我使用下面的代码在R中运行层次聚类分析后生成了树状图。我现在尝试根据另一个因子变量为标签着色,该变量保存为矢量。我实现这一目标的最接近的方法是使用ColourDendrogram
包中的sparcl
函数对分支进行颜色编码。如果可能的话,我更愿意对标签进行颜色编码。我在以下链接中找到了类似问题的答案Color branches of dendrogram using an existing column& Colouring branches in a dendrogram in R,但我无法找到如何为我的目的转换示例代码。下面是一些示例数据和代码。
> dput(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))
df.nw = df[,3:4]
labs = df$labs
d = dist(as.matrix(df.nw)) # find distance matrix
hc = hclust(d, method="complete") # apply hierarchical clustering
plot(hc, hang=-0.01, cex=0.6, labels=labs, xlab="") # plot the dendrogram
hcd = as.dendrogram(hc) # convert hclust to dendrogram
plot(hcd, cex=0.6) # plot using dendrogram object
Var = df$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
library(sparcl)
ColorDendrogram(hc, y=varCol, branchlength=0.9, labels=labs,
xlab="", ylab="", sub="")
非常感谢任何有关如何做到这一点的建议。
答案 0 :(得分:3)
尝试
# ... 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
}
plot(dendrapply(hcd, colLab))
(via)
答案 1 :(得分:2)
要为标签着色,最简单的方法是使用dendextend包中的labels_colors
功能。例如:
# install.packages("dendextend")
library(dendextend)
small_iris <- iris[c(1, 51, 101, 2, 52, 102), ]
dend <- as.dendrogram(hclust(dist(small_iris[,-5])))
# Like:
# dend <- small_iris[,-5] %>% dist %>% hclust %>% as.dendrogram
# By default, the dend has no colors to the labels
labels_colors(dend)
par(mfrow = c(1,2))
plot(dend, main = "Original dend")
# let's add some color:
colors_to_use <- as.numeric(small_iris[,5])
colors_to_use
# But sort them based on their order in dend:
colors_to_use <- colors_to_use[order.dendrogram(dend)]
colors_to_use
# Now we can use them
labels_colors(dend) <- colors_to_use
# Now each state has a color
labels_colors(dend)
plot(dend, main = "A color for every Species")
有关该软件包的更多详细信息,您可以查看at its vignette。