所有成对的python

Question

Given a list l=range(n) , how can I iterate over all distinct pairs of distinct pairs from that list.

For example, if l = [0,1,2,3] I would like [((0,1), (0,2)),((0,1),(0,3)), ((0,1),(1,2)), ((0,1), (1,3)),((0,1), (2,3)), ((0,2), (0,3)), ((0,2),(1,2)),((0,2),(1,3)),((0,2),(2,3))...

Answer 1

You can use itertools.combinations :

from itertools import combinations

for pair in combinations(combinations(l, 2), 2):
    # use pair

The first call creates the initial pairs:

>>> l = [0,1,2,3]
>>> list(combinations(l, 2))
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

The second pairs them again:

>>> list(combinations(combinations(l, 2), 2))
[((0, 1), (0, 2)), ((0, 1), (0, 3)), ((0, 1), (1, 2)), ((0, 1), (1, 3)), 
 ((0, 1), (2, 3)), ((0, 2), (0, 3)), ((0, 2), (1, 2)), ((0, 2), (1, 3)), 
 ((0, 2), (2, 3)), ((0, 3), (1, 2)), ((0, 3), (1, 3)), ((0, 3), (2, 3)), 
 ((1, 2), (1, 3)), ((1, 2), (2, 3)), ((1, 3), (2, 3))]