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Creating a parameter matrix using itertools

I am trying to create a matrix of True and False values ​​that displays all permutations for a certain number of options. Thus, for 5 options you will get the following result.

FFFFF TFFFF TTFFF TTTFF ... FTFFF ...

I was looking for the use of itertool permutations and combinations, but they do not work by position, not by value, which leads to duplication.

I'm sure there is a standard algorithm for this problem, but I'm struggling to find his name.

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3 answers

Use itertools.product :

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

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

 (False, False) (False, True) (True, False) (True, True)
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It seems you need an n-dimensional Cartesian product [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 briefly (the theft 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)]
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This is the same form as the binary representation of integers from 0 to (2 ** n) -1. If you were prone to such perversity, you could represent integers as zero binary strings using str.format (), and then force the string (of the form “00101”) to the logical list, doing something terrible:

 >>> 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, I'm sure the right way to think about it, but this way made me giggle.

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