dendextend (version 0.14.2)

cutree_1k.dendrogram: cutree for dendrogram (by 1 k value only!)

Description

Cuts a dendrogram tree into several groups by specifying the desired number of clusters k (only a single k value!). In case there exists no such k for which exists a relevant split of the dendrogram, a warning is issued to the user, and NA is returned.

Usage

cutree_1k.dendrogram(tree, k, dend_heights_per_k = NULL,
  use_labels_not_values = TRUE, order_clusters_as_data = TRUE,
  warn = TRUE, ...)

Arguments

tree

a dendrogram object

k

numeric scalar (not a vector!) with the number of clusters the tree should be cut into.

dend_heights_per_k

a named vector that resulted from running. heights_per_k.dendrogram. When running the function many times, supplying this object will help improve the running time.

use_labels_not_values

logical, defaults to TRUE. If the actual labels of the clusters do not matter - and we want to gain speed (say, 10 times faster) - then use FALSE (gives the "leaves order" instead of their labels.). This is passed to cutree_1h.dendrogram

order_clusters_as_data

logical, defaults to TRUE. There are two ways by which to order the clusters: 1) By the order of the original data. 2) by the order of the labels in the dendrogram. In order to be consistent with cutree,

warn

logical. Should the function send a warning in case the desried k is not available? (deafult is TRUE)

...

(not currently in use)

Value

• cutree_1k.dendrogram returns an integer vector with group memberships. In case there exists no such k for which exists a relevant split of the dendrogram, a warning is issued to the user, and NA is returned.

See Also

hclust, cutree, cutree_1h.dendrogram

Examples

hc <- hclust(dist(USArrests[c(1,6,13,20, 23),]), "ave")
dend <- as.dendrogram(hc)
cutree(hc, k=3) # on hclust
cutree_1k.dendrogram(dend, k=3) # on a dendrogram

labels(dend)

# the default (ordered by original data's order)
cutree_1k.dendrogram(dend, k=3, order_clusters_as_data = TRUE)

# A different order of labels - order by their order in the tree
cutree_1k.dendrogram(dend, k=3, order_clusters_as_data = FALSE)


# make it faster
require(microbenchmark)
dend_ks <- heights_per_k.dendrogram
microbenchmark(
         cutree_1k.dendrogram = cutree_1k.dendrogram(dend, k=4),
         cutree_1k.dendrogram_no_labels = cutree_1k.dendrogram(dend,
                                       k=4,use_labels_not_values = FALSE),
         cutree_1k.dendrogram_no_labels_per_k = cutree_1k.dendrogram(dend,
                                       k=4, use_labels_not_values = FALSE,
                                             dend_heights_per_k = dend_ks)
         )
         # the last one is the fastest...