如何使用itertools重复每个Python列表的元素n次？

Question

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

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] .

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

I tried to use this expression:

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

But it gives me this kind of result:

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

which is obviously not what I need.

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])) ,

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

Answer 1

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]

Answer 2

Firstly, using functions from itertools won't necessarily be faster than a list comprehension — you should benchmark both approaches. (In fact, on my machine it's the opposite).

Pure list comprehension approach:

>>> numbers = [1, 2, 3, 4]
>>> [y for x in numbers for y in (x,)*3]
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Using chain.from_iterable() with a generator expression :

>>> from itertools import chain, repeat
>>> list(chain.from_iterable(repeat(n, 3) for n in numbers))
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 3

Think you were very close, just rewriting the comprehension to a generator:

n = 3
numbers = [1, 2, 3, 4]
list(itertools.chain.from_iterable((itertools.repeat(i, n) for i in numbers)))