在列表中迭代（item，others）

Question

Suppose I have a list:

l = [0, 1, 2, 3]

How can I iterate over the list, taking each item along with its complement from the list? That is,

for item, others in ...
    print(item, others)

would print

0 [1, 2, 3]
1 [0, 2, 3]
2 [0, 1, 3]
3 [0, 1, 2]

Ideally I'm looking for a concise expression that I can use in a comprehension.

Answer 1

This is quite easy and understandable:

for index, item in enumerate(l):
    others = l[:index] + l[index+1:]

You could make an iterator out of this if you insist:

def iter_with_others(l):
    for index, item in enumerate(l):
        yield item, l[:index] + l[index+1:]

Giving it's usage:

for item, others in iter_with_others(l):
    print(item, others)

Answer 2

Answering my own question, it is possible to use itertools.combinations exploiting the fact that the result is emitted in lexicographical order:

from itertools import combinations
zip(l, combinations(reversed(l), len(l) - 1))

However, this is fairly obscure; nightcracker's solution is a lot easier to understand for the reader!

Answer 3

What about

>>> [(i, [j for j in L if j != i]) for i in L]
[(0, [1, 2, 3]), (1, [0, 2, 3]), (2, [0, 1, 3]), (3, [0, 1, 2])]

OK, that's a gazillion of tests and @nightcracker's solution is likely more efficient, but eh...