如何在单个hclust()图中绘制cutree()聚类结果

Question

我用cutree()将我的hclust()树分成了几组。现在，我想要一个函数将几个组成员作为hclust()重新聚类...

同时：

我将一棵树切成了168个组，我需要168颗hclust()树...

我的数据是一个1600*1600的矩阵。

我的数据太大了，所以我给你一个例子。

m<-matrix(1:1600,nrow=40)
#m<-as.matrix(m) // I know it isn't necessary here
m_dist<-as.dist(m,diag = FALSE )


m_hclust<-hclust(m_dist, method= "average")
plot(m_hclust)

groups<- cutree(m_hclust, k=18)

现在我想绘制18棵树...每组一个树。我尝试了很多...

- Mareaky

Answer 1

提前警告，对于这样的大树，可能大多数解决方案运行速度都会有些慢。但是这里有一种解决方案（使用 dendextend R包）：

m<-matrix(1:1600,nrow=40)
#m<-as.matrix(m) // I know it isn't necessary here
m_dist<-as.dist(m,diag = FALSE )
m_hclust<-hclust(m_dist, method= "complete")
plot(m_hclust)
groups <- cutree(m_hclust, k=18)

# Get dendextend
install.packages.2 <- function (pkg) if (!require(pkg)) install.packages(pkg);
install.packages.2('dendextend')
install.packages.2('colorspace')
library(dendextend)
library(colorspace)

# I'll do this to just 4 clusters for illustrative purposes
k <- 4
cols <- rainbow_hcl(k)
dend <- as.dendrogram(m_hclust)
dend <- color_branches(dend, k = k)
plot(dend)
labels_dend <- labels(dend)
groups <- cutree(dend, k=4, order_clusters_as_data = FALSE)
dends <- list()
for(i in 1:k) {
    labels_to_keep <- labels_dend[i != groups]
    dends[[i]] <- prune(dend, labels_to_keep)
}

par(mfrow = c(2,2))
for(i in 1:k) { 
    plot(dends[[i]], 
        main = paste0("Tree number ", i))
}
# p.s.: because we have 3 root only trees, they don't have color (due to a "missing feature" in the way R plots root only dendrograms)

让我们在一棵“更好”的树上再做一次：

m_dist<-dist(mtcars,diag = FALSE )
m_hclust<-hclust(m_dist, method= "complete")
plot(m_hclust)

# Get dendextend
install.packages.2 <- function (pkg) if (!require(pkg)) install.packages(pkg);
install.packages.2('dendextend')
install.packages.2('colorspace')
library(dendextend)
library(colorspace)

# I'll do this to just 4 clusters for illustrative purposes
k <- 4
cols <- rainbow_hcl(k)
dend <- as.dendrogram(m_hclust)
dend <- color_branches(dend, k = k)
plot(dend)
labels_dend <- labels(dend)
groups <- cutree(dend, k=4, order_clusters_as_data = FALSE)
dends <- list()
for(i in 1:k) {
    labels_to_keep <- labels_dend[i != groups]
    dends[[i]] <- prune(dend, labels_to_keep)
}

par(mfrow = c(2,2))
for(i in 1:k) { 
    plot(dends[[i]], 
        main = paste0("Tree number ", i))
}
# p.s.: because we have 3 root only trees, they don't have color (due to a "missing feature" in the way R plots root only dendrograms)

- Tal Galili