使用RColorBrewer将颜色居中调整为0

Question

所以我想用这样的颜色可视化矩阵

library(RColorBrewer)
vec = rbinom(10000,1,0.1)
n = sum(vec)
vec = ifelse(vec == 1, rnorm(n), 0)
mat = matrix(vec,100,100)
image(t(mat)[,nrow(mat):1],
      col=brewer.pal(8,"RdBu"),
      xaxt= "n", yaxt= "n", frame.plot=T,
      useRaster = TRUE
)

这给了我情节

但我希望颜色“以0为中心”。我的意思是我希望零值为白色，正/负值为红色/蓝色（或蓝色/红色无关紧要）。如果可能的话，有什么想法吗？

Answer 1

bluered包中的函数gplots执行此操作。您可以将调色板设为：

library(gplots) # not to be confused with `ggplot2`, which is a very different package
color_palette <- bluered(9) # change the number to adjust how many shades of blue/red you have.  Even numbers will assign white to two bins in the middle.

要强制它们在中间居中，您可以使用heatmap.2功能，也可以使用gplots - 只是不要让它进行任何群集：

heatmap.2(mat,
  Rowv = FALSE,
  Colv = FALSE, 
  dendrogram = 'none',
  trace = 'none',
  col = bluered, # this can take a function
  symbreaks = TRUE, # this is the key  value for symmetric breaks
)

要坚持使用image功能，您需要手动设置中断。以下代码将为您提供：

pos_breaks <- quantile(abs(mat), probs = seq(0, 1, length.out = 5))
centered_breaks <- c(rev(-pos_breaks[-1]), pos_breaks)

Answer 2

除heatmap2之外，您可以使用pheatmap：

library(pheatmap)
pheatmap(mat,
         color = brewer.pal(7,"RdBu"),
         border_color = NA,
         cluster_rows = FALSE,
         cluster_cols = FALSE)

如果您希望使用legend = FALSE，也可以隐藏图例，这会产生与图像调用类似的结果，但白色为0。

Answer 3

这是一个使用ggplot2软件包进行手动缩放（Ref）的解决方案

library(RColorBrewer)
library(ggplot2)
library(reshape2)

set.seed(2020)
vec <- rbinom(10000, 1, 0.1)
n <- sum(vec)
vec <- ifelse(vec == 1, rnorm(n), 0)
mat <- matrix(vec, 100, 100)

# convert to long format
df <- melt(mat)
summary(df)
#>       Var1             Var2            value          
#>  Min.   :  1.00   Min.   :  1.00   Min.   :-2.916137  
#>  1st Qu.: 25.75   1st Qu.: 25.75   1st Qu.: 0.000000  
#>  Median : 50.50   Median : 50.50   Median : 0.000000  
#>  Mean   : 50.50   Mean   : 50.50   Mean   : 0.000772  
#>  3rd Qu.: 75.25   3rd Qu.: 75.25   3rd Qu.: 0.000000  
#>  Max.   :100.00   Max.   :100.00   Max.   : 3.214787

### default
p1 <- ggplot(df, aes(x = Var1, y = Var2, fill = value)) +
  geom_raster() +
  theme_minimal(base_size = 16)

重新缩放

# set the limits of the palette so that zero is in the middle of the range.
limit <- max(abs(df$value)) * c(-1, 1)

p1 + 
  scale_fill_distiller(palette = 'RdBu', limit = limit)

# test with the scico package 
# https://github.com/thomasp85/scico
library(scico)
p1 + 
  scale_fill_scico(palette = "roma", limit = limit) 

# test with the rcartocolor package 
# https://github.com/Nowosad/rcartocolor
library(rcartocolor)
p1 + 
  scale_fill_carto_c(palette = 'Earth', limit = limit) 

^{由reprex package（v0.3.0）于2020-02-07创建}

Answer 4

这是一个没有任何额外包的解决方案。在您的代码中，您没有将vec变量中的值分配给八个颜色区中的任何一个。您需要将vec数组切割成八个分区，然后将每个分区分配给一个颜色，然后绘制：

library(RColorBrewer)
vec = rbinom(10000,1,0.1)
n = sum(vec)
vec = ifelse(vec == 1, rnorm(n), 0)
mat = matrix(vec,100,100)

#cut the original data into 9 groups
cutcol<-cut(vec, 9)
#Create color palette with white as the center color
colorpal<-brewer.pal(8,"RdBu")
colorpal<-c(colorpal[1:4], "#FFFFFF", colorpal[5:8])

#assign the data to the 9 color groups
color<-colorpal[cutcol]
#create the color matrix to match the original data
colormat<-matrix(color,100,100)

#plot with the assigned colors
image(t(mat)[,nrow(mat):1],
      col=colormat,
      xaxt= "n", yaxt= "n", frame.plot=T,
      useRaster = TRUE
)

#check color assignment
#hist(vec)
#hist(as.numeric(cutcol), breaks=8)