Repeat list n times, sorted by original order

Question

I cannot find a solution to the following problem I have a list of strings, say

base_list = ['d', 'ssa', 'aaa', 'bb']

and I want to repeat this list n times but I need to keep the order of the original list. For example with n=2 the result should be:

final_list =  ['d', 'd', 'ssa', 'ssa', 'aaa', 'aaa', 'bb', 'bb']

EDIT: Not sure if I should mark this question as duplicate or not, since it is in my opinion more generic than the possible duplicate Duplicate element in python list

http://stackoverflow.com/questions/14878538/duplicate-element-in-python-list — Maroun, Aug 05 '15 at 07:55
That solves it, despite that the original question asked for duplicates and not the case where `n` is arbitrary. Probably why I have not found the question in first place. — MMCM_, Aug 05 '15 at 08:08

score 3 · Answer 1 · answered Aug 05 '15 at 08:04

>>> base_list = ['d', 'ssa', 'aaa', 'bb']
>>> n = 2
>>> [i for i in base_list for _ in xrange(n)]
['d', 'd', 'ssa', 'ssa', 'aaa', 'aaa', 'bb', 'bb']

Inside list comprehension you iterate through range with n length (inside loop) and through original list with outer one.

score 0 · Answer 2 · answered Aug 05 '15 at 08:00

The following should do what you're looking for

>>> base_list = ['d', 'ssa', 'aaa', 'bb']
>>> final_list = []
>>> n = 10
>>> for item in base_list:
...     for i in range(n):
...             final_list.append(item)
... 
>>> final_list
['d', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'ssa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'aaa', 'bb', 'bb', 'bb', 'bb', 'bb', 'bb', 'bb', 'bb', 'bb', 'bb']

score 0 · Answer 3 · edited May 23 '17 at 11:51

0

It is already explained here.

Applied to your code:

>>> base_list = ['d', 'ssa', 'aaa', 'bb']
>>> from itertools import repeat
>>> final_list = [x for item in base_list for x in repeat(item, 2)]
>>> final_list
['d', 'd', 'ssa', 'ssa', 'aaa', 'aaa', 'bb', 'bb']

edited May 23 '17 at 11:51

Community

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answered Aug 05 '15 at 08:00

ezcoding

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score 0 · Answer 4 · answered Aug 05 '15 at 08:02

0

I guess one answer from the tagged question (where n=2) also applies to the more generic case:

from itertools import chain, repeat

final_list = list(chain.from_iterable(repeat(item, n) for item in base_list))

answered Aug 05 '15 at 08:02

MMCM_

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This is good one, but for small lists its kinda overkill. – Slam Aug 05 '15 at 08:08

score 0 · Answer 5 · answered Aug 05 '15 at 08:12

0

My first thought was to zip it with itself, then flatten:

import itertools
base_list = ['d', 'ssa', 'aaa', 'bb']
n = 2
list(itertools.chain.from_iterable(zip(*[base_list] * n)))

answered Aug 05 '15 at 08:12

o11c

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This probably runs fastest, without any imports: Line 1: `n = 2` Line 2: `list(sum(zip(*[base_list] * n), ()))` – The Brofessor Aug 05 '15 at 08:17
`sum` is quadratic rather than linear though. – o11c Aug 05 '15 at 18:24

Repeat list n times, sorted by original order

5 Answers5