计算方差的算法不正常工作

Question

我写了一个函数，它应该返回给定簇中一组像素的R，G，B分量的方差

def RGB_variance(img, density):
    R_total = 0
    G_total = 0
    B_total = 0
    N = len(density)

    for pixel in density:
        x = pixel[0]
        y = pixel[1]
        R_total += img[x][y][0]
        G_total += img[x][y][1]
        B_total += img[x][y][2]

    R_mean, G_mean, B_mean = R_total/N, G_total/N, B_total/N

    R_sum = 0
    G_sum = 0
    B_sum = 0

    means = [R_mean, G_mean, B_mean]

    sums = [R_sum, G_sum, B_sum]

    for pixel in density:
        x = pixel[0]
        y = pixel[1]
        for i in range(0,3):
            sums[i] += (img[x][y][i] - means[i])**2

    variances = {
    "R_variance":sums[0]/N, 
    "G_variance":sums[1]/N, 
    "B_variance":sums[2]/N,
    }

    return variances

这些照片显示了我正在处理的区域和相应的像素：

https://i.ibb.co/djd1b97/1.png

https://i.ibb.co/NxkX703/1.png

我得到的结果是，它并不完全正确：

{'R_variance': 71.56184661531957, 'G_variance': 91.47570122951443, 'B_variance': 521.035179427894}

当B分量的值范围为范围（0255）时，它的方差怎么可能是521

Answer 1

嗯，方差可以大于值的范围。下面是一个例子：

data = [10, 10, 10, 10, 0, 0, 0, 0]
range of values = [0, 10]
mean = 5
var = ((10 - 5)^2 * 4 + (0 - 5)^2 * 4) / 8 = 200 / 8 = 25

方差高于数据可以采用的最大值。也许你想要的是standard deviation，它是方差的平方根