Boolean Indexing numpy Array with or logical operator

Question

I was trying to do an or boolean logical indexing on a Numpy array but I cannot find a good way. The and operator & works properly like:

X = np.arange(25).reshape(5, 5)
# We print X
print()
print('Original X = \n', X)
print()

X[(X > 10) & (X < 17)] = -1

# We print X
print()
print('X = \n', X)
print()

Original X = 
 [[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 11 12 13 14]
 [15 16 17 18 19]
 [20 21 22 23 24]]

X = 
 [[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 -1 -1 -1 -1]
 [-1 -1 17 18 19]
 [20 21 22 23 24]]

But when I try with:

X = np.arange(25).reshape(5, 5)

# We use Boolean indexing to assign the elements that are between 10 and 17 the value of -1
X[ (X < 10) or (X > 20) ] = 0 # No or condition possible!?!

I got the error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Does exist any good way to use the or logic operator?

Enough to replace `or` by `|` – FBruzzesi May 13 '20 at 07:58 — FBruzzesi, May 13 '20 at 07:58
`|` is the counter part to `&` – hpaulj May 13 '20 at 07:58 — hpaulj, May 13 '20 at 07:58
Yes possible: `X[ (X < 10) | (X > 20) ] = 0` – Austin May 13 '20 at 07:58 — Austin, May 13 '20 at 07:58

score 0 · Answer 1 · answered May 13 '20 at 08:10

You can use numpy.logical_or for that task following way:

import numpy as np
X = np.arange(25).reshape(5,5)
X[np.logical_or(X<10,X>20)] = 0
print(X)

Output:

[[ 0  0  0  0  0]
 [ 0  0  0  0  0]
 [10 11 12 13 14]
 [15 16 17 18 19]
 [20  0  0  0  0]]

There is also numpy.logical_and, numpy.logical_xor and numpy.logical_not

score 0 · Answer 2 · answered May 13 '20 at 08:13

I would use something with np.logical_and and np.where. For your given example, I believe this would work.

X = np.arange(25).reshape(5, 5)
i = np.where(np.logical_and(X > 10 , X < 17))
X[i] = -1

This is not a very pythonic answer. But it's pretty clear

Boolean Indexing numpy Array with or logical operator

2 Answers2