dendextend (version 0.16.1)

cutree: Cut a Tree (Dendrogram/hclust/phylo) into Groups of Data

Description

Cuts a dendrogram tree into several groups by specifying the desired number of clusters k(s), or cut height(s). For hclust.dendrogram - 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(tree, k = NULL, h = NULL,...)

## S3 method for class 'hclust':
cutree(tree, k = NULL, h = NULL,
                           use_labels_not_values = TRUE,
                          order_clusters_as_data =TRUE,
                          sort_cluster_numbers = TRUE,
                          warn = TRUE,
                          ...)

## S3 method for class 'phylo':
cutree(tree, k = NULL, h = NULL,...)

## S3 method for class 'dendrogram':
cutree(tree, k = NULL, h = NULL,
                              dend_heights_per_k = NULL,
                              use_labels_not_values = TRUE,
                              order_clusters_as_data =TRUE,
                              sort_cluster_numbers = TRUE,
                              warn = TRUE,
                              try_cutree_hclust = TRUE,
                              ...)

Arguments

tree

a dendrogram object

k

numeric scalar (OR a vector) with the number of clusters the tree should be cut into.

h

numeric scalar (OR a vector) with a height where the tree should be cut.

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 if using k!=NULL .

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, this is s

sort_cluster_numbers

logical (TRUE). Should the resulting cluster id numbers be sorted? (default is TRUE in order to make the function compatible with cutree ) from {stats}, but it allows for sensible color order when us

warn

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

try_cutree_hclust

logical. default is TRUE. Since cutree for hclust is MUCH faster than for dendrogram - cutree.dendrogram will first try to change the dendrogram into an hclust object. If it will fail (for example, with unbranched trees), it will continue using the cut

...

(not currently in use)

Value

• If k or h are scalar - cutree.dendrogram returns an integer vector with group memberships. Otherwise a matrix with group memberships is returned where each column corresponds to the elements of k or h, respectively (which are also used as column names). 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.

Details

At least one of k or h must be specified, k overrides h if both are given. as opposed to cutree for hclust, cutree.dendrogram allows the cutting of trees at a given height also for non-ultrametric trees (ultrametric tree == a tree with monotone clustering heights). This sort_cluster_numbers parameter is based on the sort_levels_values function. Using that parameter with TRUE makes the clusters id's from cutree to be ordered from left to right. e.g: the left most cluster in the tree will be numbered "1", the one after it will be "2" etc...).

See Also

hclust, cutree, cutree_1h.dendrogram, cutree_1k.dendrogram,

Examples

hc <- hclust(dist(USArrests[c(1,6,13,20, 23),]), "ave")
dend <- as.dendrogram(hc)
unbranch_dend <- unbranch(dend,2)

cutree(hc, k=2:4) # on hclust
cutree(dend, k=2:4) # on dendrogram

cutree(hc, k=2) # on hclust
cutree(dend, k=2) # on dendrogram

cutree(dend, h = c(20, 25.5, 50,170))
cutree(hc, h = c(20, 25.5, 50,170))

# the default (ordered by original data's order)
cutree(dend, k=2:3, order_clusters_as_data = FALSE)
labels(dend)

# as.hclust(unbranch_dend) # ERROR - can not do this...
cutree(unbranch_dend, k = 2) # all NA's
cutree(unbranch_dend, k = 1:4)
cutree(unbranch_dend, h = c(20, 25.5, 50,170))
cutree(dend, h = c(20, 25.5, 50,170))


require(microbenchmark)
## this shows how as.hclust is expensive - but still worth it if possible
microbenchmark(
      cutree(hc, k=2:4),
      cutree(as.hclust(dend), k=2:4),
      cutree(dend, k=2:4),
      cutree(dend, k=2:4, try_cutree_hclust=FALSE)
   )
         # the dendrogram is MUCH slower...

# Unit: microseconds
##                       expr      min       lq    median        uq       max neval
##        cutree(hc, k = 2:4)   91.270   96.589   99.3885  107.5075   338.758   100
##    tree(as.hclust(dend),
##			  k = 2:4)           1701.629 1767.700 1854.4895 2029.1875  8736.591   100
##      cutree(dend, k = 2:4) 1807.456 1869.887 1963.3960 2125.2155  5579.705   100
##  cutree(dend, k = 2:4,
##	try_cutree_hclust = FALSE) 8393.914 8570.852 8755.3490 9686.7930 14194.790   100

# and trying to "hclust" is not expensive (which is nice...)
microbenchmark(
  cutree_unbranch_dend = cutree(unbranch_dend, k=2:4),
  cutree_unbranch_dend_not_trying_to_hclust =
  cutree(unbranch_dend, k=2:4, try_cutree_hclust=FALSE)
)


## Unit: milliseconds
##                   expr      min       lq   median       uq      max neval
##cutree_unbranch_dend       7.309329 7.428314 7.494107 7.752234 17.59581   100
##cutree_unbranch_dend_not
##_trying_to_hclust        6.945375 7.079198 7.148629 7.577536 16.99780   100
##There were 50 or more warnings (use warnings() to see the first 50)