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?

Answer 1

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

Answer 2

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