I just realized that permutations isn't quite what you want, as swapping the first "1" for the second "1" gives the same thing. You can use set to get the unique permutations. e.g. In [4]: set(itertools.permutations([1,1,0,0])) Out[4]: set([(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0)]) On Fri, Mar 19, 2010 at 10:17 AM, Joe Kington <jkington@wisc.edu> wrote:

See itertools.permutations (python standard library)

e.g. In [3]: list(itertools.permutations([1,1,0,0])) Out[3]: [(1, 1, 0, 0), (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 0, 1), (1, 1, 0, 0), (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1), (0, 0, 1, 1), (0, 1, 1, 0), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1),

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

Hope that helps, -Joe

On Fri, Mar 19, 2010 at 9:53 AM, gerardob <gberbeglia@gmail.com> wrote:

...

Hello, i would like to produce lists of lists 1's and 0's.

For example, to produce the list composed of:

L = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]

I just need to do the following:

n=4 numpy.eye(n,dtype=int).tolist()

I would like to know a simple way to generate a list containing all the lists having two 1's at each element.

Example, n = 4 L2 = [[1,1,0,0],[1,0,1,0],[1,0,0,1],[0,1,1,0],[0,1,0,1],[0,0,1,1]]

Any ideas? Thanks.

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