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*k*-means clustering is a partitioning method.
The function `kmeans`

partitions
data into *k* mutually exclusive clusters, and returns
the index of the cluster to which it has assigned each observation.
Unlike hierarchical clustering, *k*-means clustering
operates on actual observations (rather than the larger set of dissimilarity
measures), and creates a single level of clusters. The distinctions
mean that *k*-means clustering is often more suitable
than hierarchical clustering for large amounts of data.

`kmeans`

treats each observation in your data as an object having a
location in space. It finds a partition in which objects within each cluster are as
close to each other as possible, and as far from objects in other clusters as
possible. You can choose from five different distance metrics, depending on the kind
of data you are clustering.

Each cluster in the partition is defined by its member objects and by its centroid, or center.
The centroid for each cluster is the point to which the sum of distances from all
objects in that cluster is minimized. `kmeans`

computes cluster
centroids differently for each distance metric, to minimize the sum with respect to
the measure that you specify.

You can control the details of the minimization using several
optional input parameters to `kmeans`

, including
ones for the initial values of the cluster centroids, and for the
maximum number of iterations. By default, `kmeans`

uses
the *k*-means++ algorithm for cluster center initialization
and the squared Euclidean metric to determine distances.

The following example explores possible clustering in four-dimensional data by analyzing the results of partitioning the points into three, four, and five clusters.

Because each part of this example generates random numbers sequentially, i.e., without setting a new state, you must perform all steps in sequence to duplicate the results shown. If you perform the steps out of sequence, the answers will be essentially the same, but the intermediate results, number of iterations, or ordering of the silhouette plots may differ.

First, load some data.

rng default % For reproducibility load kmeansdata size(X)

`ans = `*1×2*
560 4

Even though these data are four-dimensional, and cannot be easily visualized,
`kmeans`

enables you to investigate whether a group structure
exists in them. Call `kmeans`

with `k`

, the
desired number of clusters, equal to `3`

. For this example, specify
the city block distance metric, and use the default *k*-means++
algorithm for cluster center initialization.

idx3 = kmeans(X,3,'Distance','cityblock');

To get an idea of how well-separated the resulting clusters
are, you can make a silhouette plot using the cluster indices output
from `kmeans`

. The silhouette plot displays a measure
of how close each point in one cluster is to points in the neighboring
clusters. This measure ranges from +1, indicating points that are
very distant from neighboring clusters, through 0, indicating points
that are not distinctly in one cluster or another, to -1, indicating
points that are probably assigned to the wrong cluster. `silhouette`

returns these values in its
first output.

figure [silh3,h] = silhouette(X,idx3,'cityblock'); h = gca; h.Children.EdgeColor = [.8 .8 1]; xlabel 'Silhouette Value' ylabel 'Cluster'

From the silhouette plot, you can see that most points in the second cluster have a large silhouette value, greater than 0.6, indicating that the cluster is somewhat separated from neighboring clusters. However, the third cluster contains many points with low silhouette values, and the first and third contain a few points with negative values, indicating that those two clusters are not well separated.

Increase the number of clusters to see if `kmeans`

can
find a better grouping of the data. This time, use the `'Display'`

name-value
pair argument to print information about each iteration.

idx4 = kmeans(X,4, 'Distance','cityblock','Display','iter');

iter phase num sum 1 1 560 1792.72 2 1 6 1771.1 Best total sum of distances = 1771.1

A silhouette plot for this solution indicates that these four clusters are better separated than the three in the previous solution.

figure [silh4,h] = silhouette(X,idx4,'cityblock'); h = gca; h.Children.EdgeColor = [.8 .8 1]; xlabel 'Silhouette Value' ylabel 'Cluster'

A more quantitative way to compare the two solutions is to look at the average silhouette values for the two cases.

cluster3 = mean(silh3)

cluster3 = 0.5352

cluster4 = mean(silh4)

cluster4 = 0.6400

Finally, try clustering the data using five clusters.

idx5 = kmeans(X,5,'Distance','cityblock','Replicates',5); figure [silh5,h] = silhouette(X,idx5,'city'); h = gca; h.Children.EdgeColor = [.8 .8 1]; xlabel 'Silhouette Value' ylabel 'Cluster'

mean(silh5)

ans = 0.5266

This silhouette plot indicates that this is probably not the
right number of clusters, since two of the clusters contain points
with mostly low silhouette values. Without some knowledge of how many
clusters are really in the data, it is a good idea to experiment with
a range of values for `k`

.

Like many other types of numerical minimizations, the solution
that `kmeans`

reaches often depends on the starting
points. It is possible for `kmeans`

to reach a
local minimum, where reassigning any one point to a new cluster would
increase the total sum of point-to-centroid distances, but where a
better solution does exist. However, you can use the `'Replicates'`

name-value
pair argument to overcome that problem.

For four clusters, specify five replicates, and use the `'Display'`

name-value
pair argument to print out the final sum of distances for each of
the solutions.

[idx4,cent4,sumdist] = kmeans(X,4,'Distance','cityblock',... 'Display','final','Replicates',5);

Replicate 1, 4 iterations, total sum of distances = 1771.1. Replicate 2, 2 iterations, total sum of distances = 1771.1. Replicate 3, 5 iterations, total sum of distances = 2303.45. Replicate 4, 4 iterations, total sum of distances = 2303.45. Replicate 5, 6 iterations, total sum of distances = 1771.1. Best total sum of distances = 1771.1

In two of the five replications, `kmeans`

found
a local (nonglobal) minimum. Since each of these five replicates begin
from a different randomly selected set of initial centroids, sometimes `kmeans`

finds
more than one local minimum. However, the final solution that `kmeans`

returns
is the one with the lowest total sum of distances, over all replicates.

sum(sumdist)

ans = 1.7711e+03