Replacing a value from one array with a value of the same index of another array?

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]

I am pretty knew to numpy and I don't really understand how the masks work, so if you can add the code it will be nice. Thanks. — Adi Koch
– Adi Koch, Commented Feb 21, 2019 at 18:42
If you don't know how masks work, how did you end up with the posted code? — Divakar
– Divakar, Commented Feb 21, 2019 at 18:45

jyalim · Accepted Answer · 2019-02-21 20:29:24Z

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]

answered Feb 21, 2019 at 20:29

jyalim

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tnknepp · Accepted Answer · 2019-02-21 20:53:46Z

0

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]

edited Feb 21, 2019 at 20:53

answered Feb 21, 2019 at 20:24

tnknepp

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6 Comments

jyalim Over a year ago

Note: your conditionals for idx fail as Non Zero and Non Zero and Zero will return False. You need to do the full negation of the original conditional statement.

tnknepp Over a year ago

@HAL9001 I don't think you are correct. There is no difference between building the index mask using == or != (other than the obvious logical difference). Short example: idx1 = (image[:, :, 0] != ]); idx2 = (image[:, :, 0] == 0); here idx1 is equal to ~idx2

jyalim Over a year ago

You are trying to identify pixels that are not black. By full negation I do not mean that you only can put a not in the front, but you have to be logically consistent (you still are using bitwise ands & instead of the proper negation). The logical difference is important...

tnknepp Over a year ago

@HAL9001 Right, black pixels will have all three levels of the array set to zero (unless I am missing something here). If the OP wants to identify all pixels that are NOT black then you need to identify pixels where all three levels are NOT zero...which is what my idx mask does.

tnknepp Over a year ago

@HAL9001 Ah! I see, thanks for catching this and setting me straight. Answer updated soon!

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