R/cutree.dendrogram.R
cutreemethods.Rd
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.
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for default
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for hclust
cutree(
tree,
k = NULL,
h = NULL,
use_labels_not_values = TRUE,
order_clusters_as_data = TRUE,
warn = dendextend_options("warn"),
NA_to_0L = TRUE,
...
)
# S3 method for phylo
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for phylo
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for agnes
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for diana
cutree(tree, k = NULL, h = NULL, ...)
# S3 method for dendrogram
cutree(
tree,
k = NULL,
h = NULL,
dend_heights_per_k = NULL,
use_labels_not_values = TRUE,
order_clusters_as_data = TRUE,
warn = dendextend_options("warn"),
try_cutree_hclust = TRUE,
NA_to_0L = TRUE,
...
)
a dendrogram object
numeric scalar (OR a vector) with the number of clusters the tree should be cut into.
numeric scalar (OR a vector) with a height where the tree should be cut.
(not currently in use)
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
.
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 set
to TRUE.
This is passed to cutree_1h.dendrogram
.
logical (default from dendextend_options("warn") is FALSE). Set if warning are to be issued, it is safer to keep this at TRUE, but for keeping the noise down, the default is FALSE. Should the function send a warning in case the desried k is not available?
logical. default is TRUE. When no clusters are possible, Should the function return 0 (TRUE, default), or NA (when set to FALSE).
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 .
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 cutree.dendrogram function. If try_cutree_hclust=FALSE, it will force to use cutree.dendrogram and not cutree.hclust.
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.
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 nonultrametric trees
(ultrametric tree == a tree with monotone clustering heights).
if (FALSE) {
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))
library(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)
# notice that if cutree can't find clusters for the desired k/h, it will produce 0's instead!
# (It will produce a warning though...)
# This is a different behaviout than stats::cutree
# For example:
cutree(as.dendrogram(hclust(dist(c(1, 1, 1, 2, 2)))),
k = 5
)
}