如何使用ggplot2和geom_tile创建自定义色阶？

Question

我想修改颜色渐变，以匹配一组预定义的阈值/切割点和颜色。我怎么能这样做？

截止值：-0.103200,0.007022,0.094090,0.548600 颜色：＆＃34;＃EDF8E9＆＃34;，＆＃34;＃BAE4B3＆＃34;，＆＃34;＃74C476＆＃34;，＆＃34;＃238B45＆＃34;

    #Create sample data

        pp <- function (n,r=4) {
              x <- seq(-r*pi, r*pi, len=n)
              df <- expand.grid(x=x, y=x)
              df$r <- sqrt(df$x^2 + df$y^2)
              df$z <- cos(df$r^2)*exp(-df$r/6)
              df
            }

            pp(20)->data

#create the plot

library(ggplo2)
            p <- ggplot(pp(20), aes(x=x,y=y))
            p + geom_tile(aes(fill=z))

#Generate custom colour ramp

library(RColorBrewer)

cols <- brewer.pal(4, "Greens")

Answer 1

您可以尝试scale_fill_brewer。首先，将您的z值加起来：

df <- pp(20)
df$z_bin <- cut(df$z, breaks = c(-Inf, -0.103200, 0.007022, 0.094090, 0.548600))

情节：

ggplot(data = df, aes(x = x, y = y, fill = z_bin)) +
  geom_tile() +
  scale_fill_brewer(palette = "Greens")

Answer 2

使用cut并将分档与您的颜色相匹配。我的代码假设-0.103200是向量的最小值（用于对bin的数量进行排序）。

trh <- c(-0.103200, 0.007022, 0.094090, 0.548600, Inf)
colors <- c("#EDF8E9", "#BAE4B3", "#74C476", "#238B45")

x <- runif(30, min = -0.103200, max = 1)
xc <- cut(x, breaks = trh)
colors[xc]