1

我有一个如下所示的主列表：

a = [ [[1,2,3],[4,5,6]], [[7,8,9],[10,11,12]], [[13,14,15],[16,17,18]] ]

我需要重新排列它以使其看起来像这样：

b = [[1,2,3,7,8,9,13,14,15], [4,5,6,10,11,12,16,17,18]]

即：获取主列表的每个子列表中的第一个子列表中的所有元素，并将它们放在0一个新列表位置的单个子列表中，然后对第二个子列表执行相同操作，并将所有元素放在单个子列表的位置1这份新名单。

我已经尝试zip结合在 Python 中从列表列表中制作一个平面列表中给出的解决方案，但我就是无法让它工作。

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4 回答 4

2

zip与*和一起使用itertools.chain.from_iterable。

>>> from itertools import chain, izip
>>> [list(chain.from_iterable(x)) for x in zip(*a)] #or `izip`
[[1, 2, 3, 7, 8, 9, 13, 14, 15], [4, 5, 6, 10, 11, 12, 16, 17, 18]]

这里zip(*a)返回：

[([1, 2, 3], [7, 8, 9], [13, 14, 15]), ([4, 5, 6], [10, 11, 12], [16, 17, 18])]

现在您可以使用chain.from_iterable(x).

时间比较：

In [1]: from itertools import izip, chain
In [2]: a = [[[1,2,3],[4,5,6]], [[7,8,9],[10,11,12]], [[13,14,15],[16,17,18]]]
In [3]: %timeit [list(chain.from_iterable(x)) for x in zip(*a)]
100000 loops, best of 3: 3.71 us per loop
In [4]: %timeit [[i for v in r for i in v] for r in zip(*a)]
100000 loops, best of 3: 2.73 us per loop

In [5]: b = a *100
In [6]: %timeit [list(chain.from_iterable(x)) for x in zip(*b)]
10000 loops, best of 3: 97.6 us per loop
In [7]: %timeit [list(chain.from_iterable(x)) for x in izip(*b)]
10000 loops, best of 3: 97.6 us per loop
In [8]: %timeit [[i for v in r for i in v] for r in zip(*b)]
10000 loops, best of 3: 144 us per loop
In [9]: %timeit [[i for v in r for i in v] for r in izip(*b)]
10000 loops, best of 3: 143 us per loop

In [10]: c = a*10000
In [11]: %timeit [list(chain.from_iterable(x)) for x in zip(*c)]
100 loops, best of 3: 12.9 ms per loop
In [12]: %timeit [list(chain.from_iterable(x)) for x in izip(*c)]
100 loops, best of 3: 12.3 ms per loop
In [13]: %timeit [[i for v in r for i in v] for r in zip(*c)]
100 loops, best of 3: 17 ms per loop
In [14]: %timeit [[i for v in r for i in v] for r in izip(*c)]
100 loops, best of 3: 17.1 ms per loop
于 2013-11-11T13:36:28.970 回答
1 
>>> a = [[[1,2,3],[4,5,6]], [[7,8,9],[10,11,12]], [[13,14,15],[16,17,18]]]
>>>
>>> [[i for v in r for i in v] for r in zip(*a)]
[[1, 2, 3, 7, 8, 9, 13, 14, 15], [4, 5, 6, 10, 11, 12, 16, 17, 18]]

看：zip()

于 2013-11-11T13:37:35.470 回答
1 
a = [ [[1,2,3],[4,5,6]], [[7,8,9],[10,11,12]], [[13,14,15],[16,17,18]] ]
b = [[item for sitems in items for item in sitems] for items in zip(*a)]
print b

输出

[[1, 2, 3, 7, 8, 9, 13, 14, 15], [4, 5, 6, 10, 11, 12, 16, 17, 18]]
于 2013-11-11T13:39:24.567 回答
0

可以使用 sum() 将空列表作为第一个元素来展平列表，因此您可以使用以下方法：

[sum((item[0] for item in a), []), sum((item[1] for item in a), [])]

或者这个，对于任意数量的内部元素：

[sum((item[i] for item in a), []) for i in range(len(a[0]))]
于 2013-11-11T16:43:16.660 回答