在python中查找（并保留）子列表的重复项

Question

I have a list of lists (sublist) that contains numbers and I only want to keep those exists in all (sub)lists.

Example:

x = [ [1, 2, 3, 4], [3, 4, 6, 7], [2, 3, 4, 6, 7]]

output => [3, 4]

How can I do this?

Answer 1

common = set(x[0])
for l in x[1:]:
    common &= set(l)
print list(common)

or:

import operator
print reduce(operator.iand, map(set, x))

Answer 2

In one liner:

>>> reduce(set.intersection, x[1:], set(x[0]))
set([3, 4])

Answer 3

def f(a, b):
    return list(set(a).intersection(set(b)))

reduce(f, x)

Answer 4

Just another way of solving, almost same as nadia but without using reduce, and i use map:

>>> x = [ [1, 2, 3, 4], [3, 4, 6, 7], [2, 3, 4, 6, 7]]
>>> set.intersection(*map(set,x))
set([3, 4])
>>>