这可能是晚了，但我只是认为我应该分享的情况下，你需要手动做(显示工作-哈哈)或当你需要所有元素出现尽可能多的次数或当你也需要它是唯一的。

请注意，还为它编写了测试。

    from nose.tools import assert_equal

    '''
    Given two lists, print out the list of overlapping elements
    '''

    def overlap(l_a, l_b):
        '''
        compare the two lists l_a and l_b and return the overlapping
        elements (intersecting) between the two
        '''

        #edge case is when they are the same lists
        if l_a == l_b:
            return [] #no overlapping elements

        output = []

        if len(l_a) == len(l_b):
            for i in range(l_a): #same length so either one applies
                if l_a[i] in l_b:
                    output.append(l_a[i])

            #found all by now
            #return output #if repetition does not matter
            return list(set(output))

        else:
            #find the smallest and largest lists and go with that
            sm = l_a if len(l_a)  len(l_b) else l_b

            for i in range(len(sm)):
                if sm[i] in lg:
                    output.append(sm[i])

            #return output #if repetition does not matter
            return list(set(output))

    ## Test the Above Implementation

    a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
    b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
    exp = [1, 2, 3, 5, 8, 13]

    c = [4, 4, 5, 6]
    d = [5, 7, 4, 8 ,6 ] #assuming it is not ordered
    exp2 = [4, 5, 6]

    class TestOverlap(object):

        def test(self, sol):
            t = sol(a, b)
            assert_equal(t, exp)
            print('Comparing the two lists produces')
            print(t)

            t = sol(c, d)
            assert_equal(t, exp2)
            print('Comparing the two lists produces')
            print(t)

            print('All Tests Passed!!')

    t = TestOverlap()
    t.test(overlap)

如果顺序不重要，你不需要担心重复，那么你可以使用set intersection:

>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]

如果你将两个列表中较大的一个转换为一个集合，你可以使用intersection()获得该集合与任何可迭代对象的交集:

a = [1,2,3,4,5]
b = [1,3,5,6]
set(a).intersection(b)

这样可以得到两个列表的交集，也可以得到公共重复项。

>>> from collections import Counter
>>> a = Counter([1,2,3,4,5])
>>> b = Counter([1,3,5,6])
>>> a &= b
>>> list(a.elements())
[1, 3, 5]

a = [1,2,3,4,5]
b = [1,3,5,6]
c = list(set(a).intersection(set(b)))

应该像做梦一样工作。并且，如果可以的话，使用集合而不是列表来避免所有这些类型更改!

你也可以使用计数器!它不会保留顺序，但会考虑副本:

>>> from collections import Counter
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> d1, d2 = Counter(a), Counter(b)
>>> c = [n for n in d1.keys() & d2.keys() for _ in range(min(d1[n], d2[n]))]
>>> print(c)
[1,3,5]