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

# Problem:  Given c1 and c2:
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]]
# how do you get c3 to be [[13, 32], [7, 13, 28], [1, 6]] ?

这里有一种不涉及集合的方法来设置c3:

c3 = []
for sublist in c2:
    c3.append([val for val in c1 if val in sublist])

但如果你喜欢只用一行,你可以这样做:

c3 = [[val for val in c1 if val in sublist]  for sublist in c2]

它是列表推导式中的列表推导式,这有点不寻常,但我认为你应该不会有太多的问题。

其他回答

我不知道我回答你的问题是否晚了。在阅读了你的问题后,我提出了一个函数intersect(),可以在列表和嵌套列表上工作。我用递归来定义这个函数,很直观。希望这是你想要的:

def intersect(a, b):
    result=[]
    for i in b:
        if isinstance(i,list):
            result.append(intersect(a,i))
        else:
            if i in a:
                 result.append(i)
    return result

例子:

>>> 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]]
>>> print intersect(c1,c2)
[[13, 32], [7, 13, 28], [1, 6]]

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

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]

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)

运算符&取两个集合的交集。

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