对我来说，使用列表推导式是一个非常明显的方法。不确定性能如何，但至少能保持列表。

[x for x in a if x in b]

或者"所有在A中的x值，如果x值在B中"

当我们使用tuple时，我们想要交叉

a=([1,2,3,4,5,20], [8,3,9,5,1,4,20])

for i in range(len(a)):

    b=set(a[i-1]).intersection(a[i])
print(b)
{1, 3, 4, 5, 20}

使用过滤器和lambda运算符可以实现函数式的方法。

list1 = [1,2,3,4,5,6]

list2 = [2,4,6,9,10]

>>> list(filter(lambda x:x in list1, list2))

[2, 4, 6]

编辑:它过滤掉了同时存在于list1和list中的x，集差异也可以使用:

>>> list(filter(lambda x:x not in list1, list2))
[9,10]

python3 filter返回一个过滤器对象，用list封装它返回输出列表。

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!

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]

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)