Documentation

cluster

Construct agglomerative clusters from linkages

Syntax

T = cluster(Z,'cutoff',c)
T = cluster(Z,'cutoff',c,'depth',d)
T = cluster(Z,'cutoff',c,'criterion',criterion)
T = cluster(Z,'maxclust',n)

Description

T = cluster(Z,'cutoff',c) constructs clusters from the agglomerative hierarchical cluster tree, Z, as generated by the linkage function. Z is a matrix of size (m – 1)-by-3, where m is the number of observations in the original data. c is a threshold for cutting Z into clusters. Clusters are formed when a node and all of its subnodes have inconsistent value less than c. All leaves at or below the node are grouped into a cluster. t is a vector of size m containing the cluster assignments of each observation.

If c is a vector, T is a matrix of cluster assignments with one column per cutoff value.

T = cluster(Z,'cutoff',c,'depth',d) evaluates inconsistent values by looking to a depth d below each node. The default depth is 2.

T = cluster(Z,'cutoff',c,'criterion',criterion) uses the specified criterion for forming clusters, where criterion is one of the strings 'inconsistent' (default) or 'distance'. The 'distance' criterion uses the distance between the two subnodes merged at a node to measure node height. All leaves at or below a node with height less than c are grouped into a cluster.

T = cluster(Z,'maxclust',n) constructs a maximum of n clusters using the 'distance' criterion. cluster finds the smallest height at which a horizontal cut through the tree leaves n or fewer clusters.

If n is a vector, T is a matrix of cluster assignments with one column per maximum value.

Examples

Compare clusters from Fisher iris data with species:

load fisheriris
d = pdist(meas);
Z = linkage(d);
c = cluster(Z,'maxclust',3:5);

crosstab(c(:,1),species)
ans =
     0     0     2
     0    50    48
    50     0     0

crosstab(c(:,2),species)
ans =
     0     0     1
     0    50    47
     0     0     2
    50     0     0

crosstab(c(:,3),species)
ans =
     0     4     0
     0    46    47
     0     0     1
     0     0     2
    50     0     0
Was this topic helpful?