Difference in numpy array masking

Question

I have three numpy arrays, one the source array, one destination array and one mask array. I want to replace the values in the destination with the same values from the source only at the places where the mask is equal to one.

My naiive try was:

import numpy as np
destination = np.arange(9).reshape((3,3))
source = np.ones((3,3))
mask = np.zeros((3,3)).astype(np.uint8)
mask[1,1]=1

destination[mask] = source[mask]

which leads me to destination being

[[1, 1, 1],
 [1, 1, 1],
 [6, 7, 8]]

whereas I expect it to be

[[0, 1, 2],
 [3, 1, 5],
 [6, 7, 8]].

I do get the correct result, when I do

destination[mask==1] = source[mask==1].

My question is: Why are these two commands not identical, or what does the first even do?

Why are using np.uint8 for a mask? Shouldn't that be a boolean array? — Divakar
– Divakar, Commented May 27, 2019 at 11:31
Fair, if I cast it to boolean, it does what it should. But my question remains, what does my first version actually do? — Dschoni
– Dschoni, Commented May 27, 2019 at 11:40
@Divakar I came from this side, and understanding what's going on is exactly the problem I have, hence asking this question. — Dschoni
– Dschoni, Commented May 27, 2019 at 11:53

Dschoni · Accepted Answer · 2019-05-27 13:52:15Z

1

First you must check inside the matrices and which matrix gives you what you want.

mask

Output

[[0, 0, 0],
 [0, 1, 0],
 [0, 0, 0]]

but destination[mask == 1] gives you a boolean matrix

mask == 1

Output

[[False, False, False],
 [False,  True, False],
 [False, False, False]]

whereas:

destination[mask]

Output

[[[0, 1, 2],
  [0, 1, 2],
  [0, 1, 2]]

[[0, 1, 2],
 [3, 4, 5],
 [0, 1, 2]],

[[0, 1, 2],
 [0, 1, 2],
 [0, 1, 2]]]

but using destination[mask == 1] gives you a single value which is 4. It's the same for the source[mask == 1] which gives you the single value 1.

and if you use destination[mask==1] = source[mask==1] instead of destination[mask] = source[mask] you will only change the value 4 in the destination matrix.

I hope my explanation is clear.

Edit:

I hope I understand your question correct:

The simple integer indexing structure x[[i]] gives you the i'th row of the matrix.

So destination[0,1,2] gives:

[[0, 1, 2],
 [3, 4, 5],
 [6, 7, 8]]

and for an understandable example the input destination[1,2,0] leads to

[[3, 4, 5],
 [6, 7, 8],
 [0, 1, 2]]

edited May 27, 2019 at 13:52

Dschoni

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answered May 27, 2019 at 12:11

Alperen Tahta

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2 Comments

Dschoni Over a year ago

Yes this helps. I still have to figure what the integer indexing together with broadcasting really is doing, but this put me in the right direction.

Dschoni Over a year ago

Thank you so much. I edited your answer for better readability but now it makes sense to me. So, I was taking lines from my original matrix and because my index-space is a 3-tuple broadcasting to a 3-D matrix happens.

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