用列表输出而不是元组压缩 [英] Zip with list output instead of tuple

问题描述

处理两个列表中的列表的最快，最优雅的方法是什么?

What is the fastest and most elegant way of doing list of lists from two lists?

我有

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

In [2]: b=[7,8,9,10,11,12]

In [3]: zip(a,b)
Out[3]: [(1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12)]

我想拥有

In [3]: some_method(a,b)
Out[3]: [[1, 7], [2, 8], [3, 9], [4, 10], [5, 11], [6, 12]]

我当时正在考虑使用map而不是zip，但我不知道是否有一些标准库方法可作为第一个参数.

I was thinking about using map instead of zip, but I don't know if there is some standard library method to put as a first argument.

我可以为此定义自己的功能，并使用map，我的问题是是否已经实现了某些功能. 否也是一个答案.

I can def my own function for this, and use map, my question is if there is already implemented something. No is also an answer.

推荐答案

如果您要压缩2个以上的列表(就此而言，甚至压缩2个)，一种可读的方式将是:

If you are zipping more than 2 lists (or even only 2, for that matter), a readable way would be:

[list(a) for a in zip([1,2,3], [4,5,6], [7,8,9])]

这使用列表推导，并将列表(元组)中的每个元素转换为列表.

This uses list comprehensions and converts each element in the list (tuples) into lists.

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