R中是否有浅色或暗色功能？ [英] Is there a is light or is dark color function in R?

问题描述

我有一张ggplot2情节，并且百分比标签为白色，但有时候条形图的颜色太浅，白色标签变得难以辨认。是否有一个函数给出一个颜色值将返回，无论它是否是是昏暗的还是轻的？然后我可以将标签颜色设置为白色或黑色......

根据@MrFlick提供的（第二个）链接中的强度等级，黑色与白色之间的关系。 博客引用了W3C出版物：用于计算颜色感知亮度的标准公式，该算法使用RGB编码颜色的算法：

（（红色值X 299）+（绿色值X 587）+（蓝色值X 114））/ 1000
<$ p
$ b> code>

col2rgb 函数提供了一个3行矩阵，我乘以该网页提供的因素。我使用了一个红色作为背景颜色的例子，所选文本将是白色。

），白）[1+（sum（col2rgb（red）* c（299,587,114））/ 1000 123）]
[1]white

 
 
 
 （b）（b）（b）（b）（b）（b）（c）（b） 
  

I have a ggplot2 plot and have percent labels in white but some times the color of the bar chart is too light and the white label becomes illegible. Is there a function that given a color value will return whether it is e.g. isDark or isLight? and then I can set the label color to white or black accordingly ...

解决方案

Here's a strategy to implement picking a text color of black vs white based on the intensity scale in the (second) link provided by @MrFlick.

The blog cited a W3C publication: a standard formula for calculating the perceived brightness of a color that used an algorithm for RGB encoded colors:

 ((Red value X 299) + (Green value X 587) + (Blue value X 114)) / 1000

The col2rgb function delivers a 3-row matrix which I multiply by the factors offered in that webpage. I used an example of "red" as a background color and the chosen text would then be "white"

 c( "black", "white")[  1+(sum( col2rgb("red") *c(299, 587,114))/1000 < 123) ]
[1] "white"

Implemented as a function:

isDark <- function(colr) { (sum( col2rgb(colr) * c(299, 587,114))/1000 < 123) }
isDark("red")
[1] TRUE

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