Question

我从来没有想过要做到这一点的好方法。我正在尝试创建一个用户函数，用于将RColorBrewer颜色映射到data_frame中的变量级别。我正在绘制igraph中的节点颜色，它需要一个颜色矢量。在得到错误之前，我已经做到了这一点。

library(tidyverse)
library(RColorBrewer)

 my_data <- data_frame(
    x = sample(c('red','white','foo'),25,T)
 ) 

my_data %>% 
  group_by(x) %>% 
  mutate(Blues = brewer.pal(n_distinct(x),'Spectral'))  ## this doesn't work

我想要一个有效的功能。如下所示：

function(df,vary){
    df %>%
       group_by(vary) %>%
       mutate(Blues = brewer.pal(n_distinct(vary),'Spectral')) %>%
       select(Blues)
    }

感谢您的帮助！

Answer 1

您不需要group_by，只需使用mutate并使用brewer.pal(n_distinct(x), 'Spectral')生成所需的颜色，然后使用match将颜色映射到变量：

_{注意：我已使用tidyeval语法捕获NSE样式的变量，因为我们使用的函数为dplyr。但如果您愿意，可以选择其他方式编程。}

var_to_color <- function(df, var) {
    var <- enquo(var)
    df %>% 
        mutate(Colors = brewer.pal(n_distinct(!!var), 'Spectral')[match(!!var, unique(!!var))]) %>%
        pull(Colors)
}

var_to_color(my_data, x)
# [1] "#FC8D59" "#FFFFBF" "#FFFFBF" "#FFFFBF" "#99D594" "#FFFFBF" "#99D594"
# [8] "#99D594" "#FC8D59" "#FFFFBF" "#FC8D59" "#FFFFBF" "#FFFFBF" "#99D594"
#[15] "#FC8D59" "#FC8D59" "#99D594" "#FFFFBF" "#FC8D59" "#FC8D59" "#FC8D59"
#[22] "#FC8D59" "#FFFFBF" "#FC8D59" "#FC8D59"

my_data$x
# [1] "white" "red"   "red"   "red"   "foo"   "red"   "foo"   "foo"   "white"
#[10] "red"   "white" "red"   "red"   "foo"   "white" "white" "foo"   "red"  
#[19] "white" "white" "white" "white" "red"   "white" "white"

不使用dplyr，一个接受一个向量然后将其映射到颜色的函数可能更有意义：

var_to_color <- function(var) brewer.pal(n_distinct(var), 'Spectral')[match(var, unique(var))]

var_to_color(my_data$x)
# [1] "#FC8D59" "#FFFFBF" "#FFFFBF" "#FFFFBF" "#99D594" "#FFFFBF" "#99D594"
# [8] "#99D594" "#FC8D59" "#FFFFBF" "#FC8D59" "#FFFFBF" "#FFFFBF" "#99D594"
#[15] "#FC8D59" "#FC8D59" "#99D594" "#FFFFBF" "#FC8D59" "#FC8D59" "#FC8D59"
#[22] "#FC8D59" "#FFFFBF" "#FC8D59" "#FC8D59"

坚持使用dpylr的mutate / summarize将颜色映射到可变级别

1 个答案: