Commutative combination of elements of two lists

Question

Just a style question: Is there a build-in method to get the combinations under the assertion of commutative property and excluding elements paired with itself?

Answer 1

itertools.combinations, if both lists are the same like here. Or in the general case itertools.product, followed by some filtering:

In [7]: a = ["1", "2", "3"]
   ...: b = ["a", "b", "c"]

In [8]: list(filter(lambda t: t[0] < t[1], product(a,b)))
Out[8]: 
[('1', 'a'),
 ('1', 'b'),
 ('1', 'c'),
 ('2', 'a'),
 ('2', 'b'),
 ('2', 'c'),
 ('3', 'a'),
 ('3', 'b'),
 ('3', 'c')]

Also, I think the term combination already means that the order of elements in the result doesn't matter.

---

Ok, Theodros is right. For compensation, here's a version which should work on a any list of lists:

l = [['1','2','3'], ['a','b'], ['x','y']] 

set(tuple(sorted(p)) for p in product(*l) if len(set(p)) > 1)

gives (appropriately sorted)

set([('1', 'a', 'x'),
     ('3', 'a', 'y'),
     ('2', 'b', 'y'),
     ('2', 'a', 'y'),
     ('1', 'a', 'y'),
     ('1', 'b', 'y'),
     ('2', 'a', 'x'),
     ('3', 'b', 'y'),
     ('1', 'b', 'x'),
     ('2', 'b', 'x'),
     ('3', 'a', 'x'),
     ('3', 'b', 'x')])

And it also works on the previous counterexample l = [[1,2,3], [1,3,4,5]]:

set([(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (2, 4), (3, 5)])