How to create a numpy array from itertools.combinations without looping

Question

Is there a way to get this result without a loop? I've made a couple attempts at fancy indexing with W[range(W.shape[0]),... but have been so far unsuccessful.

import itertools
import numpy as np
n = 4
ct = 2
one_index_tuples = list(itertools.combinations(range(n), r=ct))
W = np.zeros((len(one_index_tuples), n), dtype='int')
for row_index, col_index in enumerate(one_index_tuples):
    W[row_index, col_index] = 1
print(W)

Result:

[[1 1 0 0]
 [1 0 1 0]
 [1 0 0 1]
 [0 1 1 0]
 [0 1 0 1]
 [0 0 1 1]]

Possible duplicate of Indexing a numpy array with a list of tuples — maxymoo
– maxymoo, Commented Apr 13, 2017 at 1:50

akuiper · Accepted Answer · 2017-04-13 02:01:12Z

3

You can use fancy indexing (advanced indexing) as follows:

# reshape the row index to 2d since your column index is also 2d so that the row index and 
# column index will broadcast properly
W[np.arange(len(one_index_tuples))[:, None], one_index_tuples] = 1

W
#array([[1, 1, 0, 0],
#       [1, 0, 1, 0],
#       [1, 0, 0, 1],
#       [0, 1, 1, 0],
#       [0, 1, 0, 1],
#       [0, 0, 1, 1]])

answered Apr 13, 2017 at 2:01

akuiper

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MaThMaX · Accepted Answer · 2017-04-13 01:51:50Z

0

Try this:

[[ 1 if i in x else 0 for i in range(n) ] for x in itertools.combinations( range(n), ct )]

answered Apr 13, 2017 at 1:51

MaThMaX

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1 Comment

maxymoo Over a year ago

this is still looping tho ;)

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