从R中的切割树状图提取标签成员资格/分类（即：树状图的Cutree函数） [英] Extract labels membership / classification from a cut dendrogram in R (i.e.: a cutree function for dendrogram)

问题描述

我正在尝试从R中的树状图提取分类，该分类已在一定高度进行了切割切割。在 hclust 对象上使用 cutree 很容易做到，但是我不知道该怎么做一个树状图对象。

此外，我不能仅仅使用原始hclust中的集群，因为（令人沮丧）， cutree 中的类编号与具有 cut 的类编号不同。

  hc<-hclust（dist（USArrests）， ave）
 
 category< --cutree（hc，h = 70）
 
 dend1<-as树状图（hc）
 dend2<-cut（dend1，h = 70）
 
 
 str （dend2 $ lower [[1]]）#group 1在这里与
分类[classification == 1] #group 1在这里
  

是否有一种方法可以使分类相互映射，或者从树图对象（也许巧妙地使用了 dendrapply ？）的格式类似于 cutree 给出的格式

解决方案

我建议您使用 cutree 函数从

希望您或其他人对此答案有所帮助。

I'm trying to extract a classification from a dendrogram in R that I've cut at a certain height. This is easy to do with cutree on an hclustobject, but I can't figure out how to do it on a dendrogram object.

Further, I can't just use my clusters from the original hclust, becuase (frustratingly), the numbering of the classes from cutree is different from the numbering of classes with cut.

hc <- hclust(dist(USArrests), "ave")

classification<-cutree(hc,h=70)

dend1 <- as.dendrogram(hc)
dend2 <- cut(dend1, h = 70)


str(dend2$lower[[1]]) #group 1 here is not the same as
classification[classification==1] #group 1 here

Is there a way to either get the classifications to map to each other, or alternatively to extract lower branch memberships from the dendrogram object (perhaps with some clever use of dendrapply?) in a format more like what cutree gives?

解决方案

I would propose for you to use the cutree function from the dendextend package. It includes a dendrogram method (i.e.: dendextend:::cutree.dendrogram).

You can learn more about the package from its introductory vignette.

I should add that while your function (classify) is good, there are several advantage for using cutree from dendextend:

1. It also allows you to use a specific k (number of clusters), and not just h (a specific height).

2. It is consistent with the result you would get from cutree on hclust (classify will not be).

3. It will often be faster.

Here are examples for using the code:

# Toy data:
hc <- hclust(dist(USArrests), "ave")
dend1 <- as.dendrogram(hc)

# Get the package:
install.packages("dendextend")
library(dendextend)

# Get the package:
cutree(dend1,h=70) # it now works on a dendrogram
# It is like using:
dendextend:::cutree.dendrogram(dend1,h=70)

By the way, on the basis of this function, dendextend allows the user to do more cool things, like color branches/labels based on cutting the dendrogram:

dend1 <- color_branches(dend1, k = 4)
dend1 <- color_labels(dend1, k = 5)
plot(dend1)

Lastly, here is some more code for demonstrating my other points:

# This would also work with k:
cutree(dend1,k=4)

# and would give identical result as cutree on hclust:
identical(cutree(hc,h=70)  , cutree(dend1,h=70)  )
   # TRUE

# But this is not the case for classify:
identical(classify(dend1,70)   , cutree(dend1,h=70)  )
   # FALSE


install.packages("microbenchmark")
require(microbenchmark)
microbenchmark(classify = classify(dend1,70),
               cutree = cutree(dend1,h=70)  )
#    Unit: milliseconds
#        expr      min       lq   median       uq       max neval
#    classify  9.70135  9.94604 10.25400 10.87552  80.82032   100
#      cutree 37.24264 37.97642 39.23095 43.21233 141.13880   100
# 4 times faster for this tree (it will be more for larger trees)

# Although (if to be exact about it) if I force cutree.dendrogram to not go through hclust (which can happen for "weird" trees), the speed will remain similar:
microbenchmark(classify = classify(dend1,70),
               cutree = cutree(dend1,h=70, try_cutree_hclust = FALSE)  )
# Unit: milliseconds
#        expr       min        lq    median       uq      max neval
#    classify  9.683433  9.819776  9.972077 10.48497 29.73285   100
#      cutree 10.275839 10.419181 10.540126 10.66863 16.54034   100

If you are thinking of ways to improve this function, please patch it through here:

https://github.com/talgalili/dendextend/blob/master/R/cutree.dendrogram.R

I hope you, or others, will find this answer helpful.

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