我在我的android up中使用了这个，它似乎令人满意，尽管不建议使用RGB空间:

    public double colourDistance(int red1,int green1, int blue1, int red2, int green2, int blue2)
{
      double rmean = ( red1 + red2 )/2;
    int r = red1 - red2;
    int g = green1 - green2;
    int b = blue1 - blue2;
    double weightR = 2 + rmean/256;
    double weightG = 4.0;
    double weightB = 2 + (255-rmean)/256;
    return Math.sqrt(weightR*r*r + weightG*g*g + weightB*b*b);
}

然后我用下面的方法得到相似度的百分比:

double maxColDist = 764.8339663572415;
double d1 = colourDistance(red1,green1,blue1,red2,green2,blue2);
String s1 = (int) Math.round(((maxColDist-d1)/maxColDist)*100) + "% match";

它工作得很好。

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));
}

一个只使用RGB的简单方法是

cR=R1-R2 
cG=G1-G2 
cB=B1-B2 
uR=R1+R2 
distance=cR*cR*(2+uR/256) + cG*cG*4 + cB*cB*(2+(255-uR)/256)

我已经使用这个工具有一段时间了，它可以很好地用于大多数目的。

Android for ColorUtils API RGBToHSL 我有两个int argb颜色(color1, color2)，我想要得到两种颜色之间的距离/差异。这是我所做的;

private float getHue(int color) {
    int R = (color >> 16) & 0xff;
    int G = (color >>  8) & 0xff;
    int B = (color      ) & 0xff;
    float[] colorHue = new float[3];
    ColorUtils.RGBToHSL(R, G, B, colorHue);
    return colorHue[0];
}

然后我使用下面的代码来查找两种颜色之间的距离。

private float getDistance(getHue(color1), getHue(color2)) {
    float avgHue = (hue1 + hue2)/2;
    return Math.abs(hue1 - avgHue);
}

请参阅维基百科关于色差的文章以获得正确的线索。 基本上，你想要在多维颜色空间中计算一个距离度量。

但是RGB并不是“感知上一致的”，所以Vadim建议的欧几里得RGB距离度量将与人类感知的颜色之间的距离不匹配。首先，L*a*b*是一个感知上均匀的颜色空间，delta度量是常用的。但有更精致的色彩空间和更精致的delta公式，更接近人类的感知。

你需要学习更多关于颜色空间和光源的知识来进行转换。但如果想要一个比欧几里得RGB度量更好的快速公式，只需这样做:

假设你的RGB值在sRGB颜色空间中 找到sRGB到L*a*b*的转换公式 将sRGB颜色转换为L*a*b* 计算两个L*a*b*值之间的delta

计算成本不高，只是一些非线性公式和一些乘法和加法。

只是另一个答案，尽管它与Supr的答案相似-只是不同的颜色空间。

问题是:人类感知颜色的差异并不均匀，而RGB颜色空间忽略了这一点。因此，如果你使用RGB颜色空间，只是计算两种颜色之间的欧几里得距离，你可能会得到一个在数学上绝对正确的差异，但与人类告诉你的不一致。

This may not be a problem - the difference is not that large I think, but if you want to solve this "better" you should convert your RGB colors into a color space that was specifically designed to avoid the above problem. There are several ones, improvements from earlier models (since this is based on human perception we need to measure the "correct" values based on experimental data). There's the Lab colorspace which I think would be the best although a bit complicated to convert it to. Simpler would be the CIE XYZ one.

这里有一个网站列出了在不同颜色空间之间转换的公式，所以你可以尝试一下。