Numpy indexing with a one dimensional boolean array

Question

The post, Getting a grid of a matrix via logical indexing in Numpy, is similar, but it does not answer my question since I am working with a one dimensional boolean array.

I am attempting to recreate the following boolean indexing feature in Octave.

octave-3.2.4:6> a = rand(3,3)
a =

   0.249912   0.934266   0.371962   
   0.505791   0.813354   0.282006 
   0.439417   0.085733   0.886841  
octave-3.2.4:8> a([true false true])
ans =

    0.24991   0.43942

However, I am unable to create the same results in Python with Numpy.

>>> import numpy as np
>>> a = np.random.rand(3,3)
array([[ 0.94362993,  0.3553076 ,  0.12761322],
       [ 0.19764288,  0.35325583,  0.17034005],
       [ 0.56812424,  0.48297648,  0.64101657]])
>>> a[[True, False, True]]
array([[ 0.19764288,  0.35325583,  0.17034005],
       [ 0.94362993,  0.3553076 ,  0.12761322],
       [ 0.19764288,  0.35325583,  0.17034005]])
>>> a[np.ix_([True, False, True])]
array([[ 0.94362993,  0.3553076 ,  0.12761322],
      [ 0.56812424,  0.48297648,  0.64101657]])

How do I recreate Octave's boolean indexing on Python with Numpy?

Luke · Accepted Answer · 2013-07-22T04:05:57.827

Two problems:

Indexing with a list [True, False, True] is not the same as indexing with a boolean array array([True,False,True]). The list will instead be interpreted as integer indexes [1,0,1]

You need to specify that you only want the results from the first column:

>>> a = np.arange(9).reshape(3,3)
>>> a
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
>>> mask = np.array([True,False,True])
>>> mask.dtype ## verify we have a bool array
dtype('bool')
>>> a[mask,0]
array([0, 6])

Numpy indexing with a one dimensional boolean array

1 Answers1