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I would like to assign values of an array arr_2 to a different array arr_1. However, I want to do this based on 2 selection criteria. As a working example, I define my selection criteria as such

import numpy as np

# An array of -1 values of shape(10,): [-1 -1 -1 -1 -1 -1 -1 -1 -1 -1]
arr_1 = np.zeros(10, dtype=int) - 1
# An array of 0-9 values of shape(10,): [0 1 2 3 4 5 6 7 8 9]
arr_2 = np.arange(10)

# Create an initial selection of values we want to change
# In this example: even indices: [ T F T F T F T F T F]
selection_a = np.arange(10) % 2 == 0

# Create a second selection based on selection_a: [F F F T T]
selection_b = arr_2[selection_a] > 5

Based on these two selection criteria I would like to assign the values of arr_2 where both conditions hold to the array arr_1. I.e. equivalent to [F F F F F F T F T F].

arr_1[selection_a][selection_b] = arr_2[selection_a][selection_b]

If I inspect both sides of the equation before the assignment, they yield the values that I expect:

print(arr_1[selection_a][selection_b]) # yields [-1 -1]
print(arr_2[selection_a][selection_b]) # yields [ 6, 8]

However, the assignment itself does not assign the values, i.e. arr_1 remains unchanged. My question is, why is this the case?

NB: I know that in most (and maybe even all cases) this can be circumvented by creating a single criterion, however I want to know why using 2 separate criteria won't work.

If anything is unclear, please let me know and I'll try to clarify.

Edit

I investigated this a bit further and the problem seems to be in the left hand side of the equation as something like

arr_1[selection_a][selection_b] = 5

does not work either.

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  • I'd do something like: idx = arr_2[selection_a][selection_b]; arr_1[idx] = idx Commented Sep 18, 2018 at 9:57
  • Define selection_b as arr_2 > 5 & selection_a then use arr_1[selection_b] = 5 Commented Sep 18, 2018 at 9:57
  • @GiacomoAlzetta selection_a and selection_b have different shapes. Commented Sep 18, 2018 at 9:59
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    @PM2Ring Not if you define them in that way. What I'm saying is: instead of having selection_b depend on the result in selection_a requiring a nested selection, just define it with the same shape as selection_a and use & to "filter". You can always rewrite a nested condition [a][b][c][d] into [a& b' & c' & d'] where the primed versions b', c', d' (ecc.) have the same shape as a. Commented Sep 18, 2018 at 10:28
  • The key is that arr_1[selection_a] is a copy, not a view. The assignment is modifying a portion of that copy, not a portion of the original. numpy is not parsing the 2 criteria together. Python is performing two separate indexing operations. Commented Sep 19, 2018 at 3:03

4 Answers 4

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A single use of [...] operators creates a selection in an array, from which you can read and to which you can write. A second use of [...] will read values with no problem. But as far as writing is concerned, the second access will be to a temporary array created to match the selection rules (i.e. a temporary of shape (2,) that is a copy of the original data) in your situation.

Edit: Boolean indexing is part of so-called "Advanced Indexing". Keeping the indexing to simple slices avoids these copy issues.

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1

Adding to Pierre de Buyl solution, If you want to get your desired result of changing arr_1 array. You can use np.where as follows:

arr_1[selection_a] = np.where(arr_2[selection_a]>5,arr_2[selection_a],arr_1[selection_a])

arr_1

Which gives

array([-1, -1, -1, -1, -1, -1,  6, -1,  8, -1])
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It has to do with = in python being interpreted as .__setitem__(), which can pick the memory addresses from the assigned variable. However, it doesn't do this recursively, so while 

a[boolean_mask] = 0

works, as the indexing points to the memory in a, 

a[mask1][mask2] = 0

does not, as the indexing points to the memory in a[mask1], which is a copy.

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Since the conditions of selection_a and selection_b do not depend on each other, you could just create a combined mask

selection_c = (np.arange(10) % 2 == 0) & (arr_2 >  5)

and use it like

arr_1[selection_c] = 5
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