Change col/lwd/lty of branches based on clusters
The user supplies a dend, a vector of clusters, and what to modify (and how).
And the function returns a dendrogram with branches col/lwd/lty accordingly. (the function assumes unique labels)
branches_attr_by_clusters(
dend,
clusters,
values,
attr = c("col", "lwd", "lty"),
branches_changed_have_which_labels = c("any", "all"),
...
)dend |
a dendrogram dend |
clusters |
an integer vector of clusters. This HAS to be of the same length as the number of leaves. Items that belong to no cluster should get the value 0. The vector should be of the same order as that of the labels in the dendrogram. If you create the clusters from something like cutree you would first need to use order.dendrogram on it, before using it in the function. |
values |
the attributes to use for non 0 values. This should be of the same length as the number of unique non-0 clusters. If it is shorter, it is recycled. OR, this can also be of the same length as the number of leaves in the tree In which case, the values will be aggreagted (i.e.: tapply), to match the number of clusters. The first value of each cluster will be used as the main value. TODO: So far, the function doesn't deal well with NA values. (this might be changed in the future) |
attr |
a character with one of the following values: col/lwd/lty |
branches_changed_have_which_labels |
character with either "any" (default) or "all". Inidicates how the branches should be updated. |
... |
ignored. |
This is probably NOT a very fast implementation of the function, but it works.
This function was designed to enable the manipulation (mainly coloring) of branches, based on the results from the cutreeDynamic function.
A dendrogram with modified branches (col/lwd/lty).
## Not run:
### Getting the hc object
iris_dist <- iris[, -5] %>% dist()
hc <- iris_dist %>% hclust()
# This is how it looks without any colors:
dend <- as.dendrogram(hc)
plot(dend)
# Both functions give the same outcome
# options 1:
dend %>%
set("branches_k_color", k = 4) %>%
plot()
# options 2:
clusters <- cutree(dend, 4)[order.dendrogram(dend)]
dend %>%
branches_attr_by_clusters(clusters) %>%
plot()
# and the second option is much slower:
system.time(set(dend, "branches_k_color", k = 4)) # 0.26 sec
system.time(branches_attr_by_clusters(dend, clusters)) # 1.61 sec
# BUT, it also allows us to do more flaxible things!
#--------------------------
# Plotting dynamicTreeCut
#--------------------------
# let's get the clusters
library(dynamicTreeCut)
clusters <- cutreeDynamic(hc, distM = as.matrix(iris_dist))
# we need to sort them to the order of the dendrogram:
clusters <- clusters[order.dendrogram(dend)]
# get some functions:
library(dendextendRcpp)
library(colorspace)
no0_unique <- function(x) {
u_x <- unique(x)
u_x[u_x != 0]
}
clusters_numbers <- no0_unique(clusters)
n_clusters <- length(clusters_numbers)
cols <- rainbow_hcl(n_clusters)
dend2 <- branches_attr_by_clusters(dend, clusters, values = cols)
# dend2 <- branches_attr_by_clusters(dend, clusters)
plot(dend2)
# add colored bars:
ord_cols <- rainbow_hcl(n_clusters)[order(clusters_numbers)]
tmp_cols <- rep(1, length(clusters))
tmp_cols[clusters != 0] <- ord_cols[clusters != 0][clusters]
colored_bars(tmp_cols, y_shift = -1.1, rowLabels = "")
# all of the ordering is to handle the fact that the cluster numbers are not ascending...
# How is this compared with the usual cutree?
dend3 <- color_branches(dend, k = n_clusters)
labels(dend2) <- as.character(labels(dend2))
# this needs fixing, since the labels are not character!
# Well, both cluster solutions are not perfect, but at least they are interesting...
tanglegram(dend2, dend3,
main_left = "cutreeDynamic", main_right = "cutree",
columns_width = c(5, .5, 5),
color_lines = cols[iris[order.dendrogram(dend2), 5]]
)
# (Notice how the color_lines is of the true Species of each Iris)
# The main difference is at the bottom,
## End(Not run)Please choose more modern alternatives, such as Google Chrome or Mozilla Firefox.