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

既然定义了intersect,一个基本的列表推导式就足够了:

>>> c3 = [intersect(c1, i) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]

多亏了S. Lott的评论和TM。的相关评论:

>>> c3 = [list(set(c1).intersection(i)) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]

其他回答

from random import *

a = sample(range(0, 1000), 100)
b = sample(range(0, 1000), 100)
print(a)
print(b)
print(set(a).intersection(set(b)))

我也在寻找一种方法来做到这一点,最终结果是这样的:

def compareLists(a,b):
    removed = [x for x in a if x not in b]
    added = [x for x in b if x not in a]
    overlap = [x for x in a if x in b]
    return [removed,added,overlap]

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)

你认为[1,2]与[1,[2]]相交吗?也就是说,你只关心数字,还是也关心列表结构?

如果只有数字,研究如何“扁平化”列表,然后使用set()方法。

既然定义了intersect,一个基本的列表推导式就足够了:

>>> c3 = [intersect(c1, i) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]

多亏了S. Lott的评论和TM。的相关评论:

>>> c3 = [list(set(c1).intersection(i)) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]