I would like to generate the following lists in Python:
[1, 1, 1, 2, 2]
[1, 1, 2, 1, 2]
... etc
[2, 1, 2, 1, 1]
[2, 2, 1, 1, 1]
There are always two "2"s and three "1"s in any list.
My intuition suggests that I will need to use the itertools module to do this. However, I am not sure where to begin, though I have read the documentation and looked at examples. Any suggestions?
You can notice that the number of such lists is equal to the number of ways to place two "2"s in a sequence of length 5. This suggests the following solution:
n = 5 # total length
n2 = 2 # number of "2"s
for idx in itertools.combinations( xrange(n), n2 ):
print [ 2 if i in idx else 1 for i in xrange(n) ]
It's easy to see that the answer using permutations is iterating over n! solutions, while my solution iterates over n!/( (n-n2)! * n2!). For example if the input list is [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2], the solution using permutations is ~90,000,000 times slower (10! * 4!)