Question

我试图在R中绘制层次聚类的结果作为树状图，矩形识别聚类。

下面的代码用于垂直树形图，但对于水平树形图（horiz=TRUE），不绘制矩形。有没有办法对水平树形图做同样的事情。

library("cluster")
dst <- daisy(iris, metric = c("gower"), stand = FALSE)
hca <- hclust(dst, method = "average")
plot(as.dendrogram(hca), horiz = FALSE)
rect.hclust(hca, k = 3, border = "red")

此外，我想绘制一条线来切割所需距离值的树。如何在R中绘制它。cutree函数返回簇，但也可以绘制它。

cutree(hca, k = 3)

我正在寻找的所需输出是这样的。

dendrogram

如何在R中完成此操作？

Answer 1

jlhoward和Backlin的答案都很好。

您还可以尝试使用专为此类设计而设计的dendextend包。它有rect.dendrogram函数，其作用类似于rect.hclust，但带有horiz参数（加上对rect边缘位置的更多控制）。要查找相关高度，您可以使用heights_per_k.dendrogram函数（当使用dendextendRcpp包时速度更快）

以下是一个简单的示例，说明如何获得与上述示例相同的结果（还有彩色分支的额外奖励，只是为了好玩）：

install.packages("dendextend")
install.packages("dendextendRcpp")

library("dendextend")
library("dendextendRcpp")

# using piping to get the dend
dend <- iris[,-5] %>% dist %>% hclust %>% as.dendrogram

# plot + color the dend's branches before, based on 3 clusters:
dend %>% color_branches(k=3) %>% plot(horiz=TRUE, main = "The dendextend package \n Gives extended functionality to R's dendrogram object")

# add horiz rect
dend %>% rect.dendrogram(k=3,horiz=TRUE)

# add horiz (well, vertical) line:
abline(v = heights_per_k.dendrogram(dend)["3"] + .6, lwd = 2, lty = 2, col = "blue")

enter image description here

Answer 2

要完成工作（虽然以一种非常丑陋的方式），您只需手动将调用中的坐标交换为rect中的rect.hclust：

rhc <- function (tree, k = NULL, which = NULL, x = NULL, h = NULL, border = 2, 
    cluster = NULL) 
{
    if (length(h) > 1L | length(k) > 1L) 
        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 seq_along(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_len(border, length(which))
    retval <- list()
    for (n in seq_along(which)) {
        rect(
             ybottom = m[which[n]] + 0.66,
             xright = par("usr")[3L],
             ytop = m[which[n] + 1] + 0.33,
             xleft = mean(rev(tree$height)[(k - 1):k]),
             border = border[n])
        retval[[n]] <- which(cluster == as.integer(names(clustab)[which[n]]))
    }
    invisible(retval)
}

并拨打rhc，就像您呼叫rect.hclust：

一样

rhc(hca, k = 3, border = "red")

enter image description here

Answer 3

以下是使用ggplot和ggdendro包的解决方案。作为额外的奖励，我们可以按群集为标签着色......

library(cluster)
dst   <- daisy(iris, metric = c("gower"), stand = FALSE)
hca   <- hclust(dst, method = "average")
k     <- 3
clust <- cutree(hca,k=k)  # k clusters

library(ggplot2)
library(ggdendro)     # for dendro_data(...)
dendr    <- dendro_data(hca, type="rectangle") # convert for ggplot
clust.df <- data.frame(label=rownames(iris), cluster=factor(clust))
dendr[["labels"]]   <- merge(dendr[["labels"]],clust.df, by="label")
rect <- aggregate(x~cluster,label(dendr),range)
rect <- data.frame(rect$cluster,rect$x)
ymax <- mean(hca$height[length(hca$height)-((k-2):(k-1))])

ggplot() + 
  geom_segment(data=segment(dendr), aes(x=x, y=y, xend=xend, yend=yend)) + 
  geom_text(data=label(dendr), aes(x, y, label=label, hjust=0, color=cluster), 
            size=3) +
  geom_rect(data=rect, aes(xmin=X1-.3, xmax=X2+.3, ymin=0, ymax=ymax), 
            color="red", fill=NA)+
  geom_hline(yintercept=0.33, color="blue")+
  coord_flip() + scale_y_reverse(expand=c(0.2, 0)) + 
  theme_dendro()

树枝切割和簇周围的矩形，用于R中的水平树状图

3 个答案: