这只是我第一次想到的一个想法(如果愚蠢的话，对不起)。 颜色的三个分量可以假定为点的三维坐标，然后可以计算点之间的距离。

外汇期货

Point1 has R1 G1 B1
Point2 has R2 G2 B2

颜色之间的距离为

d=sqrt((r2-r1)^2+(g2-g1)^2+(b2-b1)^2)

比例是

p=d/sqrt((255)^2+(255)^2+(255)^2)

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。

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));
}

这只是我第一次想到的一个想法(如果愚蠢的话，对不起)。 颜色的三个分量可以假定为点的三维坐标，然后可以计算点之间的距离。

外汇期货

Point1 has R1 G1 B1
Point2 has R2 G2 B2

颜色之间的距离为

d=sqrt((r2-r1)^2+(g2-g1)^2+(b2-b1)^2)

比例是

p=d/sqrt((255)^2+(255)^2+(255)^2)

我尝试了各种方法，如LAB颜色空间，HSV比较，我发现光度在这个目的上非常有效。

这是Python版本

def lum(c):
    def factor(component):
        component = component / 255;
        if (component <= 0.03928):
            component = component / 12.92;
        else:
            component = math.pow(((component + 0.055) / 1.055), 2.4);

        return component
    components = [factor(ci) for ci in c]

    return (components[0] * 0.2126 + components[1] * 0.7152 + components[2] * 0.0722) + 0.05;

def color_distance(c1, c2):

    l1 = lum(c1)
    l2 = lum(c2)
    higher = max(l1, l2)
    lower = min(l1, l2)

    return (higher - lower) / higher


c1 = ImageColor.getrgb('white')
c2 = ImageColor.getrgb('yellow')
print(color_distance(c1, c2))

会给你

0.0687619047619048

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
}