问题描述
我有一个带有数字的列表:numbers = [1, 2, 3, 4].
I have a list with numbers:numbers = [1, 2, 3, 4].
我希望有一个列表，其中他们会重复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].
问题是我只想使用itertools，因为我对性能的限制很大.
The problem is that I would like to only use itertools for this, since I am very constrained in performance.
我尝试使用此表达式:
list(itertools.chain.from_iterable(itertools.repeat(numbers, 3)))
但这给了我这种结果:
[1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]
这显然不是我所需要的.
which is obviously not what I need.
有没有办法仅使用itertools来执行此操作，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]))，
但是它也使用列表理解，这是我想避免的.
but it also uses list comprehension, which I would like to avoid.
推荐答案
由于您不想使用列表推导，因此以下是一种纯粹的(+ 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]
这篇关于如何仅使用itertools将Python列表的每个元素重复n次?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！