颜色值有不止一个维度，所以没有内在的方法来比较两种颜色。您必须为您的用例确定颜色的含义，从而确定如何最好地比较它们。

很可能你想比较颜色的色相、饱和度和/或明度属性，而不是红/绿/蓝组件。如果你不知道如何比较它们，那就拿几对样品颜色，在心里比较一下，然后试着向自己证明/解释为什么它们相似/不同。

一旦你知道了你想要比较的颜色的哪些属性/成分，那么你就需要弄清楚如何从颜色中提取这些信息。

最有可能的是，你只需要将颜色从常见的RedGreenBlue表示转换为HueSaturationLightness，然后计算类似的东西

avghue = (color1.hue + color2.hue)/2
distance = abs(color1.hue-avghue)

这个例子会给你一个简单的标量值，指示颜色的渐变/色相彼此之间的距离。

参见维基百科上的HSL和HSV。

通过人类感知来比较两种颜色的最佳方法之一是CIE76。这个差值叫做e。当小于1时，人眼无法识别差异。

有一个很棒的颜色工具类ColorUtils(代码如下)，它包括CIE76比较方法。作者是苏黎世大学的丹尼尔·斯特雷贝尔。

从ColorUtils.class我使用的方法:

static double colorDifference(int r1, int g1, int b1, int r2, int g2, int b2)

r1,g1,b1 -第一种颜色的RGB值

r2,g2,b2 -您想比较的第二个颜色的RGB值

如果你使用Android，你可以得到这样的值:

r1 = Color.red(像素);

g1 = Color.green(像素);

b1 = Color.blue(像素);

ColorUtils.class作者:Daniel Strebel，苏黎世大学:

import android.graphics.Color;

public class ColorUtil {
public static int argb(int R, int G, int B) {
    return argb(Byte.MAX_VALUE, R, G, B);
}

public static int argb(int A, int R, int G, int B) {
    byte[] colorByteArr = {(byte) A, (byte) R, (byte) G, (byte) B};
    return byteArrToInt(colorByteArr);
}

public static int[] rgb(int argb) {
    return new int[]{(argb >> 16) & 0xFF, (argb >> 8) & 0xFF, argb & 0xFF};
}

public static int byteArrToInt(byte[] colorByteArr) {
    return (colorByteArr[0] << 24) + ((colorByteArr[1] & 0xFF) << 16)
            + ((colorByteArr[2] & 0xFF) << 8) + (colorByteArr[3] & 0xFF);
}

public static int[] rgb2lab(int R, int G, int B) {
    //http://www.brucelindbloom.com

    float r, g, b, X, Y, Z, fx, fy, fz, xr, yr, zr;
    float Ls, as, bs;
    float eps = 216.f / 24389.f;
    float k = 24389.f / 27.f;

    float Xr = 0.964221f;  // reference white D50
    float Yr = 1.0f;
    float Zr = 0.825211f;

    // RGB to XYZ
    r = R / 255.f; //R 0..1
    g = G / 255.f; //G 0..1
    b = B / 255.f; //B 0..1

    // assuming sRGB (D65)
    if (r <= 0.04045)
        r = r / 12;
    else
        r = (float) Math.pow((r + 0.055) / 1.055, 2.4);

    if (g <= 0.04045)
        g = g / 12;
    else
        g = (float) Math.pow((g + 0.055) / 1.055, 2.4);

    if (b <= 0.04045)
        b = b / 12;
    else
        b = (float) Math.pow((b + 0.055) / 1.055, 2.4);


    X = 0.436052025f * r + 0.385081593f * g + 0.143087414f * b;
    Y = 0.222491598f * r + 0.71688606f * g + 0.060621486f * b;
    Z = 0.013929122f * r + 0.097097002f * g + 0.71418547f * b;

    // XYZ to Lab
    xr = X / Xr;
    yr = Y / Yr;
    zr = Z / Zr;

    if (xr > eps)
        fx = (float) Math.pow(xr, 1 / 3.);
    else
        fx = (float) ((k * xr + 16.) / 116.);

    if (yr > eps)
        fy = (float) Math.pow(yr, 1 / 3.);
    else
        fy = (float) ((k * yr + 16.) / 116.);

    if (zr > eps)
        fz = (float) Math.pow(zr, 1 / 3.);
    else
        fz = (float) ((k * zr + 16.) / 116);

    Ls = (116 * fy) - 16;
    as = 500 * (fx - fy);
    bs = 200 * (fy - fz);

    int[] lab = new int[3];
    lab[0] = (int) (2.55 * Ls + .5);
    lab[1] = (int) (as + .5);
    lab[2] = (int) (bs + .5);
    return lab;
}

/**
 * Computes the difference between two RGB colors by converting them to the L*a*b scale and
 * comparing them using the CIE76 algorithm { http://en.wikipedia.org/wiki/Color_difference#CIE76}
 */
public static double getColorDifference(int a, int b) {
    int r1, g1, b1, r2, g2, b2;
    r1 = Color.red(a);
    g1 = Color.green(a);
    b1 = Color.blue(a);
    r2 = Color.red(b);
    g2 = Color.green(b);
    b2 = Color.blue(b);
    int[] lab1 = rgb2lab(r1, g1, b1);
    int[] lab2 = rgb2lab(r2, g2, b2);
    return Math.sqrt(Math.pow(lab2[0] - lab1[0], 2) + Math.pow(lab2[1] - lab1[1], 2) + Math.pow(lab2[2] - lab1[2], 2));
}
}

我猜你最后想分析一幅完整的图像，对吧?所以你可以检查单位颜色矩阵的最小/最大差值。

大多数处理图形的数学操作都使用矩阵，因为使用矩阵的可能算法通常比经典的逐点距离和比较计算更快。(例如，对于使用DirectX, OpenGL，…的操作)

所以我认为你应该从这里开始:

http://en.wikipedia.org/wiki/Identity_matrix

http://en.wikipedia.org/wiki/Matrix_difference_equation

…正如Beska在上面评论的那样:

这可能不会带来最好的“可见”差异……

这也意味着，如果你在处理图像，你的算法取决于你对“相似”的定义。

Actually I walked the same path a couple of months ago. There is no perfect answer to the question (that was asked here a couple of times) but there is one, more sophisticated than the sqrt(r-r) etc. answer and more easy to implement directly with RGB without moving to all kinds of alternate color spaces. I found this formula here which is a low cost approximation of the quite complicated real formula (by the CIE which is the W3C of colors, since this is a not finished quest, you can find older and simpler color difference equations there). Good Luck.

编辑:为了子孙后代，这里是相关的C代码:

typedef struct {
     unsigned char r, g, b;
} RGB;

double ColourDistance(RGB e1, RGB e2)
{
    long rmean = ( (long)e1.r + (long)e2.r ) / 2;
    long r = (long)e1.r - (long)e2.r;
    long g = (long)e1.g - (long)e2.g;
    long b = (long)e1.b - (long)e2.b;
    return sqrt((((512+rmean)*r*r)>>8) + 4*g*g + (((767-rmean)*b*b)>>8));
}

Kotlin版本与你想匹配的百分比有多少。

方法调用，参数为percent

isMatchingColor(intColor1, intColor2, 95) // should match color if 95% similar

方法体

private fun isMatchingColor(intColor1: Int, intColor2: Int, percent: Int = 90): Boolean {
    val threadSold = 255 - (255 / 100f * percent)

    val diffAlpha = abs(Color.alpha(intColor1) - Color.alpha(intColor2))
    val diffRed = abs(Color.red(intColor1) - Color.red(intColor2))
    val diffGreen = abs(Color.green(intColor1) - Color.green(intColor2))
    val diffBlue = abs(Color.blue(intColor1) - Color.blue(intColor2))

    if (diffAlpha > threadSold) {
        return false
    }

    if (diffRed > threadSold) {
        return false
    }

    if (diffGreen > threadSold) {
        return false
    }

    if (diffBlue > threadSold) {
        return false
    }

    return true
}

快速回答

我找到这个帖子是因为我需要这个问题的Swift版本。由于还没有人给出答案，我的答案是:

extension UIColor {

    var rgba: (red: CGFloat, green: CGFloat, blue: CGFloat, alpha: CGFloat) {
        var red: CGFloat = 0
        var green: CGFloat = 0
        var blue: CGFloat = 0
        var alpha: CGFloat = 0
        getRed(&red, green: &green, blue: &blue, alpha: &alpha)

        return (red, green, blue, alpha)
    }

    func isSimilar(to colorB: UIColor) -> Bool {
        let rgbA = self.rgba
        let rgbB = colorB.rgba

        let diffRed = abs(CGFloat(rgbA.red) - CGFloat(rgbB.red))
        let diffGreen = abs(rgbA.green - rgbB.green)
        let diffBlue = abs(rgbA.blue - rgbB.blue)

        let pctRed = diffRed
        let pctGreen = diffGreen
        let pctBlue = diffBlue

        let pct = (pctRed + pctGreen + pctBlue) / 3 * 100

        return pct < 10 ? true : false
    }
}

用法:

let black: UIColor = UIColor.black
let white: UIColor = UIColor.white

let similar: Bool = black.isSimilar(to: white)

我设置小于10%的差异返回相似的颜色，但你可以自定义这自己。