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

丑陋的Python,如果你不想写它:)

'''
Input a string without hash sign of RGB hex digits to compute
complementary contrasting color such as for fonts
'''
def contrasting_text_color(hex_str):
    (r, g, b) = (hex_str[:2], hex_str[2:4], hex_str[4:])
    return '000' if 1 - (int(r, 16) * 0.299 + int(g, 16) * 0.587 + int(b, 16) * 0.114) / 255 < 0.5 else 'fff'

其他回答

请注意,在谷歌闭包库中有一个算法,该算法引用了w3c推荐:http://www.w3.org/TR/AERT#color-contrast。但是,在这个API中,您提供了一个建议颜色列表作为起点。

/**
 * Find the "best" (highest-contrast) of the suggested colors for the prime
 * color. Uses W3C formula for judging readability and visual accessibility:
 * http://www.w3.org/TR/AERT#color-contrast
 * @param {goog.color.Rgb} prime Color represented as a rgb array.
 * @param {Array<goog.color.Rgb>} suggestions Array of colors,
 *     each representing a rgb array.
 * @return {!goog.color.Rgb} Highest-contrast color represented by an array.
 */
goog.color.highContrast = function(prime, suggestions) {
  var suggestionsWithDiff = [];
  for (var i = 0; i < suggestions.length; i++) {
    suggestionsWithDiff.push({
      color: suggestions[i],
      diff: goog.color.yiqBrightnessDiff_(suggestions[i], prime) +
          goog.color.colorDiff_(suggestions[i], prime)
    });
  }
  suggestionsWithDiff.sort(function(a, b) { return b.diff - a.diff; });
  return suggestionsWithDiff[0].color;
};


/**
 * Calculate brightness of a color according to YIQ formula (brightness is Y).
 * More info on YIQ here: http://en.wikipedia.org/wiki/YIQ. Helper method for
 * goog.color.highContrast()
 * @param {goog.color.Rgb} rgb Color represented by a rgb array.
 * @return {number} brightness (Y).
 * @private
 */
goog.color.yiqBrightness_ = function(rgb) {
  return Math.round((rgb[0] * 299 + rgb[1] * 587 + rgb[2] * 114) / 1000);
};


/**
 * Calculate difference in brightness of two colors. Helper method for
 * goog.color.highContrast()
 * @param {goog.color.Rgb} rgb1 Color represented by a rgb array.
 * @param {goog.color.Rgb} rgb2 Color represented by a rgb array.
 * @return {number} Brightness difference.
 * @private
 */
goog.color.yiqBrightnessDiff_ = function(rgb1, rgb2) {
  return Math.abs(
      goog.color.yiqBrightness_(rgb1) - goog.color.yiqBrightness_(rgb2));
};


/**
 * Calculate color difference between two colors. Helper method for
 * goog.color.highContrast()
 * @param {goog.color.Rgb} rgb1 Color represented by a rgb array.
 * @param {goog.color.Rgb} rgb2 Color represented by a rgb array.
 * @return {number} Color difference.
 * @private
 */
goog.color.colorDiff_ = function(rgb1, rgb2) {
  return Math.abs(rgb1[0] - rgb2[0]) + Math.abs(rgb1[1] - rgb2[1]) +
      Math.abs(rgb1[2] - rgb2[2]);
};

谢谢你写这篇文章。

对于可能感兴趣的人,这里有一个Delphi中的函数示例:

function GetContrastColor(ABGColor: TColor): TColor;
var
  ADouble: Double;
  R, G, B: Byte;
begin
  if ABGColor <= 0 then
  begin
    Result := clWhite;
    Exit; // *** EXIT RIGHT HERE ***
  end;

  if ABGColor = clWhite then
  begin
    Result := clBlack;
    Exit; // *** EXIT RIGHT HERE ***
  end;

  // Get RGB from Color
  R := GetRValue(ABGColor);
  G := GetGValue(ABGColor);
  B := GetBValue(ABGColor);

  // Counting the perceptive luminance - human eye favors green color...
  ADouble := 1 - (0.299 * R + 0.587 * G + 0.114 * B) / 255;

  if (ADouble < 0.5) then
    Result := clBlack  // bright colors - black font
  else
    Result := clWhite;  // dark colors - white font
end;

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
}

基于Gacek的答案,但直接返回颜色常数(其他修改见下文):

public Color ContrastColor(Color iColor)
{
  // Calculate the perceptive luminance (aka luma) - human eye favors green color... 
  double luma = ((0.299 * iColor.R) + (0.587 * iColor.G) + (0.114 * iColor.B)) / 255;

  // Return black for bright colors, white for dark colors
  return luma > 0.5 ? Color.Black : Color.White;
}

注意:我去掉了亮度值的反转,使明亮的颜色有一个更高的值,这对我来说似乎更自然,也是“默认”的计算方法。 (编辑:这在原来的答案中也被采用了)

我使用了与Gacek相同的常数,因为它们非常适合我。


你也可以使用下面的签名来实现这个扩展方法:

public static Color ContrastColor(this Color iColor)

然后,您可以轻松地将其称为via foregroundColor = backgroundColor.ContrastColor()。

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