Python -Intersection of multiple lists?

问题

I am playing with python and am able to get the intersection of two lists:

result = set(a).intersection(b)

Now if d is a list containing a and b and a third element c, is there an built-in function for finding the intersection of all the three lists inside d? So for instance,

d = [[1,2,3,4], [2,3,4], [3,4,5,6,7]]

then the result should be

[3,4]

回答1:

for 2.4, you can just define an intersection function.

def intersect(*d):
    sets = iter(map(set, d))
    result = sets.next()
    for s in sets:
        result = result.intersection(s)
    return result

for newer versions of python:

the intersection method takes an arbitrary amount of arguments

result = set(d[0]).intersection(*d[:1])

alternatively, you can intersect the first set with itself to avoid slicing the list and making a copy:

result = set(d[0]).intersection(*d)

I'm not really sure which would be more efficient and have a feeling that it would depend on the size of the d[0] and the size of the list unless python has an inbuilt check for it like

if s1 is s2:
    return s1

in the intersection method.

>>> d = [[1,2,3,4], [2,3,4], [3,4,5,6,7]]
>>> set(d[0]).intersection(*d)
set([3, 4])
>>> set(d[0]).intersection(*d[1:])
set([3, 4])
>>>

回答2:

set.intersection(*map(set,d))

回答3:

@user3917838

Nice and simple but needs some casting to make it work and give a list as a result. It should look like:

list(reduce(set.intersection, [set(item) for item in d ]))

where:

d = [[1,2,3,4], [2,3,4], [3,4,5,6,7]]

And result is:

[3, 4]

At least in Python 3.4

回答4:

Lambda reduce.

from functools import reduce #you won't need this in Python 2
l=[[1, 2, 3, 4], [2, 3, 4], [3, 4, 5, 6, 7]]
reduce(set.intersection, [set(l_) for l_ in l])

来源：https://stackoverflow.com/questions/3852780/python-intersection-of-multiple-lists

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