Create all combinations of boolean for a specified n

Asked Jun 09 '18 at 18:43

Active Jun 09 '18 at 18:43

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I want to replicate the cartesian product of [False,True] n times so that the output should be for example with n=3:

[(False, False, False),(False, False, True),(False, True, False),
(False, True, True),(True, False, False),(True, False, True),
(True, True, False),(True, True, True)]

I found how to do it with:

from itertools import product
list(product([False,True],[False,True],[False,True]))

But the thing is that I want to do it for bigger n without writing n times the boolean.

asked Jun 09 '18 at 18:43

Arli94

You're looking for the `repeat` flag. Try `list(product([False, True], repeat=3))` – user3483203 Jun 09 '18 at 18:45
I just want to point out that this is equivalent to counting to 8. You can just look at bits then. – Olivier Melançon Jun 09 '18 at 18:50
`def get_bits(n, k): return [bool(n >> i & 1) for i in range(k)]; for x in range(8): print(get_bits(x, 3))` – Olivier Melançon Jun 09 '18 at 18:57

Create all combinations of boolean for a specified n

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