Every permutation of list elements (without replacement)

Question

In Python 2.7 I’d like to get the self-cartesian product of the elements of a list, but without an element being paired with itself.

 In[]: foo = ['a', 'b', 'c']
 In[]: [x for x in itertools.something(foo)]
Out[]: 
       [('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'c'), ('c', 'a'), ('c', 'b')]

Currently I do:

[x for x in itertools.product(foo, repeat=2) if x[0] != x[1]]

but I suspect there’s a built-in method for this. What is it?

Note: itertools.combinations wouldn’t give me ('a', 'b') and ('b', 'a')

Asked By: LondonRob

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Source

Answer 1

You are looking for permutations instead of combinations.

from itertools import permutations

foo = ['a', 'b', 'c']
print(list(permutations(foo, 2)))

# Out: [('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'c'), ('c', 'a'), ('c', 'b')]

Answered By: Delgan

Every permutation of list elements (without replacement)

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