你应该使用这段代码(来自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks)，这段代码未经测试，但我很确定它是有效的:

def flatten(x):
    """flatten(sequence) -> list

    Returns a single, flat list which contains all elements retrieved
    from the sequence and all recursively contained sub-sequences
    (iterables).

    Examples:
    >>> [1, 2, [3,4], (5,6)]
    [1, 2, [3, 4], (5, 6)]
    >>> flatten([[[1,2,3], (42,None)], [4,5], [6], 7, MyVector(8,9,10)])
    [1, 2, 3, 42, None, 4, 5, 6, 7, 8, 9, 10]"""

    result = []
    for el in x:
        #if isinstance(el, (list, tuple)):
        if hasattr(el, "__iter__") and not isinstance(el, basestring):
            result.extend(flatten(el))
        else:
            result.append(el)
    return result

在你平摊了列表之后，你用通常的方式执行交叉:

c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]

def intersect(a, b):
     return list(set(a) & set(b))

print intersect(flatten(c1), flatten(c2))

你不需要定义交集。它已经是集合的一流部分了。

>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> set(b1).intersection(b2)
set([4, 5])

对于只想找到两个列表交集的人，Asker提供了两个方法:

B1 = [1,2,3,4,5,9,11,15] B2 = [4,5,6,7,8] B3 = [val for val in b1 if val in b2] 而且 Def相交(a, b): 返回列表(set(a) & set(b)) 打印相交(b1, b2)

但是有一种混合方法更有效，因为你只需要在list/set之间做一次转换，而不是三次:

b1 = [1,2,3,4,5]
b2 = [3,4,5,6]
s2 = set(b2)
b3 = [val for val in b1 if val in s2]

这将在O(n)中运行，而他最初的包含列表理解的方法将在O(n²)中运行

查找迭代对象之间的差异和交集的简单方法

如果重复很重要，就使用这种方法

from collections import Counter

def intersection(a, b):
    """
    Find the intersection of two iterables

    >>> intersection((1,2,3), (2,3,4))
    (2, 3)

    >>> intersection((1,2,3,3), (2,3,3,4))
    (2, 3, 3)

    >>> intersection((1,2,3,3), (2,3,4,4))
    (2, 3)

    >>> intersection((1,2,3,3), (2,3,4,4))
    (2, 3)
    """
    return tuple(n for n, count in (Counter(a) & Counter(b)).items() for _ in range(count))

def difference(a, b):
    """
    Find the symmetric difference of two iterables

    >>> difference((1,2,3), (2,3,4))
    (1, 4)

    >>> difference((1,2,3,3), (2,3,4))
    (1, 3, 4)

    >>> difference((1,2,3,3), (2,3,4,4))
    (1, 3, 4, 4)
    """
    diff = lambda x, y: tuple(n for n, count in (Counter(x) - Counter(y)).items() for _ in range(count))
    return diff(a, b) + diff(b, a)

考虑到:

> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]

> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]

我发现下面的代码工作得很好，如果使用set操作可能会更简洁:

> c3 = [list(set(f)&set(c1)) for f in c2] 

它有:

> [[32, 13], [28, 13, 7], [1, 6]]

如需订购:

> c3 = [sorted(list(set(f)&set(c1))) for f in c2] 

我们有:

> [[13, 32], [7, 13, 28], [1, 6]]

顺便说一下，对于更python的风格，这个也很好:

> c3 = [ [i for i in set(f) if i in c1] for f in c2]

如果你想:

c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [[13, 32], [7, 13, 28], [1,6]]

下面是Python 2的解决方案:

c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]

在Python 3中，filter返回一个可迭代对象而不是list，所以你需要用list()来包装filter调用:

c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]

解释:

过滤器部分获取每个子列表的项并检查它是否在源列表c1中。 对c2中的每个子列表执行列表推导式。