Question

我有两个嵌套列表：

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

我希望得到以下结果[（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成带有元组对的列表？

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list comprehension内使用zip两次：

>>> 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

您需要展开列表，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

惯用方法是使用星号表示法和itertools.chain在压缩列表之前展平列表。星号表示法将一个iterable解包为函数的参数，而itertools.chain函数将其参数中的iterables链接在一起成为一个可迭代的。



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

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

将两个嵌套列表中的项连接到元组中的对

4 个答案: