How to invert numpy.where (np.where) function

Question

I frequently use the numpy.where function to gather a tuple of indices of a matrix having some property. For example

import numpy as np
X = np.random.rand(3,3)
>>> X
array([[ 0.51035326,  0.41536004,  0.37821622],
   [ 0.32285063,  0.29847402,  0.82969935],
   [ 0.74340225,  0.51553363,  0.22528989]])
>>> ix = np.where(X > 0.5)
>>> ix
(array([0, 1, 2, 2]), array([0, 2, 0, 1]))

ix is now a tuple of ndarray objects that contain the row and column indices, whereas the sub-expression X>0.5 contains a single boolean matrix indicating which cells had the >0.5 property. Each representation has its own advantages.

What is the best way to take ix object and convert it back to the boolean form later when it is desired? For example

G = np.zeros(X.shape,dtype=np.bool)
>>> G[ix] = True

Is there a one-liner that accomplishes the same thing?

Raver

Answer 1

Something like this maybe?

mask = np.zeros(X.shape, dtype='bool')
mask[ix] = True

but if it's something simple like X > 0, you're probably better off doing mask = X > 0 unless mask is very sparse or you no longer have a reference to X.

Ballesteros

Answer 2

mask = X > 0
imask = np.logical_not(mask)

For example

Edit: Sorry for being so concise before. Shouldn't be answering things on the phone :P

As I noted in the example, it's better to just invert the boolean mask. Much more efficient/easier than going back from the result of where.

Elsewhere

Answer 3

The bottom of the np.where docstring suggests to use np.in1d for this.

>>> x = np.array([1, 3, 4, 1, 2, 7, 6])
>>> indices = np.where(x % 3 == 1)[0]
>>> indices
array([0, 2, 3, 5])
>>> np.in1d(np.arange(len(x)), indices)
array([ True, False,  True,  True, False,  True, False], dtype=bool)

(While this is a nice one-liner, it is a lot slower than @Bi Rico's solution.)

Brom