用另一个数组的相同索引的值替换一个数组中的值？

Question

I have two 3D numpy.array objects which represent two images. I have a code that replaces every black pixel in an image to white, but instead of that, I want to replace each pixel that is not black in the first image to the "parallel" pixel color in the other image. How can I do this by changing my code? Thanks!

r1, g1, b1 = 0, 0, 0  # Original value
r2, g2, b2 = 255, 255, 255  # Value that we want to replace it with

red, green, blue = image[:, :, 0], image[:, :, 1], image[:, :, 2]
mask = (red == r1) & (green == g1) & (blue == b1)
image[:, :, :3][mask] = [r2, g2, b2]

Answer 1

You can use numpy.sum to test whether the pixel is black since the rgb sum will be zero for that pixel if and only if the pixel is black. The test on that summation provides a mask that can be used to update your image.

import numpy as np
# Assume image1 and image2 exist in memory as 3-dimensional numpy.arrays
# with shapes (M,N,k) where k is the channel depth (r,g,b -> k=3)
mask = np.sum(image1,axis=-1) > 0
image1[mask] = image2[mask]

Answer 2

You can create a mask to selectively operate on items within your array. It is easier to visualize with 2d arrays, so for example sake:

import numpy as np

a = np.random.randint(0, 10, (5, 4))
b = np.random.randint(0, 10, (5, 4))

Let's see what a and b look like.

In [317]: a
Out[317]: 
array([[6, 0, 4, 0],
       [1, 9, 1, 6],
       [7, 2, 5, 0],
       [8, 3, 5, 0],
       [1, 8, 1, 6]])

In [318]: b
Out[318]: 
array([[1, 3, 2, 1],
       [9, 1, 9, 4],
       [9, 4, 5, 5],
       [6, 0, 6, 4],
       [5, 1, 1, 2]])

Suppose we want to select locations where a==0 and b==3, we build an index (mask). idx = (a==0) & (b==3)

How does idx look?

In [320]: idx
Out[320]: 
array([[False,  True, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False],
       [False, False, False, False]])

Now, if you want to operate on array a where a==0 and b==3 (suppose we want to make the a value equal the b value:

a[idx] = b[idx]

Now what does a look like?

In [322]: a
Out[322]: 
array([[6, 3, 4, 0],
       [1, 9, 1, 6],
       [7, 2, 5, 0],
       [8, 3, 5, 0],
       [1, 8, 1, 6]])

With this knowledge in hand, you can apply the same method to 3d arrays (though harder to visualize).

# identify pixels that are NOT black (i.e. not equal to 0 0 0)
idx = (image1[:, :, 0] == 0) & (image1[:, :, 1] == 0) & (image1[:, :, 2] == 0)

image1[~idx] = image2[~idx]