如何仅使用 itertools 将 Python 列表的每个元素重复 n 次?

2023-10-19Python开发问题
0

本文介绍了如何仅使用 itertools 将 Python 列表的每个元素重复 n 次?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

我有一个带有数字的列表:数字 = [1, 2, 3, 4].

I have a list with numbers: numbers = [1, 2, 3, 4].

我想要一个列表,他们像这样重复 n 次(对于 n = 3):

I would like to have a list where they repeat n times like so (for n = 3):

[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4].

问题是我只想为此使用 itertools,因为我的性能非常有限.

The problem is that I would like to only use itertools for this, since I am very constrained in performance.

我尝试使用这个表达式:

I tried to use this expression:

list(itertools.chain.from_iterable(itertools.repeat(numbers, 3)))

但它给了我这样的结果:

But it gives me this kind of result:

[1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]

这显然不是我需要的.

有没有办法只使用 itertools 来做到这一点,而不使用排序、循环和列表推导?我能得到的最接近的是:

Is there a way to do this with itertools only, without using sorting, loops and list comprehensions? The closest I could get is:

list(itertools.chain.from_iterable([itertools.repeat(i, 3) for i in numbers])),

但它也使用列表理解,我想避免.

but it also uses list comprehension, which I would like to avoid.

推荐答案

由于不想使用列表推导式,下面是纯(+zip) itertools 方法——

Since you don't want to use list comprehension, following is a pure (+zip) itertools method to do it -

from itertools import chain, repeat

list(chain.from_iterable(zip(*repeat(numbers, 3))))
# [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

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