如何使用列表理解将一个元组的元组转换为一维列表? [英] How do I convert a tuple of tuples to a one-dimensional list using list comprehension?

问题描述

我有一个元组元组-例如:

I have a tuple of tuples - for example:

tupleOfTuples = ((1, 2), (3, 4), (5,))

我想按顺序将其转换为所有元素的平面一维列表:

I want to convert this into a flat, one-dimensional list of all the elements in order:

[1, 2, 3, 4, 5]

我一直在尝试通过列表理解来实现这一目标.但我似乎无法弄清楚.我能够通过一个for-each循环来完成它:

I've been trying to accomplish this with list comprehension. But I can't seem to figure it out. I was able to accomplish it with a for-each loop:

myList = []
for tuple in tupleOfTuples:
   myList = myList + list(tuple)

但是我觉得必须有一种方法来理解列表.

But I feel like there must be a way to do this with a list comprehension.

一个简单的[list(tuple) for tuple in tupleOfTuples]只是给您一个列表列表，而不是单个元素.我以为可以在此基础上使用拆包运算符来对列表进行拆包，就像这样:

A simple [list(tuple) for tuple in tupleOfTuples] just gives you a list of lists, instead of individual elements. I thought I could perhaps build on this by using the unpacking operator to then unpack the list, like so:

[*list(tuple) for tuple in tupleOfTuples]

或

[*(list(tuple)) for tuple in tupleOfTuples]

...但是那没用.有任何想法吗?还是我应该坚持循环?

... but that didn't work. Any ideas? Or should I just stick to the loop?

推荐答案

通常称为展平嵌套结构.

it's typically referred to as flattening a nested structure.

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> [element for tupl in tupleOfTuples for element in tupl]
[1, 2, 3, 4, 5]

只是为了证明效率:

>>> import timeit
>>> it = lambda: list(chain(*tupleOfTuples))
>>> timeit.timeit(it)
2.1475738355700913
>>> lc = lambda: [element for tupl in tupleOfTuples for element in tupl]
>>> timeit.timeit(lc)
1.5745135182887857

ETA :请不要使用tuple作为变量名，因为它是内置变量.

ETA: Please don't use tuple as a variable name, it shadows built-in.

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