如何绘制具有引用相同标签的多个点列的图例

Question

假设我有这个例子data.frame:

df = data.frame(x = c(1,1,1,2,2,2,3,3,3), y = runif(9,0,0.9))

我为其分配了如下颜色（请原谅我对这部分的非复杂代码）：

df$color = rep(NA, nrow(df))
color.breaks = seq(0,0.9,0.3)
colors = c("lightgray","gray","darkgray")
df$color[which(df$x == 1)] = colors[findInterval(df$y[which(df$x == 1)], color.breaks)]
colors = c("lightblue","blue","darkblue")
df$color[which(df$x == 2)] = colors[findInterval(df$y[which(df$x == 2)], color.breaks)]
colors = c("lightgreen","green","darkgreen")
df$color[which(df$x == 3)] = colors[findInterval(df$y[which(df$x == 3)], color.breaks)]

然后我绘制：

plot(x = df$x, y = df$y, col = as.character(df$col), pch = 16)

现在我想添加一个三行四列的图例，其中第一列是color.breaks，接下来的三列是各自的颜色。

类似这样的：

知道我是如何使用legend 函数完成这项工作的吗？

Answer 1

这是一个基本解决方案，需要进行一些更精细的调整...

plot(1, type = 'n')
l <- legend(1, 1, 
       legend = rep(NA, 9), 
       ncol = 3, pch = 16, bty="n",
       col = c("lightgray","gray","darkgray", "lightblue","blue","darkblue", 
               "lightgreen","green","darkgreen"), trace = TRUE)
text(l$text$x-0.05, l$text$y, c('A', 'B', 'C', rep(NA, 6)), pos = 2)

关键步骤是：

• 使用ncol指定列数
• 省略标签（将它们设置为 NA），然后添加它们（将它们绘制到左侧）
• 使用trace 获取标签的位置。