Python

Question

这是我的两个列表；

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]]
kDash = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]]

我的输出应该如下；

[[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

我怎样才能得到这个输出？

提前致谢

Answer 1

您必须将列表转换为元组列表，然后使用交集。请注意，下面的解决方案可能包含不同顺序的元素，并且显然不会出现重复项，因为我使用的是 set.

In [1]: l1 = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]]

In [2]: l2 = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]]

In [3]: [list(x) for x in set(tuple(x) for x in l1).intersection(set(tuple(x) for x in l2))]
Out[3]: [[1, 2], [5, 6, 2], [3], [4]]

您也可以将交集保存在变量中并获得最终列表，如果顺序，则需要重复：

In [4]: intersection = set(tuple(x) for x in l1).intersection(set(tuple(x) for x in l2))

In [5]: [x for x in l1 if tuple(x) in intersection]
Out[5]: [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

还有路口，有兴趣的可以看看。

In [6]: print intersection
set([(1, 2), (5, 6, 2), (3,), (4,)])

这对于大型列表非常有效，但如果列表很小，请探索@timegb 的其他解决方案（对于较长的列表，其解决方案将非常不理想）

Answer 2

因为你的输出列表有重复的元素，你似乎真的不需要经典的交集。基本的列表理解将完成一切。

>>> k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]]
>>> kDash = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]]
>>> [x for x in k if x in kDash]
[[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

对于大型列表，我们希望获得调用 __contains__ 所需的时间为 O(1) 而不是 O(n)：

>>> stuff_in_kDash = set(map(tuple, kDash))
>>> [x for x in k if tuple(x) in stuff_in_kDash]
[[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

Answer 3

更简洁的交集写法是

{tuple(x) for x in l1} & {tuple(x) for x in l2}

一个好的选择是

{tuple(x) for x in l1}.intersection(map(tuple, l2))

Answer 4

尽管这里写的是更优雅的解决方案，但这里还有另一个

def foo(L1,L2):
    res=[]
    for lst in L1:
        if lst in L2:
            res.append(lst)
    return res

Python - 两个列表列表的交集

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