View source: R/cutree.dendrogram.R
cutree_1h.dendrogram | R Documentation |
Cuts a dendrogram tree into several groups by specifying the desired cut height (only a single height!).
cutree_1h.dendrogram(
dend,
h,
order_clusters_as_data = TRUE,
use_labels_not_values = TRUE,
warn = dendextend_options("warn"),
...
)
dend |
a dendrogram object |
h |
numeric scalar (NOT a vector) with a height where the dend should be cut. |
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. |
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.). |
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. |
... |
(not currently in use) |
cutree_1h.dendrogram
returns an integer vector with group memberships
Tal Galili
hclust
, cutree
hc <- hclust(dist(USArrests[c(1, 6, 13, 20, 23), ]), "ave")
dend <- as.dendrogram(hc)
cutree(hc, h = 50) # on hclust
cutree_1h.dendrogram(dend, h = 50) # on a dendrogram
labels(dend)
# the default (ordered by original data's order)
cutree_1h.dendrogram(dend, h = 50, order_clusters_as_data = TRUE)
# A different order of labels - order by their order in the tree
cutree_1h.dendrogram(dend, h = 50, order_clusters_as_data = FALSE)
# make it faster
## Not run:
library(microbenchmark)
microbenchmark(
cutree_1h.dendrogram(dend, h = 50),
cutree_1h.dendrogram(dend, h = 50, use_labels_not_values = FALSE)
)
# 0.8 vs 0.6 sec - for 100 runs
## End(Not run)
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