如何使用 python 计算 3d-RGB 数组的均值和标准差？

Question

我现在需要获取 10 张图片 (400px,400px) 的 RGB 值的均值和标准差。我的意思是mean_of_Red(x,y)、std_of_Red(x,y)等等……

使用 cv2.imread，我得到了 10 (400,400,3) 个形状数组。所以，我首先尝试使用 numpy.dstack 来堆叠每个 RGB 值以获得 (400,400,3,10) 形状数组。但是，它不起作用，因为数组的形状会随着迭代而改变。

所以，我终于在下面写了代码

def average_and_std_of_RGB(pic_database,start,num_past_frame):
    background = pic_database[0] #initialize background
    past_frame = pic_database[1:num_past_frame+1]
    width,height,depth = background.shape
    sumB = np.zeros(width*height)
    sumG = np.zeros(width*height)
    sumR = np.zeros(width*height)
    sumB_sq = np.zeros(width*height)
    sumG_sq = np.zeros(width*height)
    sumR_sq = np.zeros(width*height)
    for item in (past_frame):
        re_item = np.reshape(item,3*width*height) #reshape (400,400,3) to (480000,)
        itemB =[re_item[i] for i in range(3*width*height) if i%3==0] #Those divisible by 3 is Blue
        itemG =[re_item[i] for i in range(3*width*height) if i%1==0] #Those divisible by 1 is Green
        itemR =[re_item[i] for i in range(3*width*height) if i%2==0] #Those divisible by 2 is Red
        itemB_sq = [item**2 for item in itemB]
        itemG_sq = [item**2 for item in itemG]
        itemR_sq = [item**2 for item in itemR]
        sumB = [x+y for (x,y) in zip(sumB,itemB)]
        sumG = [x+y for (x,y) in zip(sumG,itemG)]
        sumR = [x+y for (x,y) in zip(sumR,itemR)]
        sumB_sq = [x+y for (x,y) in zip(sumB_sq,itemB_sq)]
        sumG_sq = [x+y for (x,y) in zip(sumG_sq,itemG_sq)]
        sumR_sq = [x+y for (x,y) in zip(sumR_sq,itemR_sq)]
    aveB = [x/num_past_frame for x in sumB]
    aveG = [x/num_past_frame for x in sumG]
    aveR = [x/num_past_frame for x in sumR]
    aveB_sq = [x/num_past_frame for x in sumB]
    aveG_sq = [x/num_past_frame for x in sumR]
    aveR_sq = [x/num_past_frame for x in sumR]
    stdB = [np.sqrt(abs(x-y**2)) for (x,y) in zip(aveB_sq,aveB)]
    stdG = [np.sqrt(abs(x-y**2)) for (x,y) in zip(aveG_sq,aveG)]
    stdR = [np.sqrt(abs(x-y**2)) for (x,y) in zip(aveR_sq,aveR)]
    return sumB,sumG,sumR,stdB,stdG,stdR

它确实有效，但看起来很残酷并且需要一些时间。 我想知道是否有更有效的方法来获得相同的结果。 请帮我一把，谢谢。

Answer 1

>>> img = cv2.imread("/home/auss/Pictures/test.png")
>>> means, stddevs  = cv2.meanStdDev(img)
>>> means
array([[ 95.84747396],
       [ 91.55859375],
       [ 96.96260851]])
>>> stddevs
array([[ 48.26534676],
       [ 48.24555701],
       [ 55.92261374]])