Creating a matrix of options using itertools

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Chapters
00:00 Question
00:48 Accepted answer (Score 8)
01:09 Answer 2 (Score 4)
01:38 Answer 3 (Score 2)
02:17 Thank you

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Full question
https://stackoverflow.com/questions/7046...

Accepted answer links:
[itertools.product]: http://docs.python.org/library/itertools...

Answer 2 links:
[Frederick]: https://stackoverflow.com/questions/7046...

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https://meta.stackexchange.com/help/lice...

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Tags
#python #itertools

#avk47

ACCEPTED ANSWER

Score 8

Use itertools.product:

itertools.product([False,True],repeat=5)

example of itertools.product([False,True],repeat=2):

(False, False)
(False, True)
(True, False)
(True, True)

ANSWER 2

Score 4

You seem to want the n-dimensional cartesian product of [False, True].

>>> print list(itertools.product(*(itertools.repeat((False, True), 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)]

Or more concisely (stealing from Frederick)

>>> print list(itertools.product((False, True), repeat=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)]

ANSWER 3

Score 2

This is the same form as the binary representation of integers between 0 to (2**n)-1. If you were inclined to such perversity, you could represent integers as zero-padded binary strings using str.format() and then coerce the string (of the form "00101") into a boolean list by doing something terrible like this:

>>> n = 5
>>> for i in xrange(2**n):
...     [bool(int(j)) for j in ("{0:0>%db}" % n).format(i)]

The n-dimensional Cartesian product above is, I'm sure, the right way to think about this, but this way made me giggle.