合并多个2d列表,按顺序考虑轴

Question

我的目的是按顺序组合多个2d列表,例如:

a = [[1,2],[3,1]]
b= [[3,6],[2,9]]
c = [[5,1],[8,10]]
Expected: [[1,2,3,6,5,1],[3,1,2,9,8,10]]

根据该网站的其他建议,我尝试使用收集模块,如下面的代码:

from collections import Counter
a = [[1,2],[3,1]]
b= [[3,6],[2,9]]
c = [[5,1],[8,10]]
d = [[k,v] for k,v in (Counter(dict(a)) + Counter(dict(b))+ Counter(dict(c))).items()]
print d

但是,结果[[1, 2], [3, 1], [3, 6], [2, 9]]不是我所期望的.

你有什么想法解决这个问题吗？也许如果有功能或模块考虑轴来组合列表.

Answer 1

您可以使用zip和列表理解:

>>> a = [[1,2],[3,1]]
>>> b = [[3,6],[2,9]]
>>> c = [[5,1],[8,10]]
>>> [x+y+z for x,y,z in zip(a, b, c)]
[[1, 2, 3, 6, 5, 1], [3, 1, 2, 9, 8, 10]]
>>>

Answer 2

你可以使用itertools.chain.from_iterable():

>>> a = [[1, 2], [3, 1]]
>>> b = [[3, 6], [2, 9]]
>>> c = [[5, 1], [8, 10]]
>>> from itertools import chain
>>> [list(chain.from_iterable(x)) for x in zip(a, b, c)]
[[1, 2, 3, 6, 5, 1], [3, 1, 2, 9, 8, 10]]

如果您有任意数量的2D列表,这可能很方便 - 例如:

>>> list_of_lists = [
...     [[1, 2], [3, 1]],
...     [[3, 6], [2, 9]],
...     [[5, 1], [8, 10]],
...     # ...
...     [[4, 7], [11, 12]]
... ]
>>> [list(chain.from_iterable(x)) for x in zip(*list_of_lists)]
[[1, 2, 3, 6, 5, 1, ..., 4, 7], [3, 1, 2, 9, 8, 10, ..., 11, 12]]

请注意*前list_of_lists一个示例中的before ,这是参数解包的示例.