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?
解决方案
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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