R ggplot在堆积条形图中更改一个变量的颜色 [英] R ggplot Changing color of one variable in stacked bar graph

问题描述

我有一个数据框 D ，如下所示：

I have a data frame D that look like this:

Track <- c(0,0,0,-1,1,1)
Length <- c(1,1,2,1,3,1)
Legend <- c("A","B","A","C","B","C")
D <- data.frame(Track,Length,Legend)

D
# Track Length Legend
#     0      1      A
#     0      1      B
#     0      2      A
#    -1      1      C
#     1      3      B
#     1      1      C

我目前使用此代码来绘制它：

I am currently using this code to plot it:

ggplot(Z, aes(x=Track, y=Length, fill=Legend)) +
      geom_bar(stat='identity') + geom_point() + expand_limits(x=0,y=0) + 
      theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(), 
            panel.background = element_blank(), axis.line = element_line(colour = "black")) 

图表看起来像这样：

And the graph looks like this:

然而，我想将白色分配给变量B.这种操作是否可行？此外，为了简单起见，我在本文中使用了虚拟数据。我将要使用的实际数据可能在Legend下有10个或更多不同的变量，因此写出每个十六进制颜色看起来不是一个干净的选项。

However I want to assign the color white to variable B. Is this manipulation possible? Additionally I am using dummy data for simplicity's sake in this post. The actual data that I will be using may have 10 or more different variables under Legend, so writing out each hex color doesn't look like a clean option.

谢谢寻求帮助

Thank you for the help

推荐答案

这是否会产生您所追求的内容？
$ b

Does this produce what you're after?

x <- length(levels(factor(Legend)))
x.colors <- hcl(h=seq(15,375,length=(x+1)),l=65,c=100)[1:x]
x.colors[x] <- "white"

ggplot(D, aes(x=Track, y=Length, fill=Legend)) +
  geom_bar(stat='identity',colour="black") + geom_point() + expand_limits(x=0,y=0) +
  scale_fill_manual(values=x.colors) +
  theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(), 
        panel.background = element_blank(), axis.line = element_line(colour ="black"))

I添加 scale_fill_manual（）给A，B和C分配颜色。我还添加了 color =black code> geom_bar（）。你可以在第二部分和没有第二部分的情况下尝试，看看你的想法。

I added scale_fill_manual() assigning colors to A, B, and C. I also added colour="black" to geom_bar(). You could try it with and without the second part and see what you think.

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