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我希望可视化聚类算法的表现如何（具有一定的距离度量）。我有样本及其相应的类。为了可视化，我进行了聚类，并希望通过聚类中的项目为树状图的分​​支着色。颜色将是层次集群中大多数项目对应的颜色（由数据\类给出）。

示例：如果我的聚类算法选择索引 1,21,24 作为某个集群（在某个级别），并且我有一个 csv 文件，其中每一行都包含一个类号，例如 1,2,1。我希望这条边的颜色为 1。

示例代码：

require(cluster)
suppressPackageStartupMessages(library(dendextend))
dir <- 'distance_metrics/'
filename <- 'aligned.csv'
my.data <- read.csv(paste(dir, filename, sep=""), header = T, row.names = 1)
my.dist <- as.dist(my.data)
real.clusters <-read.csv("clusters", header = T, row.names = 1)
clustered <- diana(my.dist)
# dend <- colour_branches(???dend, max(real.clusters)???)
plot(dend)

编辑：另一个示例部分代码

dir <- 'distance_metrics/' # csv in here contains a symmetric matrix
clust.dir <- "clusters/" #csv in here contains a column vector with classes
my.data <- read.csv(paste(dir, filename, sep=""), header = T, row.names = 1)
filename <- 'table.csv'
my.dist <- as.dist(my.data)
real.clusters <-read.csv(paste(clust.dir, filename, sep=""), header = T, row.names = 1)
clustered <- diana(my.dist)
dnd <- as.dendrogram(clustered)
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3 回答 3

1

您正在寻找的函数color_brances来自 dendextend R 包，使用参数 clusters 和 col。这是一个示例（基于 Shaun Wilkinson 的示例）：

library(cluster)
set.seed(999)
iris2 <- iris[sample(x = 1:150,size = 50,replace = F),]
clust <- diana(iris2)
dend <- as.dendrogram(clust)

temp_col <- c("red", "blue", "green")[as.numeric(iris2$Species)]
temp_col <- temp_col[order.dendrogram(dend)]
temp_col <- factor(temp_col, unique(temp_col))

library(dendextend)
dend %>% color_branches(clusters = as.numeric(temp_col), col = levels(temp_col)) %>% 
   set("labels_colors", as.character(temp_col)) %>% 
   plot

在此处输入图像描述

于 2017-08-02T08:24:45.937 回答
1

节点和边缘颜色属性都可以使用dendrapply. cluster包还具有用于“diana”类对象的as.dendrogram方法，因此对象类型之间的转换是无缝的。使用您的diana聚类并从@Edvardoss iris 示例中借用一些代码，您可以创建彩色树状图，如下所示： 

library(cluster)
set.seed(999)
iris2 <- iris[sample(x = 1:150,size = 50,replace = F),]
clust <- diana(iris2)
dnd <- as.dendrogram(clust)

## Duplicate rownames aren't allowed, so we need to set the "labels"
## attributes recursively. We also label inner nodes here. 
rectify_labels <- function(node, df){
  newlab <- df$Species[unlist(node, use.names = FALSE)]
  attr(node, "label") <- (newlab)
  return(node)
}
dnd <- dendrapply(dnd, rectify_labels, df = iris2)

## Create a color palette as a data.frame with one row for each spp
uniqspp <- as.character(unique(iris$Species))
colormap <- data.frame(Species = uniqspp, color = rainbow(n = length(uniqspp)))
colormap[, 2] <- c("red", "blue", "green")
colormap

## Now color the inner dendrogram edges
color_dendro <- function(node, colormap){
  if(is.leaf(node)){
    nodecol <- colormap$color[match(attr(node, "label"), colormap$Species)]
    attr(node, "nodePar") <- list(pch = NA, lab.col = nodecol)
    attr(node, "edgePar") <- list(col = nodecol)
  }else{
    spp <- attr(node, "label")
    dominantspp <- levels(spp)[which.max(tabulate(spp))]
    edgecol <- colormap$color[match(dominantspp, colormap$Species)]
    attr(node, "edgePar") <- list(col = edgecol)
  }
  return(node)
}
dnd <- dendrapply(dnd, color_dendro, colormap = colormap)

## Plot the dendrogram
plot(dnd)

在此处输入图像描述

于 2017-08-02T00:40:58.810 回答
0

有误解了这个问题的怀疑，但我会尝试回答：从我之前的目标被 iris 的例子重写

clrs <- rainbow(n = 3) # create palette
clrs <- clrs[iris$Species] # assign colors
plot(x = iris$Sepal.Length,y = iris$Sepal.Width,col=clrs) # simple test colors
# cluster
dt <- cbind(iris,clrs)
dt <- dt[sample(x = 1:150,size = 50,replace = F),] # create short dataset for visualization convenience
empty.labl <- gsub("."," ",dt$Species) # create a space vector with length of names intended for  reserve place to future text labels
dst <- dist(x = scale(dt[,1:4]),method = "manhattan")
hcl <- hclust(d = dst,method = "complete")
plot(hcl,hang=-1,cex=1,labels = empty.labl, xlab = NA,sub=NA)
dt <- dt[hcl$order,] # sort rows for  order objects in dendrogramm
text(x = seq(nrow(dt)), y=-.5,labels = dt$Species,srt=90,cex=.8,xpd=NA,adj=c(1,0.7),col=as.character(dt$clrs))

结果

于 2017-08-01T11:00:44.820 回答