找到可迭代对象之间的差异和交集的简单方法

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]

或

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？

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)

展平变体：

>>> [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

＆amp;运算符取两个集合。

{1,2,3}＆amp; {2,3,4} 出[1]：{2,3}

Answer 7

采用两个列表交叉的pythonic方法是：

[x for x in list1 if x in list2]

Answer 8

您应该使用此代码扁平化（取自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks），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

在您对列表进行展平后，您将以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))

Answer 9

自定义intersect以来，基本列表理解就足够了：

>>> 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

假设：

> 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]]

如果需要订单：

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

我们得到了：

> [[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]

Answer 11

我不知道我是否迟到了回答你的问题。在阅读了你的问题后，我想出了一个函数intersect（），它可以在列表和嵌套列表上工作。我使用递归来定义这个函数，它非常直观。希望这是你在寻找的东西：

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

示例：

>>> 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]

Answer 12

您认为[1,2]与[1, [2]]相交吗？也就是说，它只是你关心的数字，还是列表结构？

如果只是数字，请调查如何“展平”列表，然后使用set()方法。

Answer 13

我也在寻找一种方法来做到这一点，并最终以这样结束：

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]

Answer 14

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]

Answer 15

可以通过reduce轻松创建

列表。

您需要使用初始化器-reduce函数中的第三个参数。

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

以上代码适用于python2和python3，但是您需要将reduce模块导入为from functools import reduce。有关详细信息，请参见下面的链接。

Answer 16

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 = [list(set(c2[i]).intersection(set(c1))) for i in xrange(len(c2))]

c3
->[[32, 13], [28, 13, 7], [1, 6]]

Answer 17

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

对我来说，这是非常优雅和快捷的方式：）

Answer 18

# 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]] ?

以下是设置c3并不涉及集合的一种方法：

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

但如果你只想使用一行，你可以这样做：

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

这是列表理解中的列表理解，这有点不寻常，但我认为你不应该有太多麻烦。

Answer 19

我们可以使用set方法：

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]]

   result = [] 
   for li in c2:
       res = set(li) & set(c1)
       result.append(list(res))

   print result

Answer 20

找到可迭代对象之间的差异和交集的简单方法

如果重复很重要，请使用此方法

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)

找到两个嵌套列表的交集？

相关

20 个答案:

纯列表理解版

列表。

找到可迭代对象之间的差异和交集的简单方法