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pythoncombinatoricscartesian-product

How can I get "permutations with repetitions/replacement" from a list (Cartesian product of a list with itself)?


Suppose I have a list die_faces = [1, 2, 3, 4, 5, 6]. I want to generate all 36 possible results for rolling two dice: (1, 1), (1, 2), (2, 1) etc. If I try using permutations from the itertools standard library:

>>> import itertools
>>> die_faces = [1, 2, 3, 4, 5, 6]
>>> list(itertools.permutations(die_faces, 2))
[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)]

there are only 30 results, missing the ones where the same number comes up on both dice. It seems that it only generates permutations without repetitions. How can I fix this?


Solution

  • You are looking for the Cartesian Product.

    In mathematics, a Cartesian product (or product set) is the direct product of two sets.

    In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there:

    import itertools
    x = [1, 2, 3, 4, 5, 6]
    [p for p in itertools.product(x, repeat=2)]
    [(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), 
     (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 
     (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), 
     (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]
    

    To get a random dice roll (in a totally inefficient way):

    import random
    random.choice([p for p in itertools.product(x, repeat=2)])
    (6, 3)