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How to compute the n-fold Cartesian product on a list, that is, A × ... × A (n times), in an elegant (concise) way in Python?

Examples:

>>> l = ["a", "b", "c"]
>>> cart_prod(l, 0)
[]
>>> cart_prod(l, 1)
[('a',), ('b',), ('c',)]
>>> cart_prod(l, 2)
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'b'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('c', 'c')]
>>> cart_prod(l, 3)
[('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'a'), ('a', 'b', 'b'), ('a', 'b', 'c'), ('a', 'c', 'a'), ('a', 'c', 'b'), ('a', 'c', 'c'),
 ('b', 'a', 'a'), ('b', 'a', 'b'), ('b', 'a', 'c'), ('b', 'b', 'a'), ('b', 'b', 'b'), ('b', 'b', 'c'), ('b', 'c', 'a'), ('b', 'c', 'b'), ('b', 'c', 'c'),
 ('c', 'a', 'a'), ('c', 'a', 'b'), ('c', 'a', 'c'), ('c', 'b', 'a'), ('c', 'b', 'b'), ('c', 'b', 'c'), ('c', 'c', 'a'), ('c', 'c', 'b'), ('c', 'c', 'c')]

I came up with the following iterative solution:

def cart_prod(l, n):
    if n == 0:
        return []  # compute the result for n = 0
    # preliminarily, create a list of lists instead of a list of tuples
    res = [[x] for x in l]  # initialize list with singleton tuples (n = 1)
    for i in range(n-1):
        res = [r + [x] for r in res for x in l]  # concatenate each n-1 tuple with each element from a
    res = [tuple(el) for el in res]  # turn the list of lists into a list of tuples
    return res

This code does the job, but is there a shorter, possibly one-liner definition, maybe a nested list comprehension or a lambda expression? I am interested in more compact solutions, not necessarily more readable ones.

This question is not a duplicate of Get the cartesian product of a series of lists?. I do not want the Cartesian product of a series of lists crossed with each other. I want the Cartesian product of a single list crossed n-times with itself, where n is a parameter given to the function.

wjandrea
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Natalie Clarius
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  • It still looks like a duplicate to me. Just replicate the list and then use `itertools.product`. – bnaecker Jun 16 '20 at 21:46
  • Better yet, `itertools.tee` the list `n` times, then use `itertools.product`. – bnaecker Jun 16 '20 at 21:47
  • @bnaecker You don't need `tee`. – chepner Jun 16 '20 at 21:48
  • @chepner Nice, I've never used that kwarg before. – bnaecker Jun 16 '20 at 21:48
  • "This question is not a duplicate of Get the cartesian product of a series of lists?. I do not want the Cartesian product of a series of lists crossed with each other. I want the Cartesian product of a single list crossed n-times with itself, where n is a parameter given to the function." As it happens, we have canonical duplicates for that too. This is a good signpost, though. Thanks @wjandrea for cleaning it up. – Karl Knechtel Jul 03 '22 at 16:34

1 Answers1

7

itertools.product takes a keyword argument to indicate the given arguments should be repeated.

>>> from itertools import product
>>> list(product([1,2], repeat=0))
[()]
>>> list(product([1,2], repeat=1))
[(1,), (2,)]
>>> list(product([1,2], repeat=2))
[(1, 1), (1, 2), (2, 1), (2, 2)]

This works with multiple iterables as well.

# Equivalent to list(product([1,2], ['a', 'b'], [1,2], ['a', 'b']))
>>> list(product([1,2], ['a', 'b'], repeat=2))
[(1, 'a', 1, 'a'), (1, 'a', 1, 'b'), (1, 'a', 2, 'a'), (1, 'a', 2, 'b'), (1, 'b', 1, 'a'), (1, 'b', 1, 'b'), (1, 'b', 2, 'a'), (1, 'b', 2, 'b'), (2, 'a', 1, 'a'), (2, 'a', 1, 'b'), (2, 'a', 2, 'a'), (2, 'a', 2, 'b'), (2, 'b', 1, 'a'), (2, 'b', 1, 'b'), (2, 'b', 2, 'a'), (2, 'b', 2, 'b')]
chepner
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