Question

我正在尝试将图例文本与使用geom_text的因子变量生成的文本颜色进行颜色匹配。这是一个最小的工作示例：

df <- data.frame(a=rnorm(10),b=1:10,c=letters[1:10],d=c("one","two"))
p1 <-ggplot(data=df,aes(x=b,y=a))
p1 <- p1 + geom_text(aes(label = c, color=d, fontface="bold"))
p1 <- p1 + scale_color_hue(name="colors should match",breaks=c("one", "two"),
                 labels=c("should be pink", "should be blue"))
p1

enter image description here

我确信这是一个简单的解决方法。对先前帖子的任何建议或参考都会有所帮助。我没有发现任何特定的内容。

Answer 1

根据上面的joran评论，您可以直接编辑grobs。这是一个相当丑陋的代码，所以道歉[使用grid命令将有一个更优雅的方式来做到这一点 - 并希望有人会发布。

library(grid)

gglabcol <- function(plot1){

         g <- ggplotGrob(plot1)

         # legend grobs
         g.b <- g[["grobs"]][[which(g$layout$name=="guide-box")]]
         l <- g.b[[1]][[1]][["grobs"]]

         # get grobs for legend symbols (extract colour)
         lg <- l[sapply(l, function(i) grepl("GRID.text", i))]

         # get grobs for legend labels 
         lb <- g.b[[1]][[1]][["grobs"]][grepl("label", g.b[[1]][[1]]$layout$name)]

         # get change colour of labels to colour of symbols
         for(i in seq_along(lg)) {

           g.b[[1]][[1]][["grobs"]][grepl("label", g.b[[1]][[1]]$layout$name)][[i]][["children"]][[1]][["children"]][[1]]$gp$col <- lg[[i]]$gp$col

           }

         # overwrite original legend
         g[["grobs"]][[which(g$layout$name=="guide-box")]] <- g.b

         grid.draw(g)

         invisible(g)
   }

剧情

gglabcol(p1)

enter image description here

Answer 2

有时使用grid编辑功能编辑grob更容易 - 如果可以找到相关grob的名称。在这种情况下，可以找到它们，编辑很简单 - 将标签的颜色从黑色更改为红色或蓝色。

library(ggplot2)
library(grid)

df <- data.frame(a=rnorm(10),b=1:10,c=letters[1:10],d=c("one","two"))
p1 <-ggplot(data=df,aes(x=b,y=a))
p1 <- p1 + geom_text(aes(label = c, color=d, fontface="bold"))
p1 <- p1 + scale_color_hue(name="colors should match",breaks=c("one", "two"),
                 labels=c("should be salmon", "should be sky blue"))
p1

# Get the ggplot grob
g <- ggplotGrob(p1)

# Check out the grobs
grid.ls(grid.force(g))

查看grobs列表。我们要编辑的grob位于列表的底部，在“指南”框中。一组grobs - 名称以＆＃34;标签＆＃34;开头。有两个grob：

标签3-3.4-4-4-4
标签4-3.5-4-5-4

# Get names of 'label' grobs.
names.grobs <- grid.ls(grid.force(g))$name 
labels <- names.grobs[which(grepl("label", names.grobs))]

# Get the colours
# The colours are the same as the colours of the plotted points.
# These are available in the ggplot build data.
gt <- ggplot_build(p1)
colours <- unique(gt$data[[1]][, "colour"])

# Edit the 'label' grobs - change their colours
# Use the `editGrob` function
for(i in seq_along(labels)) {
    g <- editGrob(grid.force(g), gPath(labels[i]), grep = TRUE,  
         gp = gpar(col = colours[i]))
}

# Draw it
grid.newpage()
grid.draw(g)

如果要求键是点而不是字母怎么办？这可能很有用，因为＆＃39; a＆＃39;是图中的符号，它是图例键中的符号。这不是一个简单的编辑，如上所述。我需要一个点grob来代替文本grob。我在视口中绘制了凹凸，但如果我能找到相关视口的名称，则应该直接进行更改。

# Find the names of the relevant viewports
current.vpTree()  # Scroll out to the right to find he relevant 'key' viewports.

viewport [key-4-1-1.5-2-5-2]，viewport [key-3-1-1.4-2-4-2]，

# Well, this is convenient. The names of the viewports are the same 
# as the names of the grobs (see above). 
# Easy enough to get the names from the 'names.grobs' list (see above). 
# Get the names of 'key' viewports(/grobs)
keys <- names.grobs[which(grepl("key-[0-9]-1-1", names.grobs))]

# Insert points grobs into the viewports:
#    Push to the viewport;
#    Insert the point grob;
#    Pop the viewport.
for(i in seq_along(keys)) {
   downViewport(keys[i])
   grid.points(x = .5, y = .5, pch = 16, gp = gpar(col = colours[i]))
   popViewport()
}
popViewport(0)

# I'm not going to worry about removing the text grobs. 
# The point grobs are large enough to hide them. 

plot = grid.grab()
grid.newpage()
grid.draw(plot)

更新

考虑@ user20650的建议来更改图例键（请参阅下面的评论）：

p1 <-ggplot(data=df,aes(x=b,y=a))
p1 <- p1 + geom_text(aes(label = c, color=d, fontface="bold"))
p1 <- p1 + scale_color_hue(name="colors should match",breaks=c("one", "two"),
                 labels=c("should be salmon", "should be sky blue"))

GeomText$draw_key <- function (data, params, size) { 
   pointsGrob(0.5, 0.5, pch = 16, 
   gp = gpar(col = alpha(data$colour, data$alpha), 
   fontsize = data$size * .pt)) }

p1

然后像以前一样继续更改图例文字的颜色。

Answer 3

图中的颜色与图例中的颜色相同，但即使将绘图符号字体设置为粗体（或斜体），图例字体仍保持平滑。我不确定这是否是ggplot2设计或预期行为的疏忽。对于某些颜色，粗体字体看起来比普通字体更饱和，使其看起来像一种不同的颜色。

无论如何，这里的垃圾比乱搞更容易，但这可能会让你得到你想要的东西。将geom_text与普通字体一起使用，但连续两次或三次（或更多次）使用，这样你就会得到过量绘图。这将使符号和图例看起来与粗体字体类似，因为两者都将被过度绘制，并且图例符号将始终与绘图符号相同。

以下是一个例子：

library(ggplot2)
library(gridExtra)

# Original plot (with larger font size)
p1 <- ggplot(data=df) +
  geom_text(aes(x=b, y=a, label=c, color=d), fontface='bold', size=8)
p1 <- p1 + scale_color_hue(name="colors should match",breaks=c("one", "two"),
                              labels=c("should be pink", "should be blue")) +
           ggtitle("Original Plot with Bold Symbols and Plain Legend")

# New version with overplotting. (You don't need to specify 'plain' fontface. 
# I've just included that to emphasize what the code is doing.)
p1.overplot <- ggplot(data=df) +
  geom_text(aes(x=b, y=a, label=c, color=d), fontface='plain', size=8) +
  geom_text(aes(x=b, y=a, label=c, color=d), fontface='plain', size=8) +
  geom_text(aes(x=b, y=a, label=c, color=d), fontface='plain', size=8)
p1.overplot <- p1.overplot + 
  scale_color_hue(name="colors should match",
                  breaks=c("one", "two"),
                  labels=c("should be pink", "should be blue")) +
  ggtitle("Both symbols and legend are overplotted 3 times")

enter image description here

Answer 4

这是一个使用ggtext并且避免直接编辑grob的解决方案。（它确实涉及从图中提取颜色，但是后续步骤更加人性化。）

# Original code, but with a stripped-down call to `scale_color_hue` (since
# we're going to replace it).
library(ggplot2)
df <- data.frame(a=rnorm(10),b=1:10,c=letters[1:10],d=c("one","two"))
p1 <-ggplot(data=df,aes(x=b,y=a))
p1 <- p1 + geom_text(aes(label = c, color=d, fontface="bold"))
p1 <- p1 + scale_color_hue(breaks=c("one", "two"))

# Load the `ggtext` library, which lets us style (parts of) text labels.
library(ggtext)
# Build the plot so we can extract the colors that were actually used.  (If you
# supply colors manually instead, this step isn't necessary.)
g1 = ggplot_build(p1)
# Add a scale with labels that are colored appropriately, using <span> tags.
# Also specify that legend labels should be processed with `element_markdown`.
p1 +
  scale_color_hue(name = "colors should match",
                  breaks = c("one", "two"),
                  labels = paste("<span style='color:",
                                 unname(unlist(unique(g1$data[[1]]["colour"]))),
                                 "'>",
                                 c("should be pink", "should be blue"),
                                 "</span>",
                                 sep = "")) +
  theme(legend.text = element_markdown())

Answer 5

此答案基于Mike H.的question here和此问题的user20650给出的答案。

获得颜色：

  pGrob <- ggplotGrob(p1)
  g.b   <- pGrob[["grobs"]][[which(pGrob$layout$name=="guide-box")]]
  l     <- g.b[[1]][[1]][["grobs"]]
  # get grobs for legend symbols (extract colour)
  lg    <- l[sapply(l, function(i) grepl("GRID.text", i))]
  clr   <- mapply(FUN=function(x){x$gp$col},x=lg)

然后使用以下方法

   gb  <- which(grepl("guide-box", pGrob$layout$name))
   gb2 <- which(grepl("guides", pGrob$grobs[[gb]]$layout$name))
   label_text <- which(grepl("label",pGrob$grobs[[gb]]$grobs[[gb2]]$layout$name))

   pGrob$grobs[[gb]]$grobs[[gb2]]$grobs[label_text] <- 
   mapply(FUN = function(x, y) {x[["children"]][[1]][["children"]][[1]]$gp <- gpar(col =y); return(x)},
          x =   pGrob$grobs[[gb]]$grobs[[gb2]]$grobs[label_text],
          y =  clr, SIMPLIFY = FALSE)

之后，您将获得相同的输出文件

grid.draw(pGrob)

我提供此答案是因为在代码的第二块中，在x$gp函数中分配给mapply的参数应该是gpar对象。 Mike H.的答案中有一个错误，我正在解决。

谢谢。

将geom_text中的图例文本颜色与符号匹配

5 个答案: