# cutree

0th

Percentile

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

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 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 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, ...)
##### 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.

...

(not currently in use)

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 set to TRUE. This is passed to cutree_1h.dendrogram.

warn

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?

NA_to_0L

logical. default is TRUE. When no clusters are possible, Should the function return 0 (TRUE, default), or NA (when set to FALSE).

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 .

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 cutree.dendrogram function. If try_cutree_hclust=FALSE, it will force to use cutree.dendrogram and not cutree.hclust.

##### 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).

##### 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.

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

##### Aliases
• cutree
• cutree.default
• cutree.hclust
• cutree.phylo
• cutree.dendrogram
##### Examples
# NOT RUN {
# }
# NOT RUN {
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)

# }
# NOT RUN {

# }

Documentation reproduced from package dendextend, version 1.8.0, License: GPL-2 | GPL-3

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