k-means clustering is a partitioning method.
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 measures, 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
kmeans computes cluster centroids
differently for each distance measure, to minimize the sum with respect
to the measure that you specify.
kmeans uses an iterative algorithm that
minimizes the sum of distances from each object to its cluster centroid,
over all clusters. This algorithm moves objects between clusters until
the sum cannot be decreased further. The result is a set of clusters
that are as compact and well-separated as possible. You can control
the details of the minimization using several optional input parameters
kmeans, including ones for the initial values
of the cluster centroids, and for the maximum number of iterations.
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.
Note 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 = 560 4
Even though these data are four-dimensional, and cannot be easily
kmeans enables you to investigate
whether a group structure exists in them. Call
the desired number of clusters, equal to
this example, specify the city block distance measure, and use the
default k-means++ algorithm for cluster center
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
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
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
find a better grouping of the data. This time, use the
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 3 2 0 1771.1 Best total sum of distances = 1771.1
Notice that the total sum of distances decreases at each iteration
kmeans reassigns points between clusters and
recomputes cluster centroids. In this case, the second phase of the
algorithm did not make any reassignments, indicating that the first
phase reached a minimum after five iterations. In some problems, the
first phase might not reach a minimum, but the second phase always
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) cluster4 = mean(silh4)
cluster3 = 0.5352 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
Like many other types of numerical minimizations, the solution
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
pair argument to overcome that problem.
For four clusters, specify five replicates, and use the
pair argument to print out the final sum of distances for each of
[idx4,cent4,sumdist] = kmeans(X,4,'Distance','cityblock',... 'Display','final','Replicates',5);
Replicate 1, 5 iterations, total sum of distances = 1771.1. Replicate 2, 3 iterations, total sum of distances = 1771.1. Replicate 3, 7 iterations, total sum of distances = 2303.36. Replicate 4, 6 iterations, total sum of distances = 2303.36. Replicate 5, 7 iterations, total sum of distances = 1771.1. Best total sum of distances = 1771.1
In two of the five replications,
a local (nonglobal) minimum. Since each of these five replicates begin
from a different randomly selected set of initial centroids, sometimes
more than one local minimum. However, the final solution that
is the one with the lowest total sum of distances, over all replicates.
ans = 1.7711e+03