为ggplot2生成明亮和黑暗的颜色对 [英] Generate pairs of bright and dark colours for ggplot2

问题描述

  brewer.pal（n = 8，name =Paired）
  

最多可以创建8个颜色对，但只有少数这些颜色适合打印。
是否有更灵活的功能可以产生黑色吊坠？
颜色较深的应该是

• 在黑暗中看起来颜色相同


• 可打印


• 容易区分明亮的商品
  有没有一种色彩工具可以解决问题那？
  
  

   >黑暗（红色）
   [1]FF5555
    

  我想出了这个函数，但它只能处理单一颜色，而不能处理矢量。

    setbrightness < -  function（rgbcolour，brightness）这个函数是一个很好的解决方案， {
   ##用法：setbrightness（col2rgb（red，0.2）
   thiscolour < -  rgb2hsv（rgbcolour）
   return（hsv（h = thiscolour [1]，s = thiscolour [ 2]，v = brightness））
  } 
    

  
  

  解决方案

  这是一个可以使用的矢量化函数，其思想是将RGB转换为HSV，其中V对应亮度，然后返回RGB：
  
  

    library（grDevices）
  ＃r，g，b的每个元素必须在[0,255] 
  ＃conv [[3]]的每个元素都必须在[ 0,1]，1是最高亮度
  亮度< - 函数（r，g，b，因子）{
   conv < -  as.list（as.data.frame（t（rgb2hsv （1，conv，[3]] *因子）
   do.call（hsv，conv）
  } 
   
  ＃将亮度降低20％
  亮度（55,100,150,0.8）
  ＃[1]＃2C5078
   
  ＃增加20％
  的亮度（55,100,150,1.2）
  ＃[1]＃4278B4
   
  亮度（55,100,150 ，c（0.8,1.2））
  ＃[1]＃2C5078＃4278B4
   
  x<  -  rep（LETTERS [1：2]，5）
   qplot（x = x，geom =bar，fill = x）+ 
   scale_fill_manual（values = brightness（55,100,150，c（0.5,1.5）））
    

  $ b

  参见？rgb2hsv ，？hsv 和 wiki 了解更多详情。
  
  

  编辑：根据您的编辑，似乎您更喜欢使用颜色的名称和亮度的直接值。在这种情况下，矢量化函数看起来非常相似：
  
  

   ＃用法：brightness（red，c（0.1，0.3， （rgbcol，v）{
   conv<  -  as.list（as.data.frame（t（rgb2hsv（col2rgb（rgbcol））））） 
   conv [[3]] < -  v 
   do.call（hsv，conv）
  } 
   
   set.seed（19）
   df <  -  data.frame（a = rlnorm（100），b = 1:10，c = rep（LETTERS [1:10]，each = 10））
   ggplot（df，aes（x = b， y = a，fill = c））+ geom_area（）+ theme_bw（）+ 
   scale_fill_manual（values = brightness（red，seq（0.1，0.7，length = 10）））
    

  
  

  brewer.pal(n=8,name="Paired")
  

  Can create up to eight colour pairs, but only few of these colours are good for printing. Is there a more flexible function that will generate a dark pendant? The darker one should

  • look like the same colour in dark
  • well printable
  • easy to distinguish from the bright one

  Is there a colourbrewer tool that can already solve that?

  > dark("red")
  [1] "FF5555"
  

  I figured out this function, but it can only handle single colours, not a vector. This function would be a nice solution, if it could be "vectorized".

  setbrightness <- function(rgbcolour, brightness) {
    ## usage: setbrightness(col2rgb("red", 0.2)
    thiscolour <- rgb2hsv(rgbcolour)
    return(hsv(h=thiscolour[1], s=thiscolour[2], v=brightness))
  }
  

  解决方案

  Here is a vectorized function that you can use. The idea is to convert RGB to HSV, where V corresponds to brightness and then go back to RGB:

  library(grDevices)
  # Every element of r, g, b must be in [0, 255]
  # Every element of conv[[3]] must be in [0, 1], 1 is highest brightness
  brightness <- function(r, g, b, factor) {
    conv <- as.list(as.data.frame(t(rgb2hsv(r, g, b))))
    conv[[3]] <- pmin(1, conv[[3]] * factor)
    do.call(hsv, conv)
  }
  
  # Reducing brightness by 20%
  brightness(55, 100, 150, 0.8)
  #[1] "#2C5078"
  
  # Increasing by 20%
  brightness(55, 100, 150, 1.2)
  #[1] "#4278B4"
  
  brightness(55, 100, 150, c(0.8, 1.2))
  #[1] "#2C5078" "#4278B4"
  
  x <- rep(LETTERS[1:2], 5)
  qplot(x = x, geom = "bar", fill = x) + 
    scale_fill_manual(values = brightness(55, 100, 150, c(0.5, 1.5)))
  

  See ?rgb2hsv, ?hsv and wiki for more details.

  Edit: according to your edit it seems that you prefer using names of colours and direct values of brightness. In that case a vectorized function would look very similarly:

  # Usage: brightness("red", c(0.1, 0.3, 0.5, 1))
  brightness <- function(rgbcol, v) {
    conv <- as.list(as.data.frame(t(rgb2hsv(col2rgb(rgbcol)))))
    conv[[3]] <- v
    do.call(hsv, conv)
  }
  
  set.seed(19)
  df <- data.frame(a = rlnorm(100), b = 1:10, c = rep(LETTERS[1:10], each = 10))
  ggplot(df, aes(x = b, y = a, fill = c)) + geom_area() + theme_bw() +
    scale_fill_manual(values = brightness("red", seq(0.1, 0.7, length = 10)))
  

  这篇关于为ggplot2生成明亮和黑暗的颜色对的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！