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 high-dimensional data.
Perform hierarchical clustering on high-dimensional 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 high-dimensional data?
When analysing high-dimensional 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 K-means 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 tree-like diagram showing the degree of similarity between clusters. The dendrogram is a key feature of hierarchical clustering. This tree-shaped 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 high-dimensional 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 ‘bottom-up’ 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 ‘top-down’ 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 next-most 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 pre-defined option using the
method
argument indist()
. Methods includeeuclidean
(default),maximum
andmanhattan
. - Create a self-defined function which calculates distance from a matrix or from two vectors. The function should only contain one argument.
Of pre-defined methods of calculating the distance matrix, Euclidean is one of the most commonly used. This method calculates the shortest straight-line 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 K-means) 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 user-defined 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 crescent-shaped 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 friends-of-friends appporach - and really, it groups all close-together
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 built-in 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 intra-cluster 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 intra-cluster variance and smaller inter-cluster 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 intra-cluster 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 k-means clustering episode).
As we said before (see previous episode), clustering is a non-trivial 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) Well-separated 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):107-145.
- 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 k-means 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.