找到两个嵌套列表的交集？

Question

我知道如何获得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]

要么

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

print intersect(b1, b2)

但是当我必须找到嵌套列表的交集时,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]]

最后我想收到:

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

你能帮我个忙吗？

有关

• 在python中展平浅表

Answer 1

您不需要定义交集.它已经是集合的一流部分.

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

Answer 2

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]]

那么这是你的Python 2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iterable而不是list,所以你需要用以下方法包装filter调用list():

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

说明:

过滤器部分获取每个子列表的项目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.

Answer 3

对于那些只想找到两个列表的交集的人来说,Asker提供了两种方法:

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]

和

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

print intersect(b1, b2)

但是有一种混合方法更有效,因为你只需要在列表/集合之间进行一次转换,而不是三种:

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,而他的原始方法涉及列表理解将在O(n ^ 2)中运行

Answer 4

功能方法:

input_list = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7]]

result = reduce(set.intersection, map(set, input_list))

它可以应用于1+列表的更一般情况

Answer 5

纯列表理解版

>>> 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]]
>>> c1set = frozenset(c1)

扁平变体:

>>> [n for lst in c2 for n in lst if n in c1set]
[13, 32, 7, 13, 28, 1, 6]

嵌套变体:

>>> [[n for n in lst if n in c1set] for lst in c2]
[[13, 32], [7, 13, 28], [1, 6]]

Answer 6

&运算符取两组的交集.

{1,2,3}&{2,3,4} Out [1]:{2,3}

Answer 7

采用2个列表的pythonic方式是:

[x for x in list1 if x in list2]

Answer 8

您应该使用此代码(取自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks)展平,代码未经测试,但我确信它有效:

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

在展平列表后,以通常的方式执行交集:

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))

Answer 9

自intersect定义以来,基本的列表理解就足够了:

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

感谢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]]

Answer 10

鉴于:

> 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]]

我发现以下代码运行良好,如果使用set操作可能更简洁:

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

它得到了:

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

如果需要订购:

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

我们有:

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

顺便说一句,对于更多的python风格,这个也很好:

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