如何使用列表推导将元组的元组转换为一维列表?

How do I convert a tuple of tuples to a one-dimensional list using list comprehension?(如何使用列表推导将元组的元组转换为一维列表?)

问题描述

我有一个元组 - 例如:

I have a tuple of tuples - for example:

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

我想将其转换为按顺序排列的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点.但我似乎无法弄清楚.我能够通过 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)

但我觉得必须有一种方法可以通过列表理解来做到这一点.

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

一个简单的 [list(tuple) for tupleOfTuples] 只是给你一个列表列表，而不是单个元素.我想我也许可以通过使用解包运算符来解包列表，如下所示:

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]

...但这没有用.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

通常称为扁平化嵌套结构.

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]

只是为了展示效率:

>>> 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，它会影响内置.

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

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