Question

我想修改由hclust对象绘制的树形图中叶子的属性。最低限度，我想改变颜色，但您可以提供任何帮助。

我确实尝试过谷歌答案，但我看到的每一个解决方案似乎都比我想象的要难得多。

Answer 1

不久前，Joris Meys向我提供了这段改变叶子颜色的代码片段。修改它以反映您的属性。

clusDendro <- as.dendrogram(Clustering)
labelColors <- c("red", "blue", "darkgreen", "darkgrey", "purple")

## function to get colorlabels
colLab <- function(n) {
   if(is.leaf(n)) {
       a <- attributes(n)
       # clusMember - a vector designating leaf grouping
       # labelColors - a vector of colors for the above grouping
       labCol <- labelColors[clusMember[which(names(clusMember) == a$label)]]
       attr(n, "nodePar") <- c(a$nodePar, lab.col = labCol)
   }
   n
}

## Graph
clusDendro <- dendrapply(clusDendro, colLab)
op <- par(mar = par("mar") + c(0,0,0,2))
plot(clusDendro,
     main = "Major title",
     horiz = T, type = "triangle", center = T)

par(op)

Answer 2

以下是使用名为“dendextend”的新软件包解决此问题的方法，该软件包完全针对此类内容构建。

您可以在以下网址的“使用情况”部分中看到该套餐的演示文稿和插图中的许多示例：https://github.com/talgalili/dendextend

以下是此问题的解决方案：

# define dendrogram object to play with:
dend <- as.dendrogram(hclust(dist(USArrests[1:3,]), "ave"))
# loading the package
install.packages('dendextend') # it is now on CRAN
library(dendextend)# let's add some color:
labels_colors(dend) <- 2:4
labels_colors(dend)
plot(dend)

enter image description here

Answer 3

目前尚不清楚您想要使用它，但我经常需要在树形图中识别分支。我已经破解了rect.hclust方法来添加密度和标签输入。

你会这样称呼它：


k <- 3 # number of branches to identify
labels.to.identify <- c('1','2','3')
required.density <- 10 # the density of shading lines, in lines per inch 

rect.hclust.nice(tree, k, labels=labels.to.identify, density=density.required)

这是函数



rect.hclust.nice = function (tree, k = NULL, which = NULL, x = NULL, h = NULL, border = 2, 
    cluster = NULL,  density = NULL,labels = NULL, ...) 
{
    if (length(h) > 1 | length(k) > 1) 
        stop("'k' and 'h' must be a scalar")
    if (!is.null(h)) {
        if (!is.null(k)) 
            stop("specify exactly one of 'k' and 'h'")
        k <- min(which(rev(tree$height) < h))
        k <- max(k, 2)
    }
    else if (is.null(k)) 
        stop("specify exactly one of 'k' and 'h'")
    if (k < 2 | k > length(tree$height)) 
        stop(gettextf("k must be between 2 and %d", length(tree$height)), 
            domain = NA)
    if (is.null(cluster)) 
        cluster <- cutree(tree, k = k)
    clustab <- table(cluster)[unique(cluster[tree$order])]
    m <- c(0, cumsum(clustab))
    if (!is.null(x)) {
        if (!is.null(which)) 
            stop("specify exactly one of 'which' and 'x'")
        which <- x
        for (n in 1L:length(x)) which[n] <- max(which(m < x[n]))
    }
    else if (is.null(which)) 
        which <- 1L:k
    if (any(which > k)) 
        stop(gettextf("all elements of 'which' must be between 1 and %d", 
            k), domain = NA)
    border <- rep(border, length.out = length(which))
    labels <- rep(labels, length.out = length(which))
    retval <- list()
    for (n in 1L:length(which)) {
        rect(m[which[n]] + 0.66, par("usr")[3L], m[which[n] + 
            1] + 0.33, mean(rev(tree$height)[(k - 1):k]), border = border[n], col = border[n], density = density, ...)
        text((m[which[n]] + m[which[n] + 1]+1)/2, grconvertY(grconvertY(par("usr")[3L],"user","ndc")+0.02,"ndc","user"),labels[n])
        retval[[n]] <- which(cluster == as.integer(names(clustab)[which[n]]))
    }
    invisible(retval)
}

改变树状图离开

3 个答案: