帶有cutree的hclust（）…如何在單個hclust（）中繪制cutree（）簇

Question

我將我的hclust（）樹與cutree（）分成幾組。 現在我想要一個函數將幾個組成員hclust（）作為hclust（）...還可以：

我將一棵樹切成168組，我想要168棵hclust（）樹...我的數據是1600 * 1600矩陣。

我的數據太多了，所以我舉一個例子

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)

現在我要繪制18棵樹...一組中的一棵樹。 我已經嘗試了很多

Answer 1

我提前警告您，對於這么大的樹木，可能大多數解決方案的運行速度都會有些慢。 但是這是一種解決方案（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)

讓我們在“更細”的樹上再做一次：

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)