将两个嵌套列表中的项目连接成元组对

Question

我有两个嵌套列表：

ls1 = [["a","b"], ["c","d"]]
ls2 = [["e","f"], ["g","h"]]

我想要以下结果[（a，e），（b，f），（c，g），（d，h）]

我已经尝试过zip（a，b），如何将嵌套列表压缩到具有元组对的列表中？

Answer 1

您还可以使用itertools.chain.from_iterable和zip ：

>>> ls1 = [["a","b"], ["c","d"]]
>>> ls2 = [["e","f"], ["g","h"]]
>>> 
>>> zip(itertools.chain.from_iterable(ls1), itertools.chain.from_iterable(ls2))
[('a', 'e'), ('b', 'f'), ('c', 'g'), ('d', 'h')]

Answer 2

您可以在列表理解内两次使用zip ：

>>> ls1 = [["a","b"], ["c","d"]]
>>> ls2 = [["e","f"], ["g","h"]]
>>> [y for x in zip(ls1, ls2) for y in zip(*x)]
[('a', 'e'), ('b', 'f'), ('c', 'g'), ('d', 'h')]
>>>

Answer 3

您需要拼合列表，并可以使用reduce ：

from functools import reduce # in Python 3.x
from operator import add
zip(reduce(add, ls1), reduce(add, ls2))

Answer 4

惯用的方法是使用星号和itertools.chain压缩列表，然后再压缩它们。 星号表示法将一个可迭代对象分解为函数的参数，而itertools.chain函数将其参数中的可迭代对象链接到一个可迭代对象中。

    ls1 = [["a","b"], ["c","d"]]
    ls2 = [["e","f"], ["g","h"]]

    import itertools as it
    zip(it.chain(*ls1), it.chain(*ls2))