geom_polygon 使用来自数据的十六进制填充颜色

Question

我有一个结构如下的小标题：

library(tidyverse)
dmd <- tibble(sample = rep(c(1,2,3), each = 2),
              time   = rep(c(1,2),3),
              score  = c(0.69,0.95,1,0.90,1,0.76)) %>%
  mutate(x = map(base::strsplit(paste(sample,
                                      sample + 0.5,
                                      sample,
                                      sample - 0.5,
                                      sep = ","),
                                split = ","),
                 as.numeric)) %>%
  mutate(y = map(base::strsplit(paste(time + 0.5,
                                      time,
                                      time - 0.5,
                                      time,
                                      sep = ","),
                                split = ","),
                 as.numeric)) %>%
  unnest(c(x,y)) %>%
  group_by(sample,time) %>%
  mutate(group_id = cur_group_id()) %>%
  ungroup

# A tibble: 24 × 6
   sample  time score     x     y group_id
    <dbl> <dbl> <dbl> <dbl> <dbl>    <int>
 1      1     1  0.69   1     1.5        1
 2      1     1  0.69   1.5   1          1
 3      1     1  0.69   1     0.5        1
 4      1     1  0.69   0.5   1          1
 5      1     2  0.95   1     2.5        2
 6      1     2  0.95   1.5   2          2
 7      1     2  0.95   1     1.5        2
 8      1     2  0.95   0.5   2          2
 9      2     1  1      2     1.5        3
10      2     1  1      2.5   1          3
# … with 14 more rows

然后我使用 grDevices::rgb 基于 score 生成灰度十六进制值 dcolor。 if() 语句在这个例子中并没有做太多，但重要的是我保留这个元素以实现我的代码的功能化。

minsc <- min(dmd$score)
thrsh <- 0.75
if (minsc < thrsh) {
  floor <- minsc
} else {
  floor <- thrsh
}

dmd <- dmd %>% mutate(temp = (1 - score) / (1 - floor)) %>%
  mutate(dcolor = grDevices::rgb(temp,temp,temp)) %>%
  dplyr::select(-temp)

# A tibble: 24 × 7
   sample  time score     x     y group_id dcolor 
    <dbl> <dbl> <dbl> <dbl> <dbl>    <int> <chr>  
 1      1     1  0.69   1     1.5        1 #FFFFFF
 2      1     1  0.69   1.5   1          1 #FFFFFF
 3      1     1  0.69   1     0.5        1 #FFFFFF
 4      1     1  0.69   0.5   1          1 #FFFFFF
 5      1     2  0.95   1     2.5        2 #292929
 6      1     2  0.95   1.5   2          2 #292929
 7      1     2  0.95   1     1.5        2 #292929
 8      1     2  0.95   0.5   2          2 #292929
 9      2     1  1      2     1.5        3 #000000
10      2     1  1      2.5   1          3 #000000
# … with 14 more rows

如果我用 geom_polygon 绘制 dmd，dcolor 被解释为一个因子。

ggplot(dmd) + 
  geom_polygon(mapping = aes(x = x,
                             y = y,
                             group = group_id,
                             fill = dcolor),
               color = "black") +
  theme_void()

我想强制 ggplot 使用 dcolor 中给出的十六进制值作为每个菱形的填充颜色。使用 scale_fill_manual 让我更接近，但颜色仍然不正确（#000000 应该是黑色，#FFFFFF 应该是白色）。

pal <- dmd %>% dplyr::select(group_id, dcolor) %>%
  unique %>% dplyr::select(dcolor) %>% as_vector

ggplot(dmd) + 
  geom_polygon(mapping = aes(x = x,
                             y = y,
                             group = group_id,
                             fill = dcolor),
               color = "black") +
  scale_fill_manual(values = palette(pal)) +
  theme_void()

Answer 1

您可以使用 scale_fill_identity 达到您想要的结果：

library(tidyverse)

ggplot(dmd) + 
  geom_polygon(mapping = aes(x = x,
                             y = y,
                             group = group_id,
                             fill = dcolor),
               color = "black") +
  theme_void() +
  scale_fill_identity(guide = guide_legend())

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