R根据cutree标签从树形图中获取子树

Question

I have clustered a large dataset and found 6 clusters I am interested in analyzing more in depth.

I found the clusters using hclust with "ward.D" method, and I would like to know whether there is a way to get "sub-trees" from hclust/dendrogram objects.

For example

library(gplots)
library(dendextend)

data <- iris[,1:4]
distance <- dist(data, method = "euclidean", diag = FALSE, upper = FALSE)
hc <- hclust(distance, method = 'ward.D')
dnd <- as.dendrogram(hc)
plot(dnd) # to decide the number of clusters
clusters <- cutree(dnd, k = 6)

I used cutree to get the labels for each of the rows in my dataset.

I know I can get the data for each corresponding cluster (cluster 1 for example) with:

c1_data = data[clusters == 1,]

Is there any easy way to get the subtrees for each corresponding label as returned by dendextend::cutree ? For example, say I am interesting in getting the

I know I can access the branches of the dendrogram doing something like

subtree <- dnd[[1]][[2]

but how I can get exactly the subtree corresponding to cluster 1?

I have tried

dnd[clusters == 1]

but this of course doesn't work. So how can I get the subtree based on the labels returned by cutree?

Answer 1

================= UPDATED answer

This can now be solved using the get_subdendrograms from dendextend .

# needed packages:
# install.packages(gplots)
# install.packages(viridis)
# install.packages(devtools)
# devtools::install_github('talgalili/dendextend') # dendextend from github

# define dendrogram object to play with:
dend <- iris[,-5] %>% dist %>% hclust %>% as.dendrogram %>%  set("labels_to_character") %>% color_branches(k=5)
dend_list <- get_subdendrograms(dend, 5)

# Plotting the result
par(mfrow = c(2,3))
plot(dend, main = "Original dendrogram")
sapply(dend_list, plot)

This can also be used within a heatmap:

# plot a heatmap of only one of the sub dendrograms
par(mfrow = c(1,1))
library(gplots)
sub_dend <- dend_list[[1]] # get the sub dendrogram
# make sure of the size of the dend
nleaves(sub_dend)
length(order.dendrogram(sub_dend))
# get the subset of the data
subset_iris <- as.matrix(iris[order.dendrogram(sub_dend),-5])
# update the dendrogram's internal order so to not cause an error in heatmap.2
order.dendrogram(sub_dend) <- rank(order.dendrogram(sub_dend))
heatmap.2(subset_iris, Rowv = sub_dend, trace = "none", col = viridis::viridis(100))

================= OLDER answer

I think what can be helpful for you are these two functions:

The first one just iterates through all clusters and extracts substructure. It requires:

• the dendrogram object from which we want to get the subdendrograms
• the clusters labels (eg returned by cutree )

Returns a list of subdendrograms.

extractDendrograms <- function(dendr, clusters){
    lapply(unique(clusters), function(clust.id){
        getSubDendrogram(dendr, which(clusters==clust.id))
    })
}

The second one performs a depth-first search to determine in which subtree the cluster exists and if it matches the full cluster returns it. Here, we use the assumption that all elements of a cluster are in one subtress. It requires:

• the dendrogram object
• positions of the elements in cluster

Returns a subdendrograms corresponding to the cluster of given elements.

getSubDendrogram<-function(dendr, my.clust){
    if(all(unlist(dendr) %in% my.clust))
        return(dendr)
    if(any(unlist(dendr[[1]]) %in% my.clust ))
        return(getSubDendrogram(dendr[[1]], my.clust))
    else 
        return(getSubDendrogram(dendr[[2]], my.clust))
}

Using these two functions we can use the variables you have provided in the question and get the following output. (I think the line clusters <- cutree(dnd, k = 6) should be clusters <- cutree(hc, k = 6) )

my.sub.dendrograms <- extractDendrograms(dnd, clusters)

plotting all six elements from the list gives all subdendrograms

EDIT

As suggested in the comment, I add a function that as an input takes a dendrogram dend and the number of subtrees k , but it still uses the previously defined, recursive function getSubDendrogram :

prune_cutree_to_dendlist <- function(dend, k, order_clusters_as_data=FALSE) {
    clusters <- cutree(dend, k, order_clusters_as_data)
    lapply(unique(clusters), function(clust.id){    
        getSubDendrogram(dend, which(clusters==clust.id))
    })
}

A test case for 5 substructures:

library(dendextend)
dend <- iris[,-5] %>% dist %>% hclust %>% as.dendrogram %>% set("labels_to_character") %>% color_branches(k=5)

subdend.list <- prune_cutree_to_dendlist(dend, 5)

#plotting
par(mfrow = c(2,3))
plot(dend, main = "original dend")
sapply(prunned_dends, plot)

I have performed some benchmark using rbenchmark with the function suggested by Tal Galili (here named prune_cutree_to_dendlist2 ) and the results are quite promising for the DFS approach from the above:

library(rbenchmark)
benchmark(prune_cutree_to_dendlist(dend, 5), 
          prune_cutree_to_dendlist2(dend, 5), replications=5)

                                test replications elapsed relative user.self
1  prune_cutree_to_dendlist(dend, 5)            5    0.02        1     0.020
2 prune_cutree_to_dendlist2(dend, 5)            5   60.82     3041    60.643

Answer 2

I wrote now function prune_cutree_to_dendlist to do what you asked for. I should add it to dendextend at some point in the future.

In the meantime, here is an example of the code and output (the function is a bit slow. Making it faster relies on having prune be faster, which I won't get to fixing in the near future.)

# install.packages("dendextend")

library(dendextend)
dend <- iris[,-5] %>% dist %>% hclust %>% as.dendrogram %>% 
  set("labels_to_character")
dend <- dend %>% color_branches(k=5)

# plot(dend)

prune_cutree_to_dendlist <- function(dend, k) {
  clusters <- cutree(dend,k, order_clusters_as_data = FALSE)
  # unique_clusters <- unique(clusters) # could also be 1:k but it would be less robust
  # k <- length(unique_clusters)
  # for(i in unique_clusters) { 
  dends <- vector("list", k)
  for(i in 1:k) { 
    leves_to_prune <- labels(dend)[clusters != i]
    dends[[i]] <- prune(dend, leves_to_prune)

  }

  class(dends) <- "dendlist"

  dends
}

prunned_dends <- prune_cutree_to_dendlist(dend, 5)
sapply(prunned_dends, nleaves)

par(mfrow = c(2,3))
plot(dend, main = "original dend")
sapply(prunned_dends, plot)

Answer 3

How did you get 6 clusters using hclust? You can cut the tree at any point, so you just ask cuttree to give you more clusters:

clusters = cutree(hclusters, number_of_clusters)

If you have a lot of data this may not be very handy though. In these cases what I do is manually picking the clusters that I want to study further and then running hclust only on the data in these clusters. I don't know of any functionality in hclust that allows you to do this automatically, but it's quite easy:

good_clusters = c(which(clusters==1), 
                  which(clusters==2)) #or whichever cLusters you want
new_df = df[good_clusters,]
new_hclusters = hclust(new_df)
new_clusters = cutree(new_hclusters, new_number_of_clusters)