Permutations with repetition in Python

Question

I want to iterate over all the vertices of an n dimensional cube of size 1. I know I could do that with itertools.product as follows:

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Answer 1

An (inefficient) alternative method:

>>> ['{0:03b}'.format(x) for x in range(8)]
['000', '001', '010', '011', '100', '101', '110', '111']

Or as tuples:

>>> [tuple(int(j) for j in list('{0:03b}'.format(x))) for x in range(8)]

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (0, 1, 1),
 (1, 0, 0),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

Sorted by number of vertices:

>>> sorted(_, key=lambda x: sum(x))

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (1, 0, 0),
 (0, 1, 1),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

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Or using itertools:

>>> sorted(itertools.product((0, 1), repeat=3), key=lambda x: sum(x))

[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (1, 0, 0),
 (0, 1, 1),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]