如何根据定义的组为树形图的标签着色？ （在R中）

Question

我在R中有一个24行和10,000列的数字矩阵。该矩阵的行名基本上是文件名，我从中读取了对应于24行中每一行的数据。除此之外，我有一个单独的因子列表，包含24个entires，指定24个文件所属的组。有3组 - 醇类，碳氢化合物和酯类。它们所属的名称和相应的组如下所示：

function init(options) {

  var viewModel = {
      forename: ko.observable(options.forename),
      surname: ko.observable(options.surname),
      email: ko.observable(options.email)
  }

  viewModel.getFullName = ko.computed(function () {
      return options.forename + ' ' + options.surname;
  });

  return viewModel;
}

我想生成一个树形图来查看矩阵中的数据是如何聚类的。所以，我使用了以下命令：

> MS.mz
[1] "int-354.19" "int-361.35" "int-368.35" "int-396.38" "int-408.41" "int-410.43" "int-422.43"
[8] "int-424.42" "int-436.44" "int-438.46" "int-452.00" "int-480.48" "int-648.64" "int-312.14"
[15] "int-676.68" "int-690.62" "int-704.75" "int-312.29" "int-326.09" "int-326.18" "int-326.31"
[22] "int-340.21" "int-340.32" "int-352.35"

> MS.groups
[1] Alcohol     Alcohol     Alcohol     Alcohol     Hydrocarbon Alcohol     Hydrocarbon Alcohol    
[9] Hydrocarbon Alcohol     Alcohol     Alcohol     Ester       Alcohol     Ester       Ester      
[17] Ester       Alcohol     Alcohol     Alcohol     Alcohol     Alcohol     Alcohol     Hydrocarbon
Levels: Alcohol Ester Hydrocarbon

我得到了一个树状图。现在我想根据它们所属的组（即酒精，碳氢化合物或酯类）为树形图中的文件名着色。我查看了论坛上发布的不同示例，如

Label and color leaf dendrogram in r

Label and color leaf dendrogram in R using ape package

Clustering with bootstrapping

，但无法为我的数据实现它。我不确定如何将row.names与MS.groups相关联以获得树形图中的彩色名称。

使用dendextend生成树（如https://nycdatascience.com/wp-content/uploads/2013/09/dendextend-tutorial.pdf中所述），我得到以下树

以下是用于生成它的代码：

require(vegan)
dist.mat<-vegdist(MS.data.scaled.transposed,method="euclidean")
clust.res<-hclust(dist.mat)
plot(clust.res)

Answer 1

我怀疑您要查找的功能是color_labels还是get_leaves_branches_col。标签的第一种颜色基于cutree（如color_branches），第二种颜色允许您获取每个叶子的分支颜色，然后使用它来为树的标签着色（如果你使用不寻常的方法为分支着色（就像使用branches_attr_by_labels时那样）。例如：

# define dendrogram object to play with:
hc <- hclust(dist(USArrests[1:5,]), "ave")
dend <- as.dendrogram(hc)

library(dendextend)
par(mfrow = c(1,2), mar = c(5,2,1,0))
dend <- dend %>%
         color_branches(k = 3) %>%
         set("branches_lwd", c(2,1,2)) %>%
         set("branches_lty", c(1,2,1))

plot(dend)

dend <- color_labels(dend, k = 3)
# The same as:
# labels_colors(dend)  <- get_leaves_branches_col(dend)
plot(dend)

无论哪种方式，你都应该看一下set函数，了解你的树形图可以做些什么（这样就省去了记住所有不同函数名称的麻烦）。

Answer 2

您可以查看本教程，该教程显示了几种用于按组显示R中树状图的方法

https://rstudio-pubs-static.s3.amazonaws.com/1876_df0bf890dd54461f98719b461d987c3d.html

但是，我认为最适合您数据的解决方案是由'dendextend'软件包提供的。请参阅教程（有关'iris'数据集的示例，与您的问题类似）：https://nycdatascience.com/wp-content/uploads/2013/09/dendextend-tutorial.pdf

另见插图：http://cran.r-project.org/web/packages/dendextend/vignettes/Cluster_Analysis.html

Answer 3

你可以尝试这个解决方案，只用'MS.groups'和'var'改变'labs'，你的'MS.groups'转换为数字（也许，用as.numeric）。 它来自How to colour the labels of a dendrogram by an additional factor variable in R

## The data
df <- structure(list(labs = c("a1", "a2", "a3", "a4", "a5", "a6", "a7", 
"a8", "b1", "b2", "b3", "b4", "b5", "b6", "b7"), var = c(1L, 1L, 2L,     
1L,2L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L), td = c(13.1, 14.5, 16.7, 
12.9, 14.9, 15.6, 13.4, 15.3, 12.8, 14.5, 14.7, 13.1, 14.9, 15.6, 14.6), 
fd = c(2L, 3L, 3L, 1L, 2L, 3L, 2L, 3L, 2L, 4L, 2L, 1L, 4L, 3L, 3L)), 
.Names = c("labs", "var", "td", "fd"), class = "data.frame", row.names = 
c(NA, -15L))

## Subset for clustering
df.nw = df[,3:4]

# Assign the labs column to a vector
labs = df$labs

d = dist(as.matrix(df.nw))                          # find distance matrix 
hc = hclust(d, method="complete")                   # apply hierarchical clustering 

## plot the dendrogram

plot(hc, hang=-0.01, cex=0.6, labels=labs, xlab="") 

## convert hclust to dendrogram 
hcd = as.dendrogram(hc)                             

## plot using dendrogram object
plot(hcd, cex=0.6)                                  

Var = df$var                                        # factor variable for colours
varCol = gsub("1","red",Var)                        # convert numbers to colours
varCol = gsub("2","blue",varCol)

# colour-code dendrogram branches by a factor 

# ... your code
colLab <- function(n) {
  if(is.leaf(n)) {
    a <- attributes(n)
    attr(n, "label") <- labs[a$label]
    attr(n, "nodePar") <- c(a$nodePar, lab.col = varCol[a$label]) 
  }
  n
}

## Coloured plot
plot(dendrapply(hcd, colLab))