问题
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.
回答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]
回答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]
回答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)))
来源：https://stackoverflow.com/questions/45799233/how-to-repeat-each-of-a-python-lists-elements-n-times-with-itertools-only