training error, but they don’t minimise the test error on unseen data. First, we’ll go through an example of what exactly this means.
To compare the training and test errors for a model of methylation features and age, we’ll split the data into training and test sets, fit a linear model and calculate the errors. First, let’s calculate the training error. Let’s start by splitting the data into training and test sets:
methylation < readRDS(here::here("data/methylation.rds"))
library("SummarizedExperiment")
age < methylation$Age
methyl_mat < t(assay(methylation))
We will then subset the data to a set of CpG markers that are known to be associated with age from a previous study by Horvath et al.^{2}, described in data.
cpg_markers < c("cg16241714", "cg14424579", "cg22736354", "cg02479575", "cg00864867",
"cg25505610", "cg06493994", "cg04528819", "cg26297688", "cg20692569",
"cg04084157", "cg22920873", "cg10281002", "cg21378206", "cg26005082",
"cg12946225", "cg25771195", "cg26845300", "cg06144905", "cg27377450"
)
horvath_mat < methyl_mat[, cpg_markers]
## Generate an index to split the data
set.seed(42)
train_ind < sample(nrow(methyl_mat), 25)
## Split the data
train_mat < horvath_mat[train_ind, ]
train_age < age[train_ind]
test_mat < horvath_mat[train_ind, ]
test_age < age[train_ind]
Now let’s fit a linear model to our training data and calculate the training error. Here we use the mean of the squared difference between our predictions and the observed data, or “mean squared error” (MSE).
## Fit a linear model
# as.data.frame() converts train_mat into a data.frame
# Using the `.` syntax above together with a `data` argument will lead to
# the same result as using `train_age ~ train_mat`: R will fit a multivariate
# regression model in which each of the columns in `train_mat` is used as
# a predictor. We opted to use the `.` syntax because it will help us to
# obtain model predictions using the `predict()` function.
fit_horvath < lm(train_age ~ ., data = as.data.frame(train_mat))
## Function to calculate the (mean squared) error
mse < function(true, prediction) {
mean((true  prediction)^2)
}
## Calculate the training error
err_lm_train < mse(train_age, fitted(fit_horvath))
err_lm_train
[1] 1.319628
The training error appears very low here – on average we’re only off by about a year!
Challenge 3
For the fitted model above, calculate the mean squared error for the test set.
Solution
First, let’s find the predicted values for the ‘unseen’ test data:
pred_lm < predict(fit_horvath, newdata = as.data.frame(test_mat))
The mean squared error for the test set is the mean of the squared error between the predicted values and true test data.
err_lm < mse(test_age, pred_lm) err_lm
[1] 223.3571
Unfortunately, the test error is a lot higher than the training error. If we plot true age against predicted age for the samples in the test set, we can gain more insight into the performance of the model on the test set. Ideally, the predicted values should be close to the test data.
par(mfrow = c(1, 1))
plot(test_age, pred_lm, pch = 19)
abline(coef = 0:1, lty = "dashed")
This figure shows the predicted ages obtained from a linear model fit plotted against the true ages, which we kept in the test dataset. If the prediction were good, the dots should follow a line. Regularisation can help us to make the model more generalisable, improving predictions for the test dataset (or any other dataset that is not used when fitting our model).
Regularisation
Regularisation can be used to reduce correlation between predictors, the number of features, and improve generalisability by restricting model parameter estimates. Essentially, we add another condition to the problem we’re solving with linear regression that controls the total size of the coefficients that come out and shrinks many coefficients to zero (or near zero).
For example, we might say that the point representing the slope and intercept must fall within a certain distance of the origin, $(0, 0)$. For the 2parameter model (slope and intercept), we could visualise this constraint as a circle with a given radius. We want to find the “best” solution (in terms of minimising the residuals) that also falls within a circle of a given radius (in this case, 2).
Ridge regression
There are multiple ways to define the distance that our solution must fall in. The one we’ve plotted above controls the squared sum of the coefficients, $\beta$. This is also sometimes called the $L^2$ norm. This is defined as
[\left\lVert \beta\right\lVert_2 = \sqrt{\sum_{j=1}^p \beta_j^2}]
To control this, we specify that the solution for the equation above also has to have an $L^2$ norm smaller than a certain amount. Or, equivalently, we try to minimise a function that includes our $L^2$ norm scaled by a factor that is usually written $\lambda$.
[\sum_{i=1}^N \biggl( y_i  x’_i\beta\biggr)^2 + \lambda \left\lVert \beta \right\lVert_2 ^2]
This type of regularisation is called ridge regression. Another way of thinking about this is that when finding the best model, we’re weighing up a balance of the ordinary least squares objective and a “penalty” term that punishes models with large coefficients. The balance between the penalty and the ordinary least squares objective is controlled by $\lambda$  when $\lambda$ is large, we want to penalise large coefficients. In other words, we care a lot about the size of the coefficients and we want to reduce the complexity of our model. When $\lambda$ is small, we don’t really care a lot about shrinking our coefficients and we opt for a more complex model. When it’s zero, we’re back to just using ordinary least squares. We see how a penalty term, $\lambda$, might be chosen later in this episode.
For now, to see how regularisation might improve a model, let’s fit a model using the same set
of 20 features (stored as cpg_markers
) selected earlier in this episode (these
are a subset of the features identified by Horvarth et al), using both
regularised and ordinary least squares. To fit regularised regression models, we will use the glmnet
package.
library("glmnet")
## glmnet() performs scaling by default, supply unscaled data:
horvath_mat < methyl_mat[, cpg_markers] # select the same 20 sites as before
train_mat < horvath_mat[train_ind, ] # use the same individuals as selected before
test_mat < horvath_mat[train_ind, ]
ridge_fit < glmnet(x = train_mat, y = train_age, alpha = 0)
plot(ridge_fit, xvar = "lambda")
abline(h = 0, lty = "dashed")
This plot shows how the estimated coefficients for each CpG site change as we increase the penalty, $\lambda$. That is, as we decrease the size of the region that solutions can fall into, the values of the coefficients that we get back tend to decrease. In this case, coefficients tend towards zero but generally don’t reach it until the penalty gets very large. We can see that initially, some parameter estimates are really, really large, and these tend to shrink fairly rapidly.
We can also notice that some parameters “flip signs”; that is, they start off positive and become negative as lambda grows. This is a sign of collinearity, or correlated predictors. As we reduce the importance of one feature, we can “make up for” the loss in accuracy from that one feature by adding a bit of weight to another feature that represents similar information.
Since we split the data into test and training data, we can prove that ridge regression predicts the test data better than the model with no regularisation. Let’s generate our predictions under the ridge regression model and calculate the mean squared error in the test set:
# Obtain a matrix of predictions from the ridge model,
# where each column corresponds to a different lambda value
pred_ridge < predict(ridge_fit, newx = test_mat)
# Calculate MSE for every column of the prediction matrix against the vector of true ages
err_ridge < apply(pred_ridge, 2, function(col) mse(test_age, col))
min_err_ridge < min(err_ridge)
# Identify the lambda value that results in the lowest MSE (ie, the "best" lambda value)
which_min_err < which.min(err_ridge)
pred_min_ridge < pred_ridge[, which_min_err]
## Return errors
min_err_ridge
[1] 46.76802
This is much lower than the test error for the model without regularisation:
err_lm
[1] 223.3571
We can see where on the continuum of lambdas we’ve picked a model by plotting the coefficient paths again. In this case, we’ve picked a model with fairly modest coefficient shrinking.
chosen_lambda < ridge_fit$lambda[which.min(err_ridge)]
plot(ridge_fit, xvar = "lambda")
abline(v = log(chosen_lambda), lty = "dashed")
Challenge 4
 Which performs better, ridge or OLS?
 Plot predicted ages for each method against the true ages. How do the predictions look for both methods? Why might ridge be performing better?
Solution
Ridge regression performs significantly better on unseen data, despite being “worse” on the training data.
min_err_ridge
[1] 46.76802
err_lm
[1] 223.3571
The ridge ones are much less spread out with far fewer extreme predictions.
all < c(pred_lm, test_age, pred_min_ridge) lims < range(all) par(mfrow = 1:2) plot(test_age, pred_lm, xlim = lims, ylim = lims, pch = 19 ) abline(coef = 0:1, lty = "dashed") plot(test_age, pred_min_ridge, xlim = lims, ylim = lims, pch = 19 ) abline(coef = 0:1, lty = "dashed")
LASSO regression
LASSO is another type of regularisation. In this case we use the $L^1$ norm, or the sum of the absolute values of the coefficients.
[\left\lVert \beta \right\lVert_1 = \sum_{j=1}^p  \beta_j  ] 
This tends to produce sparse models; that is to say, it tends to remove features from the model that aren’t necessary to produce accurate predictions. This is because the region we’re restricting the coefficients to has sharp edges. So, when we increase the penalty (reduce the norm), it’s more likely that the best solution that falls in this region will be at the corner of this diagonal (i.e., one or more coefficient is exactly zero).
Challenge 5
 Use
glmnet()
to fit a LASSO model (hint: setalpha = 1
). Plot the model object. Remember that for ridge regression, we set
xvar = "lambda"
. What if you don’t set this? What’s the relationship between the two plots? How do the coefficient paths differ to the ridge case?
Solution
Fitting a LASSO model is very similar to a ridge model, we just need to change the
alpha
setting.fit_lasso < glmnet(x = methyl_mat, y = age, alpha = 1)
When
xvar = "lambda"
, the xaxis represents increasing model sparsity from left to right, because increasing lambda increases the penalty added to the coefficients. When we instead plot the L1 norm on the xaxis, increasing L1 norm means that we are allowing our coefficients to take on increasingly large values.par(mfrow = c(1, 2)) plot(fit_lasso, xvar = "lambda") plot(fit_lasso)
When using LASSO, the paths tend to go to exactly zero much more as the penalty ($ \lambda $) increases. In the ridge case, the paths tend towards zero but less commonly reach exactly zero.
Crossvalidation to find the best value of $\lambda$
Ideally, we want $\lambda$ to be large enough to reduce the complexity of our model, thus reducing the number of and correlations between the features, and improving generalisability. However, we don’t want the value of $\lambda$ to be so large that we lose a lot of the valuable information in the features.
Various methods can be used to balance this tradeoff and thus select the “best” value for $\lambda$. One method splits the data into $K$ chunks. We then use $K1$ of these as the training set, and the remaining $1$ chunk as the test set. We can repeat this until we’ve rotated through all $K$ chunks, giving us a good estimate of how well each of the lambda values work in our data. This is called crossvalidation, and doing this repeated test/train split gives us a better estimate of how generalisable our model is.
We can use this new idea to choose a lambda value by finding the lambda that minimises the error across each of the test and training splits. In R:
# fit lasso model with crossvalidation across a range of lambda values
lasso < cv.glmnet(methyl_mat, age, alpha = 1)
plot(lasso)
# Extract the coefficients from the model with the lowest mean squared error from crossvalidation
coefl < coef(lasso, lasso$lambda.min)
# select only nonzero coefficients
selection < which(coefl != 0)
# and convert to a normal matrix
selected_coefs < as.matrix(coefl)[selection, 1]
selected_coefs
(Intercept) cg02388150 cg06493994 cg22449114 cg22736354
8.4133296328 0.6966503392 0.1615535465 6.4255580409 12.0507794749
cg03330058 cg09809672 cg11299964 cg19761273 cg26162695
0.0002362055 0.7487594618 2.0399663416 5.2538055304 0.4486970332
cg09502015 cg24771684 cg08446924 cg13215762 cg24549277
1.0787003366 4.5743800395 0.5960137381 0.1481402638 0.6290915767
cg12304482 cg13131095 cg17962089 cg13842639 cg04080666
1.0167896196 2.8860222552 6.3065284096 0.1590465147 2.4889065761
cg06147194 cg03669936 cg14230040 cg19848924 cg23964682
0.6838637838 0.0352696698 0.1280760909 0.0006938337 1.3378854603
cg13578928 cg02745847 cg17410295 cg17459023 cg06223736
0.8601170264 2.2346315955 2.3008028295 0.0370389967 1.6158734083
cg06717750 cg20138604 cg12851161 cg20972027 cg23878422
2.3401693309 0.0084327521 3.3033355652 0.2442751682 1.1059030593
cg16612298 cg03762081 cg14428146 cg16908369 cg16271524
0.0050053190 6.5228858163 0.3167227488 0.2302773154 1.3787104336
cg22071651 cg04262805 cg24969251 cg11233105 cg03156032
0.3480551279 1.1841804186 8.3024629942 0.6130598151 1.1121959544
We can see that crossvalidation has selected a value of $\lambda$ resulting in 44 features and the intercept.
Elastic nets: blending ridge regression and the LASSO
So far, we’ve used ridge regression (where alpha = 0
) and LASSO regression,
(where alpha = 1
). What if alpha
is set to a value between zero and one?
Well, this actually lets us blend the properties of ridge and LASSO
regression. This allows us to have the nice properties of the LASSO, where
uninformative variables are dropped automatically, and the nice properties
of ridge regression, where our coefficient estimates are a bit more
conservative, and as a result our predictions are a bit better.
Formally, the objective function of elastic net regression is to optimise the following function:
[\left(\sum_{i=1}^N y_i  x’_i\beta\right) + \lambda \left( \alpha \left\lVert \beta \right\lVert_1 + (1\alpha) \left\lVert \beta \right\lVert_2 \right)]
You can see that if alpha = 1
, then the L1 norm term is multiplied by one,
and the L2 norm is multiplied by zero. This means we have pure LASSO regression.
Conversely, if alpha = 0
, the L2 norm term is multiplied by one, and the L1
norm is multiplied by zero, meaning we have pure ridge regression. Anything
in between gives us something in between.
The contour plots we looked at previously to visualise what this penalty looks
like for different values of alpha
.
Challenge 6
 Fit an elastic net model (hint: alpha = 0.5) without crossvalidation and plot the model object.
 Fit an elastic net model with crossvalidation and plot the error. Compare with LASSO.
 Select the lambda within one standard error of the minimum crossvalidation error (hint:
lambda.1se
). Compare the coefficients with the LASSO model. Discuss: how could we pick an
alpha
in the range (0, 1)? Could we justify choosing one a priori?Solution
Fitting an elastic net model is just like fitting a LASSO model. You can see that coefficients tend to go exactly to zero, but the paths are a bit less extreme than with pure LASSO; similar to ridge.
elastic < glmnet(methyl_mat[, 1], age, alpha = 0.5) plot(elastic)
The process of model selection is similar for elastic net models as for LASSO models.
elastic_cv < cv.glmnet(methyl_mat[, 1], age, alpha = 0.5) plot(elastic_cv)
You can see that the coefficients from these two methods are broadly similar, but the elastic net coefficients are a bit more conservative. Further, more coefficients are exactly zero in the LASSO model.
coefe < coef(elastic_cv, elastic_cv$lambda.1se) sum(coefe[, 1] == 0)
[1] 4973
sum(coefl[, 1] == 0)
[1] 4956
plot( coefl[, 1], coefe[, 1], pch = 16, xlab = "LASSO coefficients", ylab = "Elastic net coefficients" )
Error in xy.coords(x, y, xlabel, ylabel, log): 'x' and 'y' lengths differ
abline(0:1, lty = "dashed")
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...): plot.new has not been called yet
You could pick an arbitrary value of
alpha
, because arguably pure ridge regression or pure LASSO regression are also arbitrary model choices. To be rigorous and to get the bestperforming model and the best inference about predictors, it’s usually best to find the best combination ofalpha
andlambda
using a grid search approach in crossvalidation. However, this can be very computationally demanding.
The biasvariance tradeoff
When we make predictions in statistics, there are two sources of error that primarily influence the (in)accuracy of our predictions. these are bias and variance.
The total expected error in our predictions is given by the following equation:
\[E(y  \hat{y}) = \text{Bias}^2 + \text{Variance} + \sigma^2\]Here, $\sigma^2$ represents the irreducible error, that we can never overcome. Bias results from erroneous assumptions in the model used for predictions. Fundamentally, bias means that our model is misspecified in some way, and fails to capture some components of the datagenerating process (which is true of all models). If we have failed to account for a confounding factor that leads to very inaccurate predictions in a subgroup of our population, then our model has high bias.
Variance results from sensitivity to particular properties of the input data. For example, if a tiny change to the input data would result in a huge change to our predictions, then our model has high variance.
Linear regression is an unbiased model under certain conditions. In fact, the GaussMarkov theorem shows that under the right conditions, OLS is the best possible type of unbiased linear model.
Introducing penalties to means that our model is no longer unbiased, meaning that the coefficients estimated from our data will systematically deviate from the ground truth. Why would we do this? As we saw, the total error is a function of bias and variance. By accepting a small amount of bias, it’s possible to achieve huge reductions in the total expected error.
This biasvariance tradeoff is also why people often favour elastic net regression over pure LASSO regression.
Other types of outcomes
You may have noticed that
glmnet
is written asglm
, notlm
. This means we can actually model a variety of different outcomes using this regularisation approach. For example, we can model binary variables using logistic regression, as shown below. The type of outcome can be specified using thefamily
argument, which specifies the family of the outcome variable.In fact,
glmnet
is somewhat cheeky as it also allows you to model survival using Cox proportional hazards models, which aren’t GLMs, strictly speaking.For example, in the current dataset we can model smoking status as a binary variable in logistic regression by setting
family = "binomial"
.The package documentation explains this in more detail.
smoking < as.numeric(factor(methylation$smoker))  1 # binary outcome table(smoking)
smoking 0 1 30 7
fit < cv.glmnet(x = methyl_mat, nfolds = 5, y = smoking, family = "binomial")
Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground Warning in lognet(xd, is.sparse, ix, jx, y, weights, offset, alpha, nobs, : one multinomial or binomial class has fewer than 8 observations; dangerous ground
coef < coef(fit, s = fit$lambda.min) coef < as.matrix(coef) coef[which(coef != 0), 1]
[1] 1.455287
plot(fit)
In this case, the results aren’t very interesting! We select an interceptonly model. However, as highlighted by the warnings above, we should not trust this result too much as the data was too small to obtain reliable results! We only included it here to provide the code that could be used to perform penalised regression for binary outcomes (i.e. penalised logistic regression).
tidymodels
A lot of the packages for fitting predictive models like regularised regression have different user interfaces. To do predictive modelling, it’s important to consider things like choosing a good performance metric and how to run normalisation. It’s also useful to compare different model “engines”.
To this end, the
tidymodels
R framework exists. We’re not doing a course on advanced topics in predictive modelling so we are not covering this framework in detail. However, the code below would be useful to perform repeated crossvalidation. More information abouttidymodels
, including installation instructions, can be found here.library("tidymodels") all_data < as.data.frame(cbind(age = age, methyl_mat)) split_data < initial_split(all_data) norm_recipe < recipe(training(split_data)) %>% ## everything other than age is a predictor update_role(everything(), new_role = "predictor") %>% update_role(age, new_role = "outcome") %>% ## center and scale all the predictors step_center(all_predictors()) %>% step_scale(all_predictors()) ## set the "engine" to be a linear model with tunable alpha and lambda glmnet_model < linear_reg(penalty = tune(), mixture = tune()) %>% set_engine("glmnet") ## define a workflow, with normalisation recipe into glmnet engine workflow < workflow() %>% add_recipe(norm_recipe) %>% add_model(glmnet_model) ## 5fold crossvalidation repeated 5 times folds < vfold_cv(training(split_data), v = 5, repeats = 5) ## define a grid of lambda and alpha parameters to search glmn_set < parameters( penalty(range = c(5, 1), trans = log10_trans()), mixture() ) glmn_grid < grid_regular(glmn_set) ctrl < control_grid(save_pred = TRUE, verbose = TRUE) ## use the metric "rmse" (root mean squared error) to grid search for the ## best model results < workflow %>% tune_grid( resamples = folds, metrics = metric_set(rmse), control = ctrl ) ## select the best model based on RMSE best_mod < results %>% select_best("rmse") best_mod ## finalise the workflow and fit it with all of the training data final_workflow < finalize_workflow(workflow, best_mod) final_workflow final_model < final_workflow %>% fit(data = training(split_data)) ## plot predicted age against true age for test data plot( testing(split_data)$age, predict(final_model, new_data = testing(split_data))$.pred, xlab = "True age", ylab = "Predicted age", pch = 16, log = "xy" ) abline(0:1, lty = "dashed")
Further reading
 An introduction to statistical learning.
 Elements of statistical learning.
 glmnet vignette.
 tidymodels.
Footnotes

Model selection including $R^2$, AIC and BIC are covered in the additional feature selection for regression episode of this course. ↩
Key Points
Regularisation is a way to fit a model, get better estimates of effect sizes, and perform variable selection simultaneously.
Test and training splits, or crossvalidation, are a useful way to select models or hyperparameters.
Regularisation can give us a more predictive set of variables, and by restricting the magnitude of coefficients, can give us a better (and more stable) estimate of our outcome.
Regularisation is often very fast! Compared to other methods for variable selection, it is very efficient. This makes it easier to practice rigorous variable selection.
Principal component analysis
Overview
Teaching: 90 min
Exercises: 40 minQuestions
What is principal component analysis (PCA) and when can it be used?
How can we perform a PCA in R?
How many principal components are needed to explain a significant amount of variation in the data?
How to interpret the output of PCA using loadings and principal components?
Objectives
Identify situations where PCA can be used to answer research questions using highdimensional data.
Perform a PCA on highdimensional data.
Select the appropriate number of principal components.
Interpret the output of PCA.
Introduction
If a dataset contains many variables ($p$), it is likely that some of these variables will be highly correlated. Variables may even be so highly correlated that they represent the same overall effect. Such datasets are challenging to analyse for several reasons, with the main problem being how to reduce dimensionality in the dataset while retaining the important features.
In this episode we will explore principal component analysis (PCA) as a popular method of analysing highdimensional data. PCA is a statistical method which allows large datasets of correlated variables to be summarised into smaller numbers of uncorrelated principal components that explain most of the variability in the original dataset. As an example, PCA might reduce several variables representing aspects of patient health (blood pressure, heart rate, respiratory rate) into a single feature capturing an overarching “patient health” effect. This is useful from an exploratory point of view, discovering how variables might be associated and combined. The resulting principal component could also be used as an effect in further analysis (e.g. linear regression).
Challenge 1
Descriptions of three datasets and research questions are given below. For which of these might PCA be considered a useful tool for analysing data so that the research questions may be addressed?
 An epidemiologist has data collected from different patients admitted to hospital with infectious respiratory disease. They would like to determine whether length of stay in hospital differs in patients with different respiratory diseases.
 A scientist has assayed gene expression levels in 1000 cancer patients and has data from probes targeting different genes in tumour samples from patients. She would like to create new variables representing relative abundance of different groups of genes to i) find out if genes form subgroups based on biological function and ii) use these new variables in a linear regression examining how gene expression varies with disease severity.
 Both of the above.
Solution
In the first case, a regression model would be more suitable; perhaps a survival model. In the second example, PCA can help to identify modules of correlated features that explain a large amount of variation within the data.
Therefore the answer here is 2.
Principal component analysis
PCA transforms a dataset of continuous variables into a new set of uncorrelated variables called “principal components”. The first principal component derived explains the largest amount of the variability in the underlying dataset. The second principal component derived explains the second largest amount of variability in the dataset and so on. Once the dataset has been transformed into principal components, we can extract a subset of the principal components in order of the variance they explain (starting with the first principal component that by definition explains the most variability, and then the second), giving new variables that explain a lot of the variability in the original dataset. Thus, PCA helps us to produce a lower dimensional dataset while keeping most of the information in the original dataset.
To see what these new principal components may look like, the figure below shows the logtransformed prostate specific antigen (lpsa
) versus cancer
volume (lcavol
) data from the prostate
data set introduced in Episode 1. Principal components are a collection of new, artificial data points, one for each individual observation called scores.
The red line on the plot represents the line passing through the scores (points) of the first principal component for each observation.
The angle that the first principal component line passes through the data points at is set to the direction with the highest
variability. The plotted first principal components can therefore be thought of reflecting the
effect in the data that has the highest variability. The second principal component explains the next highest amount of variability
in the data and is represented by the line perpendicular to the first (the green line). The second principal component can be thought of as
capturing the overall effect in the data that has the secondhighest variability.
The animation below illustrates how principal components are calculated from data. You can imagine that the black line is a rod and each red dashed line is a spring. The energy of each spring is proportional to its squared length. The direction of the first principal component is the one that minimises the total energy of all of the springs. In the animation below, the springs pull the rod, finding the direction of the first principal component when they reach equilibrium. We then use the length of the springs from the rod as the first principal component. This is explained in more detail on this Q&A website.
Mathematical description of PCA
Mathematically, each principal component is a linear combination of the variables in the dataset. That is, the first principal component values or scores, $Z_1$, are a linear combination of variables in the dataset, $X_1…X_p$, given by
\[Z_1 = a_{11}X_1 + a_{21}X_2 +....+a_{p1}X_p,\]where $a_{11}…a_{p1}$ represent principal component loadings.
In summary, the principal components’ values are called scores. The loadings can be thought of as the degree to which each original variable contributes to the principal component scores. In this episode, we will see how to perform PCA to summarise the information in highdimensional datasets.
How do we perform a PCA?
To illustrate how to perform PCA initially, we revisit the lowdimensional prostate
dataset.
The data come from a study which examined the
correlation between the level of prostate specific antigen and a number of
clinical measures in 97 men who were about to receive a radical prostatectomy.
The data have 97 rows and 9 columns.
The variables in the dataset (columns) are:
lcavol
(logtransformed cancer volume),lweight
(logtransformed prostate weight),lbph
(logtransformed amount of benign prostate enlargement),svi
(seminal vesicle invasion),lcp
(logtransformed capsular penetration; amount of spread of cancer in outer walls of prostate),gleason
(Gleason score; grade of cancer cells),pgg45
(percentage Gleason scores 4 or 5),lpsa
(logtransformed prostate specific antigen; level of PSA in blood).age
(patient age in years).
We will perform PCA on the five continuous clinical variables in our dataset so that we can create fewer variables representing clinical markers of cancer progression.
First, we will examine the prostate
dataset (originally part of the
lasso2
package):
library("here")
prostate < readRDS(here("data/prostate.rds"))
head(prostate)
lcavol lweight age lbph svi lcp gleason pgg45 lpsa
1 0.5798185 2.769459 50 1.386294 0 1.386294 6 0 0.4307829
2 0.9942523 3.319626 58 1.386294 0 1.386294 6 0 0.1625189
3 0.5108256 2.691243 74 1.386294 0 1.386294 7 20 0.1625189
4 1.2039728 3.282789 58 1.386294 0 1.386294 6 0 0.1625189
5 0.7514161 3.432373 62 1.386294 0 1.386294 6 0 0.3715636
6 1.0498221 3.228826 50 1.386294 0 1.386294 6 0 0.7654678
Note that each row of the dataset represents a single patient.
We will create a subset of the data including only the clinical variables we want to use in the PCA.
pros2 < prostate[, c("lcavol", "lweight", "lbph", "lcp", "lpsa")]
head(pros2)
lcavol lweight lbph lcp lpsa
1 0.5798185 2.769459 1.386294 1.386294 0.4307829
2 0.9942523 3.319626 1.386294 1.386294 0.1625189
3 0.5108256 2.691243 1.386294 1.386294 0.1625189
4 1.2039728 3.282789 1.386294 1.386294 0.1625189
5 0.7514161 3.432373 1.386294 1.386294 0.3715636
6 1.0498221 3.228826 1.386294 1.386294 0.7654678
Do we need to scale the data?
PCA derives principal components based on the variance they explain in the data. Therefore, we may need to apply some preprocessing to scale variables in our dataset if we want to ensure that each variable is considered equally by the PCA. Scaling is essential if we want to avoid the PCA ignoring variables that may be important to our analysis just because they take low values and have low variance. We do not need to scale if we want variables with low variance to carry less weight in the PCA.
In this example, we want each variable to be treated equally by the PCA since variables with lower values may be just as informative as variables with higher values. Let’s therefore investigate the variables in our dataset to see if we need to scale our variables first:
apply(pros2, 2, var)
lcavol lweight lbph lcp lpsa
1.389157 0.246642 2.104840 1.955102 1.332476
par(mfrow = c(1, 2))
hist(pros2$lweight)
hist(pros2$lbph)
Note that variance is greatest for lbph
and lowest for lweight
. Since we
want each of the variables to be treated equally in our PCA, but there are large differences in the variances of the variables, we need to scale each of the variables before including
them in a PCA to ensure that differences in variances
do not drive the calculation of principal components. In this example, we
standardise all five variables to have a mean of 0 and a standard
deviation of 1.
Challenge 2
 Why might it be necessary to standardise variables before performing a PCA?
Some ideas:
 To make the results of the PCA interesting.
 If you want to ensure that variables with different ranges of values contribute equally to analysis.
 To allow the feature matrix to be calculated faster, especially in cases where there are a lot of input variables.
 To allow both continuous and categorical variables to be included in the PCA.
 All of the above.
 Can you think of datasets where it might not be necessary to standardise variables?
Solution
Scaling the data isn’t guaranteed to make the results more interesting. It also won’t affect how quickly the output will be calculated, whether continuous and categorical variables are present or not.
We use scaling to ensure that features with different ranges of values have equal weighting in the resulting principal components (point 2).
You may not want to standardise datasets which contain continuous variables all measured on the same scale (e.g. gene expression data or RNA sequencing data). In this case, variables with very little sampletosample variability may represent only random noise, and standardising the data would give these extra weight in the PCA.
Performing PCA
Throughout this episode, we will carry out PCA using the Bioconductor package PCAtools
.
This package provides several functions that are useful for exploring data via PCA and
producing useful figures and analysis tools. The package is made for the somewhat unusual
Bioconductor style of data tables (observations in columns, features in rows). When
using Bioconductor data sets and PCAtools
, it is thus not necessary to transpose the data.
You can use the help files in the package documentation to find out about the pca()
function (type help("pca")
or ?pca
in R). Although we focus on PCA implementation using this package, there are several other
options for performing PCA, including a base R function, prcomp()
. Equivalent functions and
output labels for the common implementations are summarised below.
library::command()  Principal component scores  Loadings 

stats::prcomp()  $x  $rotation 
stats::princomp()  $scores  $loadings 
PCAtools::pca()  $rotated  $loadings 
Let’s use the PCAtools
package to perform PCA on the scaled prostate
data. First, let’s load the PCAtools
package.
library("PCAtools")
The input data (pros2
) is in the form of a matrix and we need to take the transpose of this
matrix and convert it to a data frame for use within the PCAtools
package. We set the scale = TRUE
argument to standardise the variables to have a mean 0 and standard deviation of 1 as above.
pros2t < data.frame(t(pros2)) # create transposed data frame
colnames(pros2t)<seq(1:dim(pros2t)[2]) # rename columns for use within PCAtools
pmetadata= data.frame("M" = rep(1, dim(pros2t)[2]), row.names = colnames(pros2t)) # create fake metadata for use with PCA tools
## implement PCA
pca.pros < pca(pros2t, scale = TRUE, center = TRUE, metadata = pmetadata)
summary(pca.pros)
Length Class Mode
rotated 5 data.frame list
loadings 5 data.frame list
variance 5 none numeric
sdev 5 none numeric
metadata 1 data.frame list
xvars 5 none character
yvars 97 none character
components 5 none character
How many principal components do we need?
We have calculated one principal component for each variable in the original dataset. How do we choose how many of these are necessary to represent the true variation in the data, without having extra components that are unnecessary?
Let’s look at the relative importance of (percentage of the variance explained by) the components:
prop.var < round(pca.pros$variance, digits = 4)
prop.var
PC1 PC2 PC3 PC4 PC5
48.9767 27.3063 11.1094 8.0974 4.5101
This returns the proportion of variance in the data explained by each of the principal components. In this example, the first principal component, PC1, explains approximately 49% of variance in the data, PC2 27% of variance, PC3 a further 11%, PC4 approximately 8% and PC5 around 5%.
Let’s visualise this. A plot of the amount of variance accounted for by each principal component is also called a scree plot. Often, scree plots show a characteristic pattern where initially, the variance drops rapidly with each additional principal component. But then there is an “elbow” after which the variance decreases more slowly. The total variance explained up to the elbow point is sometimes interpreted as structural variance that is relevant and should be retained versus noise which may be discarded after the elbow.
We can create a scree plot using the PCAtools
function screeplot()
:
screeplot(pca.pros, axisLabSize = 5, titleLabSize = 8,
drawCumulativeSumLine = FALSE, drawCumulativeSumPoints = FALSE) +
geom_line(aes(x = 1:length(pca.pros$components), y = as.numeric(pca.pros$variance)))
The scree plot shows that the first principal component explains most of the variance in the data (>50%) and each subsequent principal component explains less and less of the total variance. The first two principal components explain >70% of variance in the data.
Selecting a subset of principal components results in loss of information from the original dataset but selecting principal components to summarise the majority of the information in the original dataset is useful from a practical dimension reduction perspective. There are no clear guidelines on how many principal components should be included in PCA. We often look at the ‘elbow’ on the scree plot as an indicator that the addition of principal components does not drastically contribute to explaining the remaining variance. In this case, the ‘elbow’ of the scree plot appears to be at the third principal component. We may therefore conclude that three principal components are sufficient to explain the majority of the variability in the original dataset. We may also choose another criterion, such as an arbitrary cutoff for the proportion of the variance explained compared to other principal components. Essentially, the criterion used to select principal components should be determined based on what is deemed a sufficient level of information retention for a specific dataset and question.
Loadings and principal component scores
Most PCA functions will produce two main output matrices: the principal component scores and the loadings. The matrix of principal component scores has as many rows as there were observations in the input matrix. These scores are what is usually visualised or used for downstream analyses. The matrix of loadings (also called rotation matrix) has as many rows as there are features in the original data. It contains information about how the (usually centered and scaled) original data relate to the principal component scores.
We can print the loadings for the PCAtools
implementation using
pca.pros$loadings
PC1 PC2 PC3 PC4 PC5
lcavol 0.5616465 0.23664270 0.01486043 0.22708502 0.75945046
lweight 0.2985223 0.60174151 0.66320198 0.32126853 0.07577123
lbph 0.1681278 0.69638466 0.69313753 0.04517286 0.06558369
lcp 0.4962203 0.31092357 0.26309227 0.72394666 0.25253840
lpsa 0.5665123 0.01680231 0.10141557 0.56487128 0.59111493
The principal component scores are obtained by carrying out matrix multiplication of the (usually centered and scaled) original data times the loadings. The following callout demonstrates this.
Computing a PCA “by hand”
The rotation matrix obtained in a PCA is identical to the eigenvectors of the covariance matrix of the data. Multiplying these with the (centered and scaled) data yields the principal component scores:
pros2.scaled < scale(pros2) # centre and scale the prostate data pros2.cov < cov(pros2.scaled) # generate covariance matrix pros2.cov
lcavol lweight lbph lcp lpsa lcavol 1.0000000 0.1941283 0.027349703 0.675310484 0.7344603 lweight 0.1941283 1.0000000 0.434934636 0.100237795 0.3541204 lbph 0.0273497 0.4349346 1.000000000 0.006999431 0.1798094 lcp 0.6753105 0.1002378 0.006999431 1.000000000 0.5488132 lpsa 0.7344603 0.3541204 0.179809410 0.548813169 1.0000000
pros2.eigen < eigen(pros2.cov) # preform eigen decomposition pros2.eigen # The slot $vectors = rotation of the PCA
eigen() decomposition $values [1] 2.4488355 1.3653171 0.5554705 0.4048702 0.2255067 $vectors [,1] [,2] [,3] [,4] [,5] [1,] 0.5616465 0.23664270 0.01486043 0.22708502 0.75945046 [2,] 0.2985223 0.60174151 0.66320198 0.32126853 0.07577123 [3,] 0.1681278 0.69638466 0.69313753 0.04517286 0.06558369 [4,] 0.4962203 0.31092357 0.26309227 0.72394666 0.25253840 [5,] 0.5665123 0.01680231 0.10141557 0.56487128 0.59111493
# generate principal component scores by by hand, using matrix multiplication my.pros2.pcs < pros2.scaled %*% pros2.eigen$vectors # compare results par(mfrow=c(1,2)) plot(pca.pros$rotated[,1:2], main="PCAtools") abline(h=0, v=0, lty=2) plot(my.pros2.pcs[,1:2], main="\"By hand\"", xlab="PC1", ylab="PC2") abline(h=0, v=0, lty=2)
par(mfrow=c(1,1)) # Note that the axis orientations may be swapped but the relative positions of the dots should be the same in both plots.
One way to visualise how principal components relate to the original variables
is by creating a biplot. Biplots usually show two principal components plotted
against each other. Observations are sometimes labelled with numbers. The
contribution of each original variable to the principal components displayed
is then shown by arrows (generated from those two columns of the rotation matrix that
correspond to the principal components shown). See help("PCAtools::biplot")
for
arguments and their meaning. For instance, lab
or colBy
may be useful.
Note that there are several biplot
implementations in different R libraries. It is thus a good idea to specify
the desired package when calling biplot()
. A biplot of the first two principal
components can be generated as follows:
PCAtools::biplot(pca.pros)
This biplot shows the position of each patient on a 2dimensional plot where
loadings can be observed via the red arrows associated with each of
the variables. The variables lpsa
, lcavol
and lcp
are associated with
positive values on PC1 while positive values on PC2 are associated with the
variables lbph
and lweight
. The length of the arrows indicates how much
each variable contributes to the calculation of each principal component.
The left and bottom axes show normalised principal component scores. The axes
on the top and right of the plot are used to interpret the loadings, where
loadings are scaled by the standard deviation of the principal components
(pca.pros$sdev
) times the square root the number of observations.
Advantages and disadvantages of PCA
Advantages:
 It is a relatively easy to use and popular method.
 Various software/packages are available to run a PCA.
 The calculations used in a PCA are simple to understand compared to other methods for dimension reduction.
Disadvantages:
 It assumes that variables in a dataset are correlated.
 It is sensitive to the scale at which input variables are measured. If input variables are measured at different scales, the variables with large variance relative to the scale of measurement will have greater impact on the principal components relative to variables with smaller variance. In many cases, this is not desirable.
 It is not robust against outliers, meaning that very large or small data points can have a large effect on the output of the PCA.
 PCA assumes a linear relationship between variables which is not always a realistic assumption.
 It can be difficult to interpret the meaning of the principal components, especially when including them in further analyses (e.g. inclusion in a linear regression).
Using PCA to analyse gene expression data
The dataset we will be analysing in this lesson includes two subsets of data:
 a matrix of gene expression data showing microarray results for different probes used to examine gene expression profiles in 91 different breast cancer patient samples.
 metadata associated with the gene expression results detailing information from patients from whom samples were taken.
In this section, we will work through Challenges and you will apply your own PCA to the highdimensional gene expression data using what we have learnt so far.
We will first load the microarray breast cancer gene expression data and associated metadata, downloaded from the Gene Expression Omnibus.
library("SummarizedExperiment")
cancer < readRDS(here("data/cancer_expression.rds"))
mat < assay(cancer)
metadata < colData(cancer)
head(mat)
head(metadata)
# Check that column names and row names match, if they do should return TRUE
all(colnames(mat) == rownames(metadata))
[1] TRUE
The mat
variable contains a matrix of gene expression profiles for each sample.
Rows represent gene expression measurements and columns represent samples. The
metadata
variable contains the metadata associated with the gene expression
data including the name of the study from which data originate, the age of the
patient from which the sample was taken, whether or not an oestrogen receptor
was involved in their cancer and the grade and size of the cancer for each
sample (represented by rows).
Microarray data are difficult to analyse for several reasons. Firstly, formulating a research question using microarray data can be difficult, especially if not much is known a priori about which genes code for particular phenotypes of interest. Secondly, they are typically highdimensional and therefore are subject to the same difficulties associated with analysing highdimensional data. Finally, exploratory analysis, which can be used to help formulate research questions and display relationships, is difficult using microarray data due to the number of potentially interesting response variables (i.e. expression data from probes targeting different genes).
If researchers hypothesise that groups of genes (e.g. biological pathways) may be associated with different phenotypic characteristics of cancers (e.g. histologic grade, tumour size), using statistical methods that reduce the number of columns in the microarray matrix to a smaller number of dimensions representing groups of genes would help visualise the data and address research questions regarding the effect different groups of genes have on disease progression.
Using PCAtools
, we will perform PCA on the cancer
gene expression data, plot the amount of variation in the data explained by
each principal component and plot the most important principal components
against each other as well as understanding what each principal component
represents.
Challenge 3
Apply a PCA to the cancer gene expression data using the
pca()
function fromPCAtools
.Let us assume we only care about the principal components accounting for the top 80% of the variance in the dataset. Use the
removeVar
argument inpca()
to remove the principal components accounting for the bottom 20%.As in the example using
prostate
data above, examine the first 5 rows and columns of rotated data and loadings from your PCA.Solution
pc < pca(mat, metadata = metadata) # implement a PCA # We can check the scree plot to see that many principal components explain a very small amount of # the total variance in the data # Let's remove the principal components with lowest 20% of the variance pc < pca(mat, metadata = metadata, removeVar = 0.2) # Explore other arguments provided in pca pc$rotated[1:5, 1:5] # obtain the first 5 principal component scores for the first 5 observations
PC1 PC2 PC3 PC4 PC5 GSM65752 29.79105 43.866788 3.255903 40.663138 15.3427597 GSM65753 37.33911 15.244788 4.948201 6.182795 9.4725870 GSM65755 29.41462 7.846858 22.880525 16.149669 22.3821009 GSM65757 33.35286 1.343573 22.579568 2.200329 15.0082786 GSM65758 40.51897 8.491125 5.288498 14.007364 0.8739772
pc$loadings[1:5, 1:5] # obtain the first 5 principal component loadings for the first 5 features
PC1 PC2 PC3 PC4 PC5 206378_at 0.0024680993 0.053253543 0.004068209 0.04068635 0.015078376 205916_at 0.0051557973 0.001315022 0.009836545 0.03992371 0.038552048 206799_at 0.0005684075 0.050657061 0.009515725 0.02610233 0.006208078 205242_at 0.0130742288 0.028876408 0.007655420 0.04449641 0.001061205 206509_at 0.0019031245 0.054698479 0.004667356 0.01566468 0.001306807
which.max(pc$loadings[, 1]) # index of the maximal loading for the first principal component
[1] 49
# principal component loadings for the feature # with the maximal loading on the first principal component: pc$loadings[which.max(pc$loadings[, 1]), ]
PC1 PC2 PC3 PC4 PC5 215281_x_at 0.03752947 0.007369379 0.006243377 0.008242589 0.004783206 PC6 PC7 PC8 PC9 PC10 215281_x_at 0.01194012 0.002822407 0.01216792 0.001137451 0.0009056616 PC11 PC12 PC13 PC14 PC15 215281_x_at 0.00196034 0.0001676705 0.00699201 0.002897995 0.0005044658 PC16 PC17 PC18 PC19 PC20 215281_x_at 0.0004547916 0.002277035 0.006199078 0.002708574 0.006217326 PC21 PC22 PC23 PC24 PC25 215281_x_at 0.00516745 0.007625912 0.003434534 0.005460017 0.001477415 PC26 PC27 PC28 PC29 PC30 215281_x_at 0.002350428 0.0007183107 0.0006195515 0.0006349803 0.00413627 PC31 PC32 PC33 PC34 PC35 215281_x_at 0.0001322301 0.003182956 0.002123462 0.001042769 0.001729869 PC36 PC37 PC38 PC39 PC40 215281_x_at 0.006556369 0.005766949 0.002537993 0.0002846248 0.00018195 PC41 PC42 PC43 PC44 PC45 215281_x_at 0.0007970789 0.003888626 0.008210075 0.0009570174 0.0007998935 PC46 PC47 PC48 PC49 PC50 215281_x_at 0.0006931441 0.005717836 0.005189649 0.002591188 0.0007810259 PC51 PC52 PC53 PC54 PC55 215281_x_at 0.006610815 0.005371134 0.001704796 0.002286475 0.001365417 PC56 PC57 PC58 PC59 PC60 215281_x_at 0.003529892 0.0003375981 0.009895923 0.001564423 0.006989092 PC61 PC62 PC63 PC64 PC65 215281_x_at 0.000971273 0.001345406 0.003575415 0.0005588113 0.006516669 PC66 PC67 PC68 PC69 PC70 215281_x_at 0.008770186 0.006699641 0.01284606 0.005041574 0.007845653 PC71 PC72 PC73 PC74 PC75 215281_x_at 0.003964697 0.01104367 0.001506485 0.001583824 0.003798343 PC76 PC77 PC78 PC79 PC80 215281_x_at 0.004817252 0.001290033 0.004402926 0.003440367 0.0001646198 PC81 PC82 PC83 PC84 PC85 215281_x_at 0.003923775 0.003179556 0.0004388192 9.664648e05 0.003501335 PC86 PC87 PC88 PC89 PC90 215281_x_at 0.00112973 0.006489667 0.0005039785 0.004296355 0.002751513 PC91 215281_x_at 0.00383085
which.max(pc$loadings[, 2]) # index of the maximal loading for the second principal component
[1] 27
# principal component loadings for the feature # with the maximal loading on the second principal component: pc$loadings[which.max(pc$loadings[, 2]), ]
PC1 PC2 PC3 PC4 PC5 211122_s_at 0.01649085 0.05090275 0.003378728 0.05178144 0.003742393 PC6 PC7 PC8 PC9 PC10 211122_s_at 0.00543753 0.03522848 0.006333521 0.01575401 0.004732546 PC11 PC12 PC13 PC14 PC15 211122_s_at 0.004687599 0.01349892 0.005207937 0.01731898 0.02323893 PC16 PC17 PC18 PC19 PC20 211122_s_at 0.02069509 0.01477432 0.005658529 0.02667751 0.01333503 PC21 PC22 PC23 PC24 PC25 211122_s_at 0.003254036 0.003572342 0.01416779 0.005511838 0.02582847 PC26 PC27 PC28 PC29 PC30 PC31 211122_s_at 0.03405417 0.01797345 0.01826328 0.005123959 0.01300763 0.0127127 PC32 PC33 PC34 PC35 PC36 211122_s_at 0.002477672 0.01933214 0.03017661 0.01935071 0.01960912 PC37 PC38 PC39 PC40 PC41 211122_s_at 0.004411188 0.01263612 0.02019279 0.01441513 0.0310399 PC42 PC43 PC44 PC45 PC46 211122_s_at 0.02540426 0.0007949801 0.00200195 0.01748543 0.006881834 PC47 PC48 PC49 PC50 PC51 211122_s_at 0.006690698 0.004000732 0.02747926 0.006963189 0.02232332 PC52 PC53 PC54 PC55 PC56 211122_s_at 0.0003089115 0.01604491 0.005649511 0.02629501 0.02332997 PC57 PC58 PC59 PC60 PC61 211122_s_at 0.01248022 0.01563245 0.005369433 0.009445262 0.005209349 PC62 PC63 PC64 PC65 PC66 211122_s_at 0.01787645 0.01629425 0.02457665 0.02384242 0.002814479 PC67 PC68 PC69 PC70 PC71 211122_s_at 0.0004584731 0.007939733 0.009554166 0.003967123 0.01825668 PC72 PC73 PC74 PC75 PC76 211122_s_at 0.00580374 0.02236727 0.001295688 0.02264723 0.006855855 PC77 PC78 PC79 PC80 PC81 211122_s_at 0.004995447 0.008404118 0.00442875 0.001027912 0.006104406 PC82 PC83 PC84 PC85 PC86 211122_s_at 0.01988441 0.009667348 0.008248781 0.01198369 0.01221713 PC87 PC88 PC89 PC90 PC91 211122_s_at 0.003864842 0.02876816 0.01771452 0.02164973 0.004593411
The function
pca()
is used to perform PCA, and uses as input a matrix (mat
) containing continuous numerical data in which rows are data variables and columns are samples, andmetadata
associated with the matrix in which rows represent samples and columns represent data variables. It has options to centre or scale the input data before a PCA is performed, although in this case gene expression data do not need to be transformed prior to PCA being carried out as variables are measured on a similar scale (values are comparable between rows). The output of thepca()
function includes a lot of information such as loading values for each variable (loadings
), principal component scores (rotated
) and the amount of variance in the data explained by each principal component.Rotated data shows principal component scores for each sample and each principal component. Loadings the contribution each variable makes to each principal component.
Scaling variables for PCA
When running
pca()
above, we kept the default setting,scale=FALSE
. That means genes with higher variation in their expression levels should have higher loadings, which is what we are interested in. Whether or not to scale variables for PCA will depend on your data and research question.Note that this is different from normalising gene expression data. Gene expression data have to be normalised before donwstream analyses can be carried out. This is to reduce technical and other potentially confounding factors. We assume that the expression data we use had been normalised previously.
Choosing how many components are important to explain the variance in the data
As in the example using the prostate
dataset, we can use a scree plot to
compare the proportion of variance in the data explained by each principal
component. This allows us to understand how much information in the microarray
dataset is lost by projecting the observations onto the first few principal
components and whether these principal components represent a reasonable
amount of the variation. The proportion of variance explained should sum to one.
Challenge 4
This time using the
screeplot()
function inPCAtools
, create a scree plot to show proportion of variance explained by the first 20 principal component (hint:components = 1:20
). Explain the output of the scree plot in terms of proportion of the variance in the data explained by each principal component.Solution
pc < pca(mat, metadata = metadata) screeplot(pc, components = 1:20) + ylim(0, 80)
The first principal component explains around 33% of the variance in the microarray data, the first 4 principal components explain around 50%, and 20 principal components explain around 75%. Many principal components explain very little variation. A first ‘elbow’ appears to be around 45 principal components, indicating that this may be a suitable number of principal components. However, these principal components cumulatively explain only 5155% of the variance in the dataset. Although the fact we are able to summarise most of the information in the complex dataset in 45 principal components may be a useful result, we may opt to retain more principal components (for example, 20) to capture more of the variability in the dataset depending on research question. A second ‘elbow’ around 12 principal components may provide a good middleground. Note that first principal component (PC1) explains more variation than other principal components (which is always the case in PCA). The scree plot shows that the first principal component only explains ~33% of the total variation in the microarray data and many principal components explain very little variation. The red line shows the cumulative percentage of explained variation with increasing principal components.
Note that in this case 18 principal components are needed to explain over 75% of variation in the data. This is not an unusual result for complex biological datasets including genetic information as clear relationships between groups are sometimes difficult to observe in the data. The scree plot shows that using a PCA we have reduced 91 predictors to 18 in order to explain a significant amount of variation in the data. See additional arguments in scree plot function for improving the appearance of the plot.
Investigating the principal components
Once the most important principal components have been identified using
screeplot()
and explored, these can be explored in more detail by plotting principal components
against each other and highlighting points based on variables in the metadata.
This will allow any potential clustering of points according to demographic or
phenotypic variables to be seen.
We can use biplots to look for patterns in the output from the PCA. Note that there
are two functions called biplot()
, one in the package PCAtools
and one in
stats
. Both functions produce biplots but their scales are different!
Challenge 5
Create a biplot of the first two principal components from your PCA using
biplot()
function inPCAtools
. Seehelp("PCAtools::biplot")
for arguments and their meaning. For instance,lab
orcolby
may be useful.Examine whether the data appear to form clusters. Explain your results.
Solution
PCAtools::biplot(pc, lab = NULL, colby = 'Grade', legendPosition = 'top')
The biplot shows the position of patient samples relative to PC1 and PC2 in a 2dimensional plot. Note that two groups are apparent along the PC1 axis according to expressions of different genes while no separation can be seem along the PC2 axis. Labels of patient samples are automatically added in the biplot. Labels for each sample are added by default, but can be removed if there is too much overlap in names. Note that
PCAtools
does not scale biplot in the same way as biplot using thestats
package.
Let’s consider this biplot in more detail, and also display the loadings:
Challenge 6
Use
colby
andlab
arguments inbiplot()
to explore whether these two groups may cluster by patient age or by whether or not the sample expresses the oestrogen receptor gene (ER+ or ER).Note: You may see a warning about
ggrepel
. This happens when there are many labels but little space for plotting. This is not usually a serious problem  not all labels will be shown.Solution
PCAtools::biplot(pc, lab = paste0(pc$metadata$Age,'years'), colby = 'ER', legendPosition = 'right')
Warning: ggrepel: 36 unlabeled data points (too many overlaps). Consider increasing max.overlaps
It appears that one cluster has more ER+ samples than the other group.
So far, we have only looked at a biplot of PC1 versus PC2 which only gives part
of the picture. The pairplots()
function in PCAtools
can be used to create
multiple biplots including different principal components.
pairsplot(pc)
The plots show two apparent clusters involving the first principal component
only. No other clusters are found involving other principal components. Each dot
is coloured differently along a gradient of blues. This can potentially help identifying
the same observation/individual in several panels. Here too, the argument colby
allows
you to set custom colours.
Finally, it can sometimes be of interest to compare how certain variables contribute
to different principal components. This can be visualised with plotloadings()
from
the PCAtools
package. The function checks the range of loadings for each
principal component specified (default: first five principal components). It then selects the features
in the top and bottom 5% of these ranges and displays their loadings. This behaviour
can be adjusted with the rangeRetain
argument, which has 0.1 as the default value (i.e.
5% on each end of the range). NB, if there are too many labels to be plotted, you will see
a warning. This is not a serious problem.
plotloadings(pc, c("PC1"), rangeRetain = 0.1)
plotloadings(pc, c("PC2"), rangeRetain = 0.1)
plotloadings(pc, c("PC1", "PC2"), rangeRetain = 0.1)
You can see how the third code line produces more dots, some of which do not have extreme loadings. This is because all loadings selected for any principal component are shown for all other principal components. For instance, it is plausible that features which have high loadings on PC1 may have lower ones on PC2.
Using PCA output in further analysis
The output of PCA can be used to interpret data or can be used in further analyses. For example, the PCA outputs new variables (principal components) which represent several variables in the original dataset. These new variables are useful for further exploring data, for example, comparing principal component scores between groups or including the new variables in linear regressions. Because the principal components are uncorrelated (and independent) they can be included together in a single linear regression.
Principal component regression
PCA is often used to reduce large numbers of correlated variables into fewer uncorrelated variables that can then be included in linear regression or other models. This technique is called principal component regression (PCR) and it allows researchers to examine the effect of several correlated explanatory variables on a single response variable in cases where a high degree of correlation initially prevents them from being included in the same model. Principal component regression (PCR) is just one example of how principal components can be used in further analysis of data. When carrying out PCR, the variable of interest (response/dependent variable) is regressed against the principal components calculated using PCA, rather than against each individual explanatory variable from the original dataset. As there as many principal components created from PCA as there are variables in the dataset, we must select which principal components to include in PCR. This can be done by examining the amount of variation in the data explained by each principal component (see above).
Further reading
 James, G., Witten, D., Hastie, T. & Tibshirani, R. (2013) An Introduction to Statistical Learning with Applications in R. Chapter 6.3 (Dimension Reduction Methods), Chapter 10 (Unsupervised Learning).
 Jolliffe, I.T. & Cadima, J. (2016) Principal component analysis: a review and recent developments. Phil. Trans. R. Soc A 374..
 Johnstone, I.M. & Titterington, D.M. (2009) Statistical challenges of highdimensional data. Phil. Trans. R. Soc A 367.
 PCA: A Practical Guide to Principal Component Analysis, Analytics Vidhya.
 A OneStop Shop for Principal Component Analysis, Towards Data Science.
Key Points
A principal component analysis is a statistical approach used to reduce dimensionality in highdimensional datasets (i.e. where $p$ is equal or greater than $n$).
PCA may be used to create a lowdimensional set of features from a larger set of variables. Examples of when a PCA may be useful include reducing highdimensional datasets to fewer variables for use in a linear regression and for identifying groups with similar features.
PCA is a useful dimensionality reduction technique used in the analysis of complex biological datasets (e.g. high throughput data or genetics data).
The first principal component represents the dimension along which there is maximum variation in the data. Subsequent principal components represent dimensions with progressively less variation.
Scree plots and biplots may be used to show: 1. how much variation in the data is explained by each principal component and 2. how data points cluster according to principal component scores and which variables are associated with these scores.
Factor analysis
Overview
Teaching: 30 min
Exercises: 10 minQuestions
What is factor analysis and when can it be used?
What are communality and uniqueness in factor analysis?
How to decide on the number of factors to use?
How to interpret the output of factor analysis?
Objectives
Perform a factor analysis on highdimensional data.
Select an appropriate number of factors.
Interpret the output of factor analysis.
Introduction
Biologists often encounter highdimensional datasets from which they wish to extract underlying features – they need to carry out dimensionality reduction. The last episode dealt with one method to achieve this, called principal component analysis (PCA), which expressed new dimensionreduced components as linear combinations of the original features in the dataset. Principal components can therefore be difficult to interpret. Here, we introduce a related but more interpretable method called factor analysis (FA), which constructs new components, called factors, that explicitly represent underlying (latent) constructs in our data. Like PCA, FA uses linear combinations, but uses latent constructs instead. FA is therefore often more interpretable and useful when we would like to extract meaning from our dimensionreduced set of variables.
There are two types of FA, called exploratory and confirmatory factor analysis (EFA and CFA). Both EFA and CFA aim to reproduce the observed relationships among a group of features with a smaller set of latent variables. EFA is used in a descriptive (exploratory) manner to uncover which measured variables are reasonable indicators of the various latent dimensions. In contrast, CFA is conducted in an a priori, hypothesistesting manner that requires strong empirical or theoretical foundations. We will mainly focus on EFA here, which is used to group features into a specified number of latent factors.
Unlike with PCA, researchers using FA have to specify the number of latent variables (factors) at the point of running the analysis. Researchers may use exploratory data analysis methods (including PCA) to provide an initial estimate of how many factors adequately explain the variation observed in a dataset. In practice, a range of different values is usually tested.
Motivating example: student scores
One scenario for using FA would be whether student scores in different subjects can be summarised by certain subject categories. Take a look at the hypothetical dataset below. If we were to run and EFA on this, we might find that the scores can be summarised well by two factors, which we can then interpret. We have labelled these hypothetical factors “mathematical ability” and “writing ability”.
So, EFA is designed to identify a specified number of unobservable factors from observable features contained in the original dataset. This is slightly different from PCA, which does not do this directly. Just to recap, PCA creates as many principal components as there are features in the dataset, each component representing a different linear combination of features. The principal components are ordered by the amount of variance they account for.
Prostate cancer patient data
We revisit the [prostate
]((https://carpentriesincubator.github.io/highdimensionalstatsr/data/index.html)
dataset of 97 men who have prostate cancer.
Although not strictly a highdimensional dataset, as with other episodes,
we use this dataset to explore the method.
In this example, we use the clinical variables to identify factors representing various clinical variables from prostate cancer patients. Two principal components have already been identified as explaining a large proportion of variance in the data when these data were analysed in the PCA episode. We may expect a similar number of factors to exist in the data.
Let’s subset the data to just include the logtransformed clinical variables for the purposes of this episode:
library("here")
prostate < readRDS(here("data/prostate.rds"))
nrow(prostate)
[1] 97
head(prostate)
lcavol lweight age lbph svi lcp gleason pgg45 lpsa
1 0.5798185 2.769459 50 1.386294 0 1.386294 6 0 0.4307829
2 0.9942523 3.319626 58 1.386294 0 1.386294 6 0 0.1625189
3 0.5108256 2.691243 74 1.386294 0 1.386294 7 20 0.1625189
4 1.2039728 3.282789 58 1.386294 0 1.386294 6 0 0.1625189
5 0.7514161 3.432373 62 1.386294 0 1.386294 6 0 0.3715636
6 1.0498221 3.228826 50 1.386294 0 1.386294 6 0 0.7654678
#select five logtransformed clinical variables for further analysis
pros2 < prostate[, c("lcavol", "lweight", "lbph", "lcp", "lpsa")]
head(pros2)
lcavol lweight lbph lcp lpsa
1 0.5798185 2.769459 1.386294 1.386294 0.4307829
2 0.9942523 3.319626 1.386294 1.386294 0.1625189
3 0.5108256 2.691243 1.386294 1.386294 0.1625189
4 1.2039728 3.282789 1.386294 1.386294 0.1625189
5 0.7514161 3.432373 1.386294 1.386294 0.3715636
6 1.0498221 3.228826 1.386294 1.386294 0.7654678
Performing exploratory factor analysis
EFA may be implemented in R using the factanal()
function
from the stats
package (which is a builtin package in base R). This
function fits a factor analysis by maximising the loglikelihood using a
data matrix as input. The number of factors to be fitted in the analysis
is specified by the user using the factors
argument.
Challenge 1 (3 mins)
Use the
factanal()
function to identify the minimum number of factors necessary to explain most of the variation in the dataSolution
# Include one factor only pros_fa < factanal(pros2, factors = 1) pros_fa
Call: factanal(x = pros2, factors = 1) Uniquenesses: lcavol lweight lbph lcp lpsa 0.149 0.936 0.994 0.485 0.362 Loadings: Factor1 lcavol 0.923 lweight 0.253 lbph lcp 0.718 lpsa 0.799 Factor1 SS loadings 2.074 Proportion Var 0.415 Test of the hypothesis that 1 factor is sufficient. The chi square statistic is 29.81 on 5 degrees of freedom. The pvalue is 1.61e05
# pvalue <0.05 suggests that one factor is not sufficient # we reject the null hypothesis that one factor captures full # dimensionality in the dataset # Include two factors pros_fa < factanal(pros2, factors = 2) pros_fa
Call: factanal(x = pros2, factors = 2) Uniquenesses: lcavol lweight lbph lcp lpsa 0.121 0.422 0.656 0.478 0.317 Loadings: Factor1 Factor2 lcavol 0.936 lweight 0.165 0.742 lbph 0.586 lcp 0.722 lpsa 0.768 0.307 Factor1 Factor2 SS loadings 2.015 0.992 Proportion Var 0.403 0.198 Cumulative Var 0.403 0.601 Test of the hypothesis that 2 factors are sufficient. The chi square statistic is 0.02 on 1 degree of freedom. The pvalue is 0.878
# pvalue >0.05 suggests that two factors is sufficient # we cannot reject the null hypothesis that two factors captures # full dimensionality in the dataset #Include three factors pros_fa < factanal(pros2, factors = 3)
Error in factanal(pros2, factors = 3): 3 factors are too many for 5 variables
# Error shows that fitting three factors are not appropriate # for only 5 variables (number of factors too high)
The output of factanal()
shows the loadings for each of the input variables
associated with each factor. The loadings are values between 1 and 1 which
represent the relative contribution each input variable makes to the factors.
Positive values show that these variables are positively related to the
factors, while negative values show a negative relationship between variables
and factors. Loading values are missing for some variables because R does not
print loadings less than 0.1.
How many factors do we need?
There are numerous ways to select the “best” number of factors. One is to use
the minimum number of features that does not leave a significant amount of
variance unaccounted for. In practice, we repeat the factor
analysis for different numbers of factors (by specifying different values
in the factors
argument). If we have an idea of how many factors there
will be before analysis, we can start with that number. The final
section of the analysis output then shows the results of
a hypothesis test in which the null hypothesis is that the number of factors
used in the model is sufficient to capture most of the variation in the
dataset. If the pvalue is less than our significance level (for example 0.05),
we reject the null hypothesis that the number of factors is sufficient and we repeat the analysis with
more factors. When the pvalue is greater than our significance level, we do not reject
the null hypothesis that the number of factors used captures variation
in the data. We may therefore conclude that
this number of factors is sufficient.
Like PCA, the fewer factors that can explain most of the variation in the dataset, the better. It is easier to explore and interpret results using a smaller number of factors which represent underlying features in the data.
Variance accounted for by factors  communality and uniqueness
The communality of a variable is the sum of its squared loadings. It represents the proportion of the variance in a variable that is accounted for by the FA model.
Uniqueness is the opposite of communality and represents the amount of variation in a variable that is not accounted for by the FA model. Uniqueness is calculated by subtracting the communality value from 1. If uniqueness is high for a given variable, that means this variable is not well explained/accounted for by the factors identified.
apply(pros_fa$loadings^2, 1, sum) #communality
lcavol lweight lbph lcp lpsa
0.8793780 0.5782317 0.3438669 0.5223639 0.6832788
1  apply(pros_fa$loadings^2, 1, sum) #uniqueness
lcavol lweight lbph lcp lpsa
0.1206220 0.4217683 0.6561331 0.4776361 0.3167212
Visualising the contribution of each variable to the factors
Similar to a biplot as we produced in the PCA episode, we can “plot the loadings”. This shows how each original variable contributes to each of the factors we chose to visualise.
#First, carry out factor analysis using two factors
pros_fa < factanal(pros2, factors = 2)
#plot loadings for each factor
plot(
pros_fa$loadings[, 1],
pros_fa$loadings[, 2],
xlab = "Factor 1",
ylab = "Factor 2",
ylim = c(1, 1),
xlim = c(1, 1),
main = "Factor analysis of prostate data"
)
abline(h = 0, v = 0)
#add column names to each point
text(
pros_fa$loadings[, 1]  0.08,
pros_fa$loadings[, 2] + 0.08,
colnames(pros2),
col = "blue"
)
Challenge 2 (3 mins)
Use the output from your factor analysis and the plots above to interpret the results of your analysis.
What variables are most important in explaining each factor? Do you think this makes sense biologically? Consider or discuss in groups.
Solution
This plot suggests that the variables
lweight
andlbph
are associated with high values on factor 2 (but lower values on factor 1) and the variables lcavol, lcp and lpsa are associated with high values on factor 1 (but lower values on factor 2). There appear to be two ‘clusters’ of variables which can be represented by the two factors.The grouping of weight and enlargement (lweight and lbph) makes sense biologically, as we would expect prostate enlargement to be associated with greater weight. The groupings of lcavol, lcp, and lpsa also make sense biologically, as larger cancer volume may be expected to be associated with greater cancer spread and therefore higher PSA in the blood.
Advantages and disadvantages of Factor Analysis
There are several advantages and disadvantages of using FA as a dimensionality reduction method.
Advantages:
 FA is a useful way of combining different groups of data into known representative factors, thus reducing dimensionality in a dataset.
 FA can take into account researchers’ expert knowledge when choosing the number of factors to use, and can be used to identify latent or hidden variables which may not be apparent from using other analysis methods.
 It is easy to implement with many software tools available to carry out FA.
 Confirmatory FA can be used to test hypotheses.
Disadvantages:
 Justifying the choice of number of factors to use may be difficult if little is known about the structure of the data before analysis is carried out.
 Sometimes, it can be difficult to interpret what factors mean after analysis has been completed.
 Like PCA, standard methods of carrying out FA assume that input variables are continuous, although extensions to FA allow ordinal and binary variables to be included (after transforming the input matrix).
Further reading
 Gundogdu et al. (2019) Comparison of performances of Principal Component Analysis (PCA) and Factor Analysis (FA) methods on the identification of cancerous and healthy colon tissues. International Journal of Mass Spectrometry 445:116204.
 Kustra et al. (2006) A factor analysis model for functional genomics. BMC Bioinformatics 7: doi:10.1186/14712105721.
 Yong, A.G. & Pearce, S. (2013) A beginner’s guide to factor analysis: focusing on exploratory factor analysis. Tutorials in Quantitative Methods for Psychology 9(2):7994.
 Confirmatory factor analysis can be carried out with the package Lavaan.
 A more sophisticated implementation of EFA is available in the packages EFA.dimensions and psych.
Key Points
Factor analysis is a method used for reducing dimensionality in a dataset by reducing variation contained in multiple variables into a smaller number of uncorrelated factors.
PCA can be used to identify the number of factors to initially use in factor analysis.
The
factanal()
function in R can be used to fit a factor analysis, where the number of factors is specified by the user.Factor analysis can take into account expert knowledge when deciding on the number of factors to use, but a disadvantage is that the output requires careful interpretation.
Kmeans
Overview
Teaching: 60 min
Exercises: 20 minQuestions
How do we detect real clusters in highdimensional data?
How does Kmeans work and when should it be used?
How can we perform Kmeans in
R
?How can we appraise a clustering and test cluster robustness?
Objectives
Understand what clusters look like in in highdimensional data.
Understand and perform Kmeans clustering in
R
.Assess clustering performance using silhouette scores.
Assess cluster robustness using bootstrapping.
Introduction
As we saw in previous episodes, visualising highdimensional data with a large amount of features is difficult and can limit our understanding of the data and associated processes. In some cases, a known grouping causes this heterogeneity (sex, treatment groups, etc). In other cases, heterogeneity may arise from the presence of unknown subgroups in the data. While PCA can be used to reduce the dimension of the dataset into a smaller set of uncorrelated variables and factor analysis can be used to identify underlying factors, clustering is a set of techniques that allow us to discover unknown groupings.
Cluster analysis involves finding groups of observations that are more similar to each other (according to some feature) than they are to observations in other groups and are thus likely to represent the same source of heterogeneity. Once groups (or clusters) of observations have been identified using cluster analysis, further analyses or interpretation can be carried out on the groups, for example, using metadata to further explore groups.
Cluster analysis is commonly used to discover unknown groupings in fields such as bioinformatics, genomics, and image processing, in which large datasets that include many features are often produced.
There are various ways to look for clusters of observations in a dataset using different clustering algorithms. One way of clustering data is to minimise distance between observations within a cluster and maximise distance between proposed clusters. Using this process, we can also iteratively update clusters so that we become more confident about the shape and size of the clusters.
What is Kmeans clustering?
Kmeans clustering groups data points into a userdefined number of distinct, nonoverlapping clusters. To create clusters of ‘similar’ data points, Kmeans clustering creates clusters that minimise the withincluster variation and thus the amount that data points within a cluster differ from each other. The distance between data points within a cluster is used as a measure of withincluster variation.
To carry out Kmeans clustering, we first pick $k$ initial points as centres or “centroids” of our clusters. There are a few ways to choose these initial “centroids” and this is discussed below. Once we have picked intitial points, we then follow these two steps until appropriate clusters have been formed:
 Assign each data point to the cluster with the closest centroid
 Update centroid positions as the average of the points in that cluster.
We can see this process in action in this animation:
While Kmeans has some advantages over other clustering methods (easy to implement and to understand), it does have some disadvantages, particularly difficulties in identifying initial clusters which observations belong to and the need for the user to specify the number of clusters that the data should be partitioned into.
Initialisation
The algorithm used in Kmeans clustering finds a local rather than a global optimum, so that results of clustering are dependent on the initial cluster that each observation is randomly assigned to. This initial configuration can have a significant effect on the final configuration of the clusters, so dealing with this limitation is an important part of Kmeans clustering. Some strategies to deal with this problem are:
 Choose $K$ points at random from the data as the cluster centroids.
 Randomly split the data into $K$ groups, and then average these groups.
 Use the Kmeans++ algorithm to choose initial values.
These each have advantages and disadvantages. In general, it’s good to be aware of this limitation of Kmeans clustering and that this limitation can be addressed by choosing a good initialisation method, initialising clusters manually, or running the algorithm from multiple different starting points.
Believing in clusters
When using clustering, it’s important to realise that data may seem to group together even when these groups are created randomly. It’s especially important to remember this when making plots that add extra visual aids to distinguish clusters. For example, if we cluster data from a single 2D normal distribution and draw ellipses around the points, these clusters suddenly become almost visually convincing. This is a somewhat extreme example, since there is genuinely no heterogeneity in the data, but it does reflect what can happen if you allow yourself to read too much into faint signals.
Let’s explore this further using an example. We create two columns of data
(‘x’ and ‘y’) and partition these data into three groups (‘a’, ‘b’, ‘c’)
according to data values. We then plot these data and their allocated clusters
and put ellipses around the clusters using the stat_ellipse()
function
in ggplot
.
The randomly created data used here appear to form three clusters when we plot the data. Putting ellipses around the clusters can further convince us that the clusters are ‘real’. But how do we tell if clusters identified visually are ‘real’?
Initialisation
The algorithm used in Kmeans clustering finds a local rather than a global optimum, so that results of clustering are dependent on the initial cluster that each observation is randomly assigned to. This initial configuration can have a significant effect on the final configuration of the clusters, so dealing with this limitation is an important part of Kmeans clustering. Some strategies to deal with this problem are:
 Choose $k$ points at random from the data as the cluster centroids.
 Randomly split the data into $k$ groups, and then average these groups.
 Use the Kmeans++ algorithm to choose initial values.
These each have advantages and disadvantages. In general, it’s good to be aware of this limitation of Kmeans clustering and that this limitation can be addressed by choosing a good initialisation method, initialising clusters manually, or running the algorithm from multiple different starting points.
Kmeans clustering applied to the singlecell RNA sequencing data
Let’s carry out Kmeans clustering in R
using some real highdimensional data.
We’re going to work with singlecell RNA sequencing data,
scRNAseq, in these clustering challenges,
which is often very highdimensional. Commonly, experiments profile the
expression level of 10,000+ genes in thousands of cells. Even after filtering
the data to remove low quality observations, the dataset we’re using in this
episode contains measurements for over 9,000 genes in over 3,000 cells.
One way to get a handle on a dataset of this size is to use something we covered earlier in the course  dimensionality reduction. Dimensionality reduction allows us to visualise this incredibly complex data in a small number of dimensions. In this case, we’ll be using principal component analysis (PCA) to compress the data by identifying the major axes of variation in the data, before running our clustering algorithms on this lowerdimensional data to explore the similarity of features.
The scater
package has some easytouse tools to calculate a PCA for
SummarizedExperiment
objects.
Let’s load the scRNAseq data and calculate some principal components.
library("SingleCellExperiment")
library("scater")
scrnaseq < readRDS(here::here("data/scrnaseq.rds"))
scrnaseq < runPCA(scrnaseq, ncomponents = 15)
pcs < reducedDim(scrnaseq)[, 1:2]
The first two principal components capture almost 50% of the variation within the data. For now, we’ll work with just these two principal components, since we can visualise those easily, and they’re a quantitative representation of the underlying data, representing the two largest axes of variation.
We can now run Kmeans clustering on the first and second principal components
of the scRNAseq data using the kmeans()
function.
set.seed(42)
cluster < kmeans(pcs, centers = 4)
scrnaseq$kmeans < as.character(cluster$cluster)
plotReducedDim(scrnaseq, "PCA", colour_by = "kmeans")
We can see that this produces a sensiblelooking partition of the data. However, is it totally clear whether there might be more or fewer clusters here?
Challenge 1
Perform clustering to group the data into $k=5$ clusters, and plot it using
plotReducedDim()
. Save this with a variable name that’s different to what we just used, because we’ll use this again later.Solution
set.seed(42) cluster5 < kmeans(pcs, centers = 5) scrnaseq$kmeans5 < as.character(cluster5$cluster) plotReducedDim(scrnaseq, "PCA", colour_by = "kmeans5")
Kmedoids (PAM)
One problem with Kmeans is that using the mean to define cluster centroids means that clusters can be very sensitive to outlying observations. Kmedoids, also known as “partitioning around medoids (PAM)” is similar to Kmeans, but uses the median rather than the mean as the method for defining cluster centroids. Using the median rather than the mean reduces sensitivity of clusters to outliers in the data. Kmedioids has had popular application in genomics, for example the wellknown PAM50 gene set in breast cancer, which has seen some prognostic applications. The following example shows how cluster centroids differ when created using medians rather than means.
x < rnorm(20) y < rnorm(20) x[10] < x[10] + 10 plot(x, y, pch = 16) points(mean(x), mean(y), pch = 16, col = "firebrick") points(median(x), median(y), pch = 16, col = "dodgerblue")
PAM can be carried out using
pam()
from thecluster
package.
Cluster separation
When performing clustering, it is important for us to be able to measure how well our clusters are separated. One measure to test this is silhouette width, which measures how similar a data point is to points in the same cluster compared to other clusters. The silhouette width is computed for every observation and can range from 1 to 1. A high silhouette width means an observation is closer to other observations within the same cluster. For each cluster, the silhouette widths can then be averaged or an overall average can be taken.
More detail on silhouette widths
In more detail, each observation’s silhouette width is computed as follows:
 Compute the average distance between the focal observation and all other observations in the same cluster.
 For each of the other clusters, compute the average distance between focal observation and all observations in the other cluster. Keep the smallest of these average distances.
 Subtract (1.)(2.) then divivde by whichever is smaller (1.) or (2).
Ideally, we would have only large positive silhouette widths, indicating that each data point is much more similar to points within its cluster than it is to the points in any other cluster. However, this is rarely the case. Often, clusters are very fuzzy, and even if we are relatively sure about the existence of discrete groupings in the data, observations on the boundaries can be difficult to confidently place in either cluster.
Here we use the silhouette()
function from the cluster
package to calculate the
silhouette width of our Kmeans clustering using a distance matrix of distances
between points in the clusters.
library("cluster")
dist_mat < dist(pcs)
sil < silhouette(cluster$cluster, dist = dist_mat)
plot(sil, border = NA)
Let’s plot the silhouette score on the original dimensions used to cluster the data. Here, we’re mapping cluster membership to point shape, and silhouette width to colour.
pc < as.data.frame(pcs)
colnames(pc) < c("x", "y")
pc$sil < sil[, "sil_width"]
pc$clust < factor(cluster$cluster)
mean(sil[, "sil_width"])
[1] 0.7065662
ggplot(pc) +
aes(x, y, shape = clust, colour = sil) +
geom_point() +
scale_colour_gradient2(
low = "dodgerblue", high = "firebrick"
) +
scale_shape_manual(
values = setNames(1:4, 1:4)
)
This plot shows that silhouette values for individual observations tends to be very high in the centre of clusters, but becomes quite low towards the edges. This makes sense, as points that are “between” two clusters may be more similar to points in another cluster than they are to the points in the cluster one they belong to.
Challenge 2
Calculate the silhouette width for the k of 5 clustering we did earlier. Are 5 clusters appropriate? Why/why not?
Can you identify where the differences lie?
Solution
sil5 < silhouette(cluster5$cluster, dist = dist_mat) scrnaseq$kmeans5 < as.character(cluster5$cluster) plotReducedDim(scrnaseq, "PCA", colour_by = "kmeans5")
mean(sil5[, "sil_width"])
[1] 0.5849979
The average silhouette width is lower when k=5.
plot(sil5, border = NA)
This seems to be because some observations in clusters 3 and 5 seem to be more similar to other clusters than the one they have been assigned to. This may indicate that k is too high.
Gap statistic
Another measure of how good our clustering is is the “gap statistic”. This compares the observed squared distance between observations in a cluster and the centre of the cluster to an “expected” squared distances. The expected distances are calculated by randomly distributing cells within the range of the original data. Larger values represent lower squared distances within clusters, and thus better clustering. We can see how this is calculated in the following example.
library("cluster") gaps < clusGap(pcs, kmeans, K.max = 20, iter.max = 20) best_k < maxSE(gaps$Tab[, "gap"], gaps$Tab[, "SE.sim"]) best_k plot(gaps$Tab[,"gap"], xlab = "Number of clusters", ylab = "Gap statistic") abline(v = best_k, col = "red")
Cluster robustness: bootstrapping
When we cluster data, we want to be sure that the clusters we identify are not a result of the exact properties of the input data. That is, we want to ensure that the clusters identified do not change substantially if the observed data change slightly. This makes it more likely that the clusters can be reproduced.
To assess this, we can bootstrap. We first resample the data with replacement to reproduce a ‘new’ data set. We can then calculate new clusters for this data set and compare these to those on the original data set, thus helping us to see how the clusters may change for small changes in the data. This is maybe easier to see with an example. First, we define some data:
data < 1:5
Then, we can take a sample from this data without replacement:
sample(data, 5)
[1] 4 1 3 5 2
This sample is a subset of the original data, and points are only present once. This is the case every time even if we do it many times:
## Each column is a sample
replicate(10, sample(data, 5))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 5 2 5 2 3 1 3 5 5 3
[2,] 4 5 4 5 4 4 1 3 1 2
[3,] 2 1 1 3 2 5 2 2 3 4
[4,] 1 4 2 1 5 3 5 1 2 5
[5,] 3 3 3 4 1 2 4 4 4 1
However, if we sample with replacement, then sometimes individual data points are present more than once.
replicate(10, sample(data, 5, replace = TRUE))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 3 1 2 2 1 3 3 2 4 2
[2,] 1 3 2 4 2 5 2 1 2 5
[3,] 5 5 4 4 2 2 1 1 1 3
[4,] 1 1 4 2 1 4 4 5 5 4
[5,] 3 1 2 1 4 2 5 3 3 2
Bootstrapping
Bootstrapping is a powerful and common statistical technique.
We would like to know about the sampling distribution of a statistic, but we don’t have any knowledge of its behaviour under the null hypothesis.
For example, we might want to understand the uncertainty around an estimate of the mean of our data. To do this, we could resample the data with replacement and calculate the mean of each average.
boots < replicate(1000, mean(sample(data, 5, replace = TRUE)))
hist(boots,
breaks = "FD",
main = "1,000 bootstrap samples",
xlab = "Mean of sample"
)
In this case, the example is simple, but it’s possible to devise more complex statistical tests using this kind of approach.
The bootstrap, along with permutation testing, can be a very flexible and general solution to many statistical problems.
In applying the bootstrap to clustering, we want to see two things:
 Will observations within a cluster consistently cluster together in different bootstrap replicates?
 Will observations frequently swap between clusters?
In the plot below, the diagonal of the plot shows how often the clusters are reproduced in boostrap replicates. High scores on the diagonal mean that the clusters are consistently reproduced in each boostrap replicate. Similarly, the offdiagonal elements represent how often observations swap between clusters in bootstrap replicates. High scores indicate that observations rarely swap between clusters.
library("pheatmap")
library("bluster")
library("viridis")
km_fun < function(x) {
kmeans(x, centers = 4)$cluster
}
ratios < bootstrapStability(pcs, FUN = km_fun, clusters = cluster$cluster)
pheatmap(ratios,
cluster_rows = FALSE, cluster_cols = FALSE,
col = viridis(10),
breaks = seq(0, 1, length.out = 10)
)
Yellow boxes indicate values slightly greater than 1, which may be observed. These are “good” (despite missing in the colour bar).
Challenge 3
Repeat the bootstrapping process with k=5. Do the results appear better or worse? Can you identify where the differences occur on the
plotReducedDim()
?Solution
km_fun5 < function(x) { kmeans(x, centers = 5)$cluster } set.seed(42) ratios5 < bootstrapStability(pcs, FUN = km_fun5, clusters = cluster5$cluster) pheatmap(ratios5, cluster_rows = FALSE, cluster_cols = FALSE, col = viridis(10), breaks = seq(0, 1, length.out = 10) )
When k=5, we can see that the values on the diagonal of the matrix are smaller, indicating that the clusters aren’t exactly reproducible in the bootstrap samples.
Similarly, the offdiagonal elements are considerably lower for some elements. This indicates that observations are “swapping” between these clusters in bootstrap replicates.
Consensus clustering
One useful and generic method of clustering is consensus clustering. This method can use kmeans or other clustering methods.
The idea behind this is to bootstrap the data repeatedly, and cluster it each time, perhaps using different numbers of clusters. If a pair of data points always end up in the same cluster, it’s likely that they really belong to the same underlying cluster.
This is really computationally demanding but has been shown to perform very well in some situations. It also allows you to visualise how cluster membership changes over different values of k.
Speed
It’s worth noting that a lot of the methods we’ve discussed here are very computationally demanding. When clustering data, we may have to compare points to each other many times. This becomes more and more difficult when we have many observations and many features. This is especially problematic when we want to do things like bootstrapping that requires us to cluster the data over and over.
As a result, there are a lot of approximate methods for finding clusters in the data. For example, the mbkmeans package includes an algorithm for clustering extremely large data. The idea behind this algorithm is that if the clusters we find are robust, we don’t need to look at all of the data every time. This is very helpful because it reduces the amount of data that needs to be held in memory at once, but also because it minimises the computational cost.
Similarly, approximate nearest neighbour methods like Annoy can be used to identify what the $k$ closest points are in the data, and this can be used in some clustering methods (for example, graphbased clustering).
Generally, these methods sacrifice a bit of accuracy for a big gain in speed.
Further reading
 Wu, J. (2012) Cluster analysis and Kmeans clustering: An Introduction. In: Advances in Kmeans Clustering. Springer Berlin, Heidelberg..
 Modern statistics for modern biology, Susan Holmes and Wolfgang Huber (Chapter 5).
 Understanding Kmeans clustering in machine learning, Towards Data Science.
Key Points
Kmeans is an intuitive algorithm for clustering data.
Kmeans has various advantages but can be computationally intensive.
Apparent clusters in highdimensional data should always be treated with some scepticism.
Silhouette width and bootstrapping can be used to assess how well our clustering algorithm has worked.
Hierarchical clustering
Overview
Teaching: 70 min
Exercises: 20 minQuestions
What is hierarchical clustering and how does it differ from other clustering methods?
How do we carry out hierarchical clustering in R?
What distance matrix and linkage methods should we use?
How can we validate identified clusters?
Objectives
Understand when to use hierarchical clustering on highdimensional data.
Perform hierarchical clustering on highdimensional data and evaluate dendrograms.
Understand and explore different distance matrix and linkage methods.
Use the Dunn index to validate clustering methods.
Why use hierarchical clustering on highdimensional data?
When analysing highdimensional data in the life sciences, it is often useful to identify groups of similar data points to understand more about the relationships within the dataset. General clustering algorithms group similar data points (or observations) into groups (or clusters). This results in a set of clusters, where each cluster is distinct, and the data points within each cluster have similar characteristics. The clustering algorithm works by iteratively grouping data points so that different clusters may exist at different stages of the algorithm’s progression.
Here, we describe hierarchical clustering. Unlike Kmeans clustering, hierarchical clustering does not require the number of clusters $k$ to be specified by the user before the analysis is carried out.
Hierarchical clustering also provides an attractive dendrogram, a treelike diagram showing the degree of similarity between clusters. The dendrogram is a key feature of hierarchical clustering. This treeshaped graph allows the similarity between data points in a dataset to be visualised and the arrangement of clusters produced by the analysis to be illustrated. Dendrograms are created using a distance (or dissimilarity) that quantify how different are pairs of observations, and a clustering algorithm to fuse groups of similar data points together.
In this episode we will explore hierarchical clustering for identifying clusters in highdimensional data. We will use agglomerative hierarchical clustering (see box) in this episode.
Agglomerative and Divisive hierarchical clustering
There are two main methods of carrying out hierarchical clustering: agglomerative clustering and divisive clustering. The former is a ‘bottomup’ approach to clustering whereby the clustering approach begins with each data point (or observation) being regarded as being in its own separate cluster. Pairs of data points are merged as we move up the tree. Divisive clustering is a ‘topdown’ approach in which all data points start in a single cluster and an algorithm is used to split groups of data points from this main group.
The agglomerative hierarchical clustering algorithm
To start with, we measure distance (or dissimilarity) between pairs of observations. Initially, and at the bottom of the dendrogram, each observation is considered to be in its own individual cluster. We start the clustering procedure by fusing the two observations that are most similar according to a distance matrix. Next, the nextmost similar observations are fused so that the total number of clusters is number of observations minus 2 (see panel below). Groups of observations may then be merged into a larger cluster (see next panel below, green box). This process continues until all the observations are included in a single cluster.
A motivating example
To motivate this lesson, let’s first look at an example where hierarchical
clustering is really useful, and then we can understand how to apply it in more
detail. To do this, we’ll return to the large methylation
dataset we worked
with in the regression lessons. Let’s load and visually inspect the data:
library("minfi")
library("here")
library("ComplexHeatmap")
methyl < readRDS(here("data/methylation.rds"))
# transpose this Bioconductor dataset to show features in columns
methyl_mat < t(assay(methyl))
Looking at a heatmap of these data, we may spot some patterns – many columns appear to have a similar methylation levels across all rows. However, they are all quite jumbled at the moment, so it’s hard to tell how many line up exactly.
Looking at a heatmap of these data, we may spot some patterns – looking at the vertical stripes, many columns appear to have a similar methylation levels across all rows. However, they are all quite jumbled at the moment, so it’s hard to tell how many line up exactly. In addition, it is challenging to tell how many groups containing similar methylation levels we may have or what the similarities and differences are between groups.
We can order these data to make the patterns more clear using hierarchical
clustering. To do this, we can change the arguments we pass to
Heatmap()
from the ComplexHeatmap
package. Heatmap()
groups features based on dissimilarity (here, Euclidean distance) and orders
rows and columns to show clustering of features and observations.
Heatmap(methyl_mat,
name = "Methylation level",
cluster_rows = TRUE, cluster_columns = TRUE,
row_dend_width = unit(0.2, "npc"),
column_dend_height = unit(0.2, "npc"),
show_row_names = FALSE, show_column_names = FALSE,
row_title="Individuals", column_title = "Methylation sites"
)
We can see that clustering the features (CpG sites) results in an overall gradient of high to low methylation levels from left to right. Maybe more interesting is the fact that the rows (corresponding to individuals) are now grouped according to methylation patterns. For example, 12 samples seem to have lower methylation levels for a small subset of CpG sites in the middle, relative to all the other samples. It’s not clear without investigating further what the cause of this is – it could be a batch effect, or a known grouping (e.g., old vs young samples). However, clustering like this can be a useful part of exploratory analysis of data to build hypotheses.
Hierarchical clustering
Now, let’s cover the inner workings of hierarchical clustering in more detail. Hierarchical clustering is a type of clustering that also allows us to estimate the number of clusters. There are two things to consider before carrying out clustering:
 how to define dissimilarity between observations using a distance matrix, and
 how to define dissimilarity between clusters and when to fuse separate clusters.
Defining the dissimilarity between observations: creating the distance matrix
Agglomerative hierarchical clustering is performed in two steps: calculating the distance matrix (containing distances between pairs of observations) and iteratively grouping observations into clusters using this matrix.
There are different ways to specify a distance matrix for clustering:
 Specify distance as a predefined option using the
method
argument indist()
. Methods includeeuclidean
(default),maximum
andmanhattan
.  Create a selfdefined function which calculates distance from a matrix or from two vectors. The function should only contain one argument.
Of predefined methods of calculating the distance matrix, Euclidean is one of the most commonly used. This method calculates the shortest straightline distances between pairs of observations.
Another option is to use a correlation matrix as the input matrix to the clustering algorithm. The type of distance matrix used in hierarchical clustering can have a big effect on the resulting tree. The decision of which distance matrix to use before carrying out hierarchical clustering depends on the type of data and question to be addressed.
Defining the dissimilarity between clusters: Linkage methods
The second step in performing hierarchical clustering after defining the distance matrix (or another function defining similarity between data points) is determining how to fuse different clusters.
Linkage is used to define dissimilarity between groups of observations (or clusters) and is used to create the hierarchical structure in the dendrogram. Different linkage methods of creating a dendrogram are discussed below.
hclust()
supports various linkage methods (e.g complete
,
single
, ward D
, ward D2
, average
, median
) and these are also supported
within the Heatmap()
function. The method used to perform hierarchical
clustering in Heatmap()
can be specified by the arguments
clustering_method_rows
and clustering_method_columns
. Each linkage method
uses a slightly different algorithm to calculate how clusters are fused together
and therefore different clustering decisions are made depending on the linkage
method used.
Complete linkage (the default in hclust()
) works by computing all pairwise
dissimilarities between data points in different clusters. For each pair of two clusters,
it sets their dissimilarity to the maximum dissimilarity value observed
between any of these clusters’ constituent points. The two clusters
with smallest dissimilarity value are then fused.
Computing a dendrogram
Dendrograms are useful tools that plot the grouping of points and clusters into bigger clusters.
We can create and plot dendrograms in R using hclust()
which takes
a distance matrix as input and creates a tree using hierarchical
clustering. Here we create some example data to carry out hierarchical
clustering.
Let’s generate 20 data points in 2D space. Each point is generated to belong to one of three classes/groups. Suppose we did not know which class data points belonged to and we want to identify these via cluster analysis. Let’s first generate and plot our data:
#First, create some example data with two variables x1 and x2
set.seed(450)
example_data < data.frame(
x1 = rnorm(20, 8, 4.5),
x2 = rnorm(20, 6, 3.4)
)
#plot the data and name data points by row numbers
plot(example_data$x1, example_data$x2, type = "n")
text(
example_data$x1,
example_data$x2,
labels = rownames(example_data),
cex = 0.7
)
## calculate distance matrix using euclidean distance
dist_m < dist(example_data, method = "euclidean")
Hierarchical clustering carried out on the data can be used to produce a dendrogram showing how the data is partitioned into clusters. But how do we interpret this dendrogram? Let’s explore this using our example data in a Challenge.
Challenge 1
Use
hclust()
to implement hierarchical clustering using the distance matrixdist_m
and thecomplete
linkage method and plot the results as a dendrogram usingplot()
. Why is hierarchical clustering and the resulting dendrogram useful for performing clustering this case?Solution:
clust < hclust(dist_m, method = "complete") plot(clust)
Hierarchical clustering is particularly useful (compared to Kmeans) when we do not know the number of clusters before we perform clustering. It is useful in this case since we have assumed we do not already know what a suitable number of clusters may be.
A dendrogram, such as the one generated in Challenge 1, shows similarities/differences in distances between data points. Each vertical line at the bottom of the dendrogram (‘leaf’) represents one of the 20 data points. These leaves fuse into fewer vertical lines (‘branches’) as the height increases. Observations that are similar fuse into the same branches. The height at which any two data points fuse indicates how different these two points are. Points that fuse at the top of the tree are very different from each other compared with two points that fuse at the bottom of the tree, which are quite similar. You can see this by comparing the position of similar/dissimilar points according to the scatterplot with their position on the tree.
Identifying the number of clusters
To identify the number of clusters, we can make a horizontal cut through the dendrogram at a userdefined height. The sets of observations beneath this cut and single points representing clusters above can be thought of as distinct clusters. Equivalently, we can count the vertical lines we encounter crossing the horizontal cut and the number of single points above the cut. For example, a cut at height 10 produces 3 clusters for the dendrogram in Challenge 1, while a cut at height 4 produces 8 clusters.
Dendrogram visualisation
We can first visualise cluster membership by highlight branches in dendrograms.
In this example, we calculate a distance matrix between
samples in the methyl_mat
dataset. We then draw boxes round clusters obtained with cutree()
.
## create a distance matrix using euclidean distance
distmat < dist(methyl_mat)
## hierarchical clustering using complete method
clust < hclust(distmat)
## plot resulting dendrogram
plot(clust)
## draw border around three clusters
rect.hclust(clust, k = 3, border = 2:6) #border argument specifies the colours
## draw border around two clusters
rect.hclust(clust, k = 2, border = 2:6)
We can also colour clusters downstream of a specified cut using color_branches()
from the dendextend
package.
## cut tree at height = 4
cut < cutree(clust, h = 50)
library("dendextend")
avg_dend_obj < as.dendrogram(clust)
## colour branches of dendrogram depending on clusters
plot(color_branches(avg_dend_obj, h = 50))
Numerical visualisation
In addition to visualising clusters directly on the dendrogram, we can cut
the dendrogram to determine number of clusters at different heights
using cutree()
. This function cuts a dendrogram into several
groups (or clusters) where the number of desired groups is controlled by the
user, by defining either k
(number of groups) or h
(height at which tree is
cut). The function outputs the cluster labels of each data point in order.
## k is a user defined parameter determining
## the desired number of clusters at which to cut the treee
as.numeric(cutree(clust, k = 9))
[1] 1 2 1 3 2 2 4 3 1 4 4 5 4 4 6 7 1 5 4 5 4 3 5 7 4 8 4 1 8 9 5 2 8 4 1 4 2
## h is a user defined parameter determining
## the numeric height at which to cut the tree
as.numeric(cutree(clust, h = 36))
[1] 1 2 1 3 2 2 4 3 1 4 4 5 4 4 6 7 1 5 4 5 4 3 5 7 4 8 4 1 8 9 5 2 8 4 1 4 2
## both give same results
four_cut < cutree(clust, h = 50)
## we can produce the cluster each observation belongs to
## using the mutate and count functions
library(dplyr)
example_cl < mutate(data.frame(methyl_mat), cluster = four_cut)
count(example_cl, cluster)
cluster n
1 1 12
2 2 6
3 3 6
4 4 13
#plot cluster each point belongs to on original scatterplot
library(ggplot2)
ggplot(example_cl, aes(x = cg01644850, y = cg01656216, color = factor(cluster))) + geom_point()
Note that this cut produces 4 clusters (zero before the cut and another four downstream of the cut).
Challenge 2:
Identify the value of
k
incutree()
that gives the same output ash = 36
Solution:
plot(clust) ## create horizontal line at height = 45 abline(h = 45, lty = 2)
cutree(clust, h = 45)
201868500150_R01C01 201868500150_R03C01 201868500150_R05C01 201868500150_R07C01 1 2 1 3 201868500150_R08C01 201868590193_R02C01 201868590193_R03C01 201868590193_R06C01 2 2 4 3 201868590206_R02C01 201868590206_R04C01 201868590206_R05C01 201868590206_R06C01 1 4 4 5 201868590206_R08C01 201868590243_R03C01 201868590243_R04C01 201868590243_R06C01 4 4 5 4 201868590243_R07C01 201868590243_R08C01 201868590267_R06C01 201868590267_R07C01 1 5 4 5 201868590267_R08C01 201869680009_R06C01 201869680009_R07C01 201869680009_R08C01 4 3 5 4 201869680030_R01C01 201869680030_R02C01 201869680030_R06C01 201870610056_R02C01 4 3 4 1 201870610056_R04C01 201870610056_R07C01 201870610056_R08C01 201870610111_R01C01 3 2 5 2 201870610111_R02C01 201870610111_R04C01 201870610111_R05C01 201870610111_R06C01 3 4 1 4 201870610111_R07C01 2
cutree(clust, k = 5)
201868500150_R01C01 201868500150_R03C01 201868500150_R05C01 201868500150_R07C01 1 2 1 3 201868500150_R08C01 201868590193_R02C01 201868590193_R03C01 201868590193_R06C01 2 2 4 3 201868590206_R02C01 201868590206_R04C01 201868590206_R05C01 201868590206_R06C01 1 4 4 5 201868590206_R08C01 201868590243_R03C01 201868590243_R04C01 201868590243_R06C01 4 4 5 4 201868590243_R07C01 201868590243_R08C01 201868590267_R06C01 201868590267_R07C01 1 5 4 5 201868590267_R08C01 201869680009_R06C01 201869680009_R07C01 201869680009_R08C01 4 3 5 4 201869680030_R01C01 201869680030_R02C01 201869680030_R06C01 201870610056_R02C01 4 3 4 1 201870610056_R04C01 201870610056_R07C01 201870610056_R08C01 201870610111_R01C01 3 2 5 2 201870610111_R02C01 201870610111_R04C01 201870610111_R05C01 201870610111_R06C01 3 4 1 4 201870610111_R07C01 2
five_cut < cutree(clust, h = 45) library(dplyr) example_cl < mutate(data.frame(methyl_mat), cluster = five_cut) count(example_cl, cluster)
cluster n 1 1 6 2 2 6 3 3 6 4 4 13 5 5 6
library(ggplot2) ggplot(example_cl, aes(x=cg01644850, y = cg01656216, color = factor(cluster))) + geom_point()
Five clusters (
k = 5
) gives similar results toh = 45
. You can plot a horizontal line on the dendrogram ath = 45
to help identify corresponding value ofk
.
The effect of different linkage methods
Now let us look into changing the default behaviour of hclust()
. First, we’ll load a synthetic dataset, comprised of two variables representing two crescentshaped point clouds:
cres < readRDS(here::here("data/cres.rds"))
plot(cres)
We might expect that the crescents are resolved into separate clusters. But if we run hierarchical clustering with the default arguments, we get this:
cresClass < cutree(hclust(dist(cres)), k=2) # save partition for colouring
plot(cres, col=cresClass) # colour scatterplot by partition
Challenge 3
Carry out hierarchical clustering on the
cres
data that we generated above. Try out different linkage methods and usecutree()
to split each resulting dendrogram into two clusters. Plot the results colouring the dots according to their inferred cluster identity.Which method(s) give you the expected clustering outcome?
Hint: Check
?hclust
to see the possible values of the argumentmethod
(the linkage method used).Solution:
#?hclust # "complete", "single", "ward.D", "ward.D2", "average", "mcquitty", "median" or "centroid" cresClassSingle < cutree(hclust(dist(cres),method = "single"), k=2) plot(cres, col=cresClassSingle, main="single")
cresClassWard.D < cutree(hclust(dist(cres), method="ward.D"), k=2) plot(cres, col=cresClassWard.D, main="ward.D")
cresClassWard.D2 < cutree(hclust(dist(cres), method="ward.D2"), k=2) plot(cres, col=cresClassWard.D2, main="ward.D2")
cresClassAverage < cutree(hclust(dist(cres), method="average"), k=2) plot(cres, col=cresClassAverage, main="average")
cresClassMcquitty < cutree(hclust(dist(cres), method="mcquitty"), k=2) plot(cres, col=cresClassMcquitty, main="mcquitty")
cresClassMedian< cutree(hclust(dist(cres), method="median"), k=2) plot(cres, col=cresClassMedian, main="median")
cresClassCentroid< cutree(hclust(dist(cres), method="centroid"), k=2) plot(cres, col=cresClassCentroid, main="centroid")
The linkage methods
single
,ward.D
, andaverage
resolve each crescent as a separate cluster.
The help page of hclust()
gives some intuition on linkage methods. It describes complete
(the default) and single
as opposite ends of a spectrum with all other methods in between.
When using complete linkage, the distance between two clusters is assumed to be the distance
between both clusters’ most distant points. This opposite it true for single linkage, where
the minimum distance between any two points, one from each of two clusters is used. Single
linkage is described as friendsoffriends appporach  and really, it groups all closetogether
points into the same cluster (thus resolving one cluster per crescent). Complete linkage on the
other hand recognises that some points a the tip of a crescent are much closer to points in the
other crescent and so it splits both crescents.
Using different distance methods
So far, we’ve been using Euclidean distance to define the dissimilarity or distance between observations. However, this isn’t always the best metric for how dissimilar the observations are. Let’s make an example to demonstrate. Here, we’re creating two samples each with ten observations of random noise:
set.seed(20)
cor_example < data.frame(
sample_a = rnorm(10),
sample_b = rnorm(10)
)
rownames(cor_example) < paste(
"Feature", 1:nrow(cor_example)
)
Now, let’s create a new sample that has exactly the same pattern across all
our features as sample_a
, just offset by 5:
cor_example$sample_c < cor_example$sample_a + 5
You can see that this is a lot like the assay()
of our methylation object
from earlier, where columns are observations or samples, and rows are features:
head(cor_example)
sample_a sample_b sample_c
Feature 1 1.1626853 0.02013537 6.162685
Feature 2 0.5859245 0.15038222 4.414076
Feature 3 1.7854650 0.62812676 6.785465
Feature 4 1.3325937 1.32322085 3.667406
Feature 5 0.4465668 1.52135057 4.553433
Feature 6 0.5696061 0.43742787 5.569606
If we plot a heatmap of this, we can see that sample_a
and sample_b
are
grouped together because they have a small distance from each other, despite
being quite different in their pattern across the different features.
In contrast, sample_a
and sample_c
are very distant, despite having
exactly the same pattern across the different features.
Heatmap(as.matrix(cor_example))
We can see that more clearly if we do a line plot:
## create a blank plot (type = "n" means don't draw anything)
## with an x range to hold the number of features we have.
## the range of y needs to be enough to show all the values for every feature
plot(
1:nrow(cor_example),
rep(range(cor_example), 5),
type = "n",
xlab = "Feature number",
ylab = "Value"
)
## draw a red line for sample_a
lines(cor_example$sample_a, col = "firebrick")
## draw a blue line for sample_b
lines(cor_example$sample_b, col = "dodgerblue")
## draw a green line for sample_c
lines(cor_example$sample_c, col = "forestgreen")
We can see that sample_a
and sample_c
have exactly the same pattern across
all of the different features. However, due to the overall difference between
the values, they have a high distance to each other.
We can see that if we cluster and plot the data ourselves using Euclidean
distance:
clust_dist < hclust(dist(t(cor_example)))
plot(clust_dist)
In some cases, we might want to ensure that samples that have similar patterns,
whether that be of gene expression, or DNA methylation, have small distances
to each other. Correlation is a measure of this kind of similarity in pattern.
However, high correlations indicate similarity, while for a distance measure
we know that high distances indicate dissimilarity. Therefore, if we wanted
to cluster observations based on the correlation, or the similarity of patterns,
we can use 1  cor(x)
as the distance metric.
The input to hclust()
must be a dist
object, so we also need to call
as.dist()
on it before passing it in.
cor_as_dist < as.dist(1  cor(cor_example))
clust_cor < hclust(cor_as_dist)
plot(clust_cor)
Now, sample_a
and sample_c
that have identical patterns across the features
are grouped together, while sample_b
is seen as distant because it has a
different pattern, even though its values are closer to sample_a
.
Using your own distance function is often useful, especially if you have missing
or unusual data. It’s often possible to use correlation and other custom
dissimilarity measures in functions that perform hierarchical clustering, such as
pheatmap()
and stats::heatmap()
:
## pheatmap allows you to select correlation directly
pheatmap(as.matrix(cor_example), clustering_distance_cols = "correlation")
## Using the builtin stats::heatmap
heatmap(
as.matrix(cor_example),
distfun = function(x) as.dist(1  cor(t(x)))
)
Validating clusters
Now that we know how to carry out hierarchical clustering, how do we know how many clusters are optimal for the dataset?
Hierarchical clustering carried out on any dataset will produce clusters, even when there are no ‘real’ clusters in the data! We need to be able to determine whether identified clusters represent true groups in the data, or whether clusters have been identified just due to chance. In the last episode, we have introduced silhouette scores as a measure of cluster compactness and bootstrapping to assess cluster robustness. Such tests can be used to compare different clustering algorithms, for example, those fitted using different linkage methods.
Here, we introduce the Dunn index, which is a measure of cluster compactness. The Dunn index is the ratio of the smallest distance between any two clusters and to the largest intracluster distance found within any cluster. This can be seen as a family of indices which differ depending on the method used to compute distances. The Dunn index is a metric that penalises clusters that have larger intracluster variance and smaller intercluster variance. The higher the Dunn index, the better defined the clusters.
Let’s calculate the Dunn index for clustering carried out on the
methyl_mat
dataset using the clValid
package.
## calculate dunn index
## (ratio of the smallest distance between obs not in the same cluster
## to the largest intracluster distance)
library("clValid")
## calculate euclidean distance between points
distmat < dist(methyl_mat)
clust < hclust(distmat, method = "complete")
plot(clust)
cut < cutree(clust, h = 50)
## retrieve Dunn's index for given matrix and clusters
dunn(distance = distmat, cut)
[1] 0.8823501
The value of the Dunn index has no meaning in itself, but is used to compare between sets of clusters with larger values being preferred.
Challenge 4
Examine how changing the
h
ork
arguments incutree()
affects the value of the Dunn index.Solution:
library("clValid") distmat < dist(methyl_mat) clust < hclust(distmat, method = "complete") #Varying h ## Obtaining the clusters cut_h_20 < cutree(clust, h = 20) cut_h_30 < cutree(clust, h = 30) ## How many clusters? length(table(cut_h_20))
[1] 36
length(table(cut_h_30))
[1] 14
dunn(distance = distmat, cut_h_20)
[1] 1.61789
dunn(distance = distmat, cut_h_30)
[1] 0.8181846
#Varying k ## Obtaining the clusters cut_k_10 < cutree(clust, k = 10) cut_k_5 < cutree(clust, k = 5) ## How many clusters? length(table(cut_k_5))
[1] 5
length(table(cut_k_10))
[1] 10
dunn(distance = distmat, cut_k_5)
[1] 0.8441528
dunn(distance = distmat, cut_k_10)
[1] 0.7967132
The figures below show in a more systematic way how changing the values of k
and
h
using cutree()
affect the Dunn index.
h_seq < 70:10
h_dunn < sapply(h_seq, function(x) dunn(distance = distmat, cutree(clust, h = x)))
k_seq < seq(2, 10)
k_dunn < sapply(k_seq, function(x) dunn(distance = distmat, cutree(clust, k = x)))
plot(h_seq, h_dunn, xlab = "Height (h)", ylab = "Dunn index")
grid()
You can see that at low values of h
, the Dunn index can be high. But this
is not very useful  cutting the given tree at a low h
value like 15 leads to allmost all observations
ending up each in its own cluster. More relevant is the second maximum in the plot, around h=55
.
Looking at the dendrogram, this corresponds to k=4
.
plot(k_seq, k_dunn, xlab = "Number of clusters (k)", ylab = "Dunn index")
grid()
For the given range of k
values explored, we obtain the highest Dunn index with k=4
.
This is in agreement with the previous plot.
There have been criticisms of the use of the Dunn index in validating clustering results, due to its high sensitivity to noise in the dataset. An alternative is to use silhouette scores (see the kmeans clustering episode).
As we said before (see previous episode), clustering is a nontrivial task. It is important to think about the nature of your data and your expectations rather than blindly using a some algorithm for clustering or cluster validation.
Further reading
 Dunn, J. C. (1974) Wellseparated clusters and optimal fuzzy partitions. Journal of Cybernetics 4(1):95–104.
 Halkidi, M., Batistakis, Y. & Vazirgiannis, M. (2001) On clustering validation techniques. Journal of Intelligent Information Systems 17(2/3):107145.
 James, G., Witten, D., Hastie, T. & Tibshirani, R. (2013) An Introduction to Statistical Learning with Applications in R. Section 10.3.2 (Hierarchical Clustering).
 Understanding the concept of Hierarchical clustering Technique. towards data science blog.
Key Points
Hierarchical clustering uses an algorithm to group similar data points into clusters. A dendrogram is used to plot relationships between clusters (using the
hclust()
function in R).Hierarchical clustering differs from kmeans clustering as it does not require the user to specify expected number of clusters
The distance (dissimilarity) matrix can be calculated in various ways, and different clustering algorithms (linkage methods) can affect the resulting dendrogram.
The Dunn index can be used to validate clusters using the original dataset.