我知道如何得到两个平面列表的交集:

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]

or

def intersect(a, b):
    return list(set(a) & set(b))
 
print intersect(b1, b2)

但当我必须为嵌套列表找到交集时,我的问题就开始了:

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]]

最后我希望收到:

c3 = [[13,32],[7,13,28],[1,6]]

你们能帮我一下吗?

相关的

在python中扁平一个浅列表


当前回答

要定义正确考虑元素基数的交集,请使用Counter:

from collections import Counter

>>> c1 = [1, 2, 2, 3, 4, 4, 4]
>>> c2 = [1, 2, 4, 4, 4, 4, 5]
>>> list((Counter(c1) & Counter(c2)).elements())
[1, 2, 4, 4, 4]

其他回答

要定义正确考虑元素基数的交集,请使用Counter:

from collections import Counter

>>> c1 = [1, 2, 2, 3, 4, 4, 4]
>>> c2 = [1, 2, 4, 4, 4, 4, 5]
>>> list((Counter(c1) & Counter(c2)).elements())
[1, 2, 4, 4, 4]

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

通过简化可以很容易地做出平面列表。

你只需要在reduce函数中使用初始化器-第三个参数。

reduce(
   lambda result, _list: result.append(
       list(set(_list)&set(c1)) 
     ) or result, 
   c2, 
   [])

上面的代码适用于python2和python3,但是你需要像从functools import reduce那样导入reduce模块。详情请参考下面的链接。

对于python2 对于python3

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。

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))