给定一个系统(例如一个网站),允许用户自定义某些部分的背景色,但不允许自定义字体颜色(以保持选项的数量最小化),是否有一种方法可以通过编程来确定“浅色”或“深色”字体颜色是必要的?

我相信有一些算法,但我对颜色、光度等了解不够,无法自己找出答案。


当前回答

iOS Swift 3.0 (UIColor扩展):

func isLight() -> Bool
{
    if let components = self.cgColor.components, let firstComponentValue = components[0], let secondComponentValue = components[1], let thirdComponentValue = components[2] {
        let firstComponent = (firstComponentValue * 299)
        let secondComponent = (secondComponentValue * 587)
        let thirdComponent = (thirdComponentValue * 114)
        let brightness = (firstComponent + secondComponent + thirdComponent) / 1000

        if brightness < 0.5
        {
            return false
        }else{
            return true
        }
    }  

    print("Unable to grab components and determine brightness")
    return nil
}

其他回答

objective-c的实现

+ (UIColor*) getContrastColor:(UIColor*) color {
    CGFloat red, green, blue, alpha;
    [color getRed:&red green:&green blue:&blue alpha:&alpha];
    double a = ( 0.299 * red + 0.587 * green + 0.114 * blue);
    return (a > 0.5) ? [[UIColor alloc]initWithRed:0 green:0 blue:0 alpha:1] : [[UIColor alloc]initWithRed:255 green:255 blue:255 alpha:1];
}

作为Kotlin / Android扩展:

fun Int.getContrastColor(): Int {
    // Counting the perceptive luminance - human eye favors green color...
    val a = 1 - (0.299 * Color.red(this) + 0.587 * Color.green(this) + 0.114 * Color.blue(this)) / 255
    return if (a < 0.5) Color.BLACK else Color.WHITE
}

颤振实现

Color contrastColor(Color color) {
  if (color == Colors.transparent || color.alpha < 50) {
    return Colors.black;
  }
  double luminance = (0.299 * color.red + 0.587 * color.green + 0.114 * color.blue) / 255;
  return luminance > 0.5 ? Colors.black : Colors.white;
}

基于R版本的@Gacek的答案来获得亮度(你可以很容易地应用你自己的阈值)

# vectorized
luminance = function(col) c(c(.299, .587, .114) %*% col2rgb(col)/255)

用法:

luminance(c('black', 'white', '#236FAB', 'darkred', '#01F11F'))
# [1] 0.0000000 1.0000000 0.3730039 0.1629843 0.5698039

谢谢@Gacek。下面是一个Android版本:

@ColorInt
public static int getContrastColor(@ColorInt int color) {
    // Counting the perceptive luminance - human eye favors green color...
    double a = 1 - (0.299 * Color.red(color) + 0.587 * Color.green(color) + 0.114 * Color.blue(color)) / 255;

    int d;
    if (a < 0.5) {
        d = 0; // bright colors - black font
    } else {
        d = 255; // dark colors - white font
    }

    return Color.rgb(d, d, d);
}

还有一个改进的(更短的)版本:

@ColorInt
public static int getContrastColor(@ColorInt int color) {
    // Counting the perceptive luminance - human eye favors green color...
    double a = 1 - (0.299 * Color.red(color) + 0.587 * Color.green(color) + 0.114 * Color.blue(color)) / 255;
    return a < 0.5 ? Color.BLACK : Color.WHITE;
}