从dendextend :: circlize_dendrogram()的输出中提取簇颜色

Al-*_*aad 10 r hierarchical-clustering dendextend

我试图提取聚类中使用的颜色circlize_dendrogram.这是一个示例代码:

library(magrittr)
library(dendextend)

cols <- c("#009000", "#FF033E", "#CB410B", "#3B444B", "#007FFF")
dend <- iris[1:40,-5] %>% dist %>% hclust %>% as.dendrogram    

dend <- color_branches(dend, k = 5, col = cols)
dend %<>% set("labels_col", value = cols, k= 5)
dend %<>% set("labels_cex", .8)
dend %<>% set("branches_lwd", 2)

circlize_dendrogram(dend)
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以便使用提取列表簇cutree(dend, k = 5).有没有办法根据cols给定的?提取树形图中簇的颜色?我需要它来使用grid包在图中插入图例.

示例,图例:群集1 - #009000; 集群2 - #FF033E; 集群3 - #CB410B; 集群4 - #3B444B; 集群5 - #007FFF.问题circlize_dendrogram是用于群集的颜色的顺序是不同的.

虽然我可以手动执行此操作,但如果我可以自动执行此操作会很有效.如果我可以提取簇的颜色,这是可能的.

Fel*_*lix 8

好的,这是一个非常hacky的解决方案.我确信有更好的,但这是第一次刺,所以忍受我.

我们的想法是搜索dend相应元素名称的对象(内部列表)(在本例中只是数字)并提取相应的颜色,将其保存在数据框中并将其用于图例.

# First we'll extract the elements and corresponding categories...
categories <- cutree(dend, k = 5)

# ... and save them in a data frame
categories_df <- data.frame(elements = as.numeric(names(categories)),
       categories = categories, 
       color = NA)

# now here's a little function that extracts the color for each element
# from the 'dend' object. It uses the list.search() function from the
# 'rlist' package

library(rlist)

extract_color <- function(element_no, dend_obj) {
  dend.search <- list.search(dend_obj, all(. == element_no))
  color <- attr(dend.search[[1]], "edgePar")$col
  return(color)
}

# I use 'dplyr' to manipulate the data
library(dplyr)

categories_df <- categories_df %>% 
  group_by(elements) %>% 
  mutate(color = extract_color(elements, dend))
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现在,这给了我们以下数据框:

> categories_df
Source: local data frame [40 x 3]
Groups: elements [40]

   elements categories   color
      (dbl)      (int)   (chr)
1         1          1 #CB410B
2         2          1 #CB410B
3         3          1 #CB410B
4         4          1 #CB410B
5         5          1 #CB410B
6         6          2 #009000
7         7          1 #CB410B
8         8          1 #CB410B
9         9          3 #007FFF
10       10          1 #CB410B
..      ...        ...     ...
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我们可以将其总结为仅具有类别颜色的数据框,例如

legend_data <- categories_df %>% 
  group_by(categories) %>% 
  summarise(color = unique(color))

> legend_data
Source: local data frame [5 x 2]

  categories   color
       (int)   (chr)
1          1 #CB410B
2          2 #009000
3          3 #007FFF
4          4 #FF033E
5          5 #3B444B
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现在生成图例很容易:

circlize_dendrogram(dend)
legend(-1.05, 1.05, legend = legend_data$categories, fill = legend_data$color, cex = 0.7)
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哪个给你:

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您可以使用cutree(dend, k = 5)确认类别颜色的数字对应于每个元素的类别.


Al-*_*aad 5

除了Felix的解决方案,我想发表自己的答案:

library(magrittr)
library(grid)
library(gridExtra)
library(dendextend)

cols <- c("#009000", "#FF033E", "#CB410B", "#3B444B", "#007FFF")
dend <- iris[1:40,-5] %>% dist %>% hclust %>% as.dendrogram    

dend <- color_branches(dend, k = 5, col = cols)
dend %<>% set("labels_col", value = cols, k= 5)
dend %<>% set("labels_cex", .8)
dend %<>% set("branches_lwd", 2)

clust <- cutree(dend, k = 5)
colors <- labels_colors(dend)[clust %>% sort %>% names]
clust_labs <- colors %>% unique

circlize_dendrogram(dend)
grid.circle(x = .95, y = .9, r = .02, gp = gpar(fill = clust_labs[1])) 
grid.circle(x = .95, y = .85, r = .02, gp = gpar(fill = clust_labs[2]))
grid.circle(x = .95, y = .8, r = .02, gp = gpar(fill = clust_labs[3]))
grid.circle(x = .95, y = .75, r = .02, gp = gpar(fill = clust_labs[4]))
grid.circle(x = .95, y = .7, r = .02, gp = gpar(fill = clust_labs[5]))

grid.text(x = .95, y = .9, label = expression(bold(1)), gp = gpar(fontsize = 9, col = "white"))
grid.text(x = .95, y = .85, label = expression(bold(2)), gp = gpar(fontsize = 9, col = "white"))
grid.text(x = .95, y = .8, label = expression(bold(3)), gp = gpar(fontsize = 9, col = "white"))
grid.text(x = .95, y = .75, label = expression(bold(4)), gp = gpar(fontsize = 9, col = "white"))
grid.text(x = .95, y = .7, label = expression(bold(5)), gp = gpar(fontsize = 9, col = "white"))
grid.text(x = .91, y = .8, label = "CLUSTERS", rot = 90, gp = gpar(fontsize = 9))
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