python - make a list with repetitive elements

Question

I know if I want to create a list like this:

[0 1 2 0 1 2 0 1 2 0 1 2]

I can use this command:

range(3) * 4

Is there a similar way to create a list like this:

[0 0 0 0 1 1 1 1 2 2 2 2]

I mean a way without loops

As a curiousity, all the current answers contain a loop (even if it is part of a list comprehension). I'd be interested to see if there's a way to do it without using for/in — ptr, Feb 03 '14 at 12:16
`sum(map(list, zip(*[ range(4) ] * 4)), [])` is without obvious loops ;-) — Alfe, Feb 03 '14 at 12:20

score 7 · Answer 1 · edited May 23 '17 at 10:31

7

Integer division can help:

[x/4 for x in range(12)]

Same thing through map:

map(lambda x: x/4, range(12))

In python 3 integer division is done with //.

Beware that multiplication of a list will likely lead to a result you probably don't expect.

edited May 23 '17 at 10:31

Community

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answered Feb 03 '14 at 12:15

sashkello

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score 6 · Answer 2 · answered Feb 03 '14 at 12:14

6

Yes, you can.

>>> [e for e in range(3) for _ in [0]*4]
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2]

answered Feb 03 '14 at 12:14

dersvenhesse

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Nigel Tufnel · Accepted Answer · 2014-02-03T12:29:50.637

3

itertools module is always an option:

>>> from itertools import chain, repeat
>>> list(chain(repeat(0, 4), repeat(1, 4), repeat(2, 4)))
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2]

More general way is:

def done(group_count, repeat_count):
    return list(chain(*map(lambda i: repeat(i, repeat_count),
                           range(group_count))))
>>> done(3, 4)
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2]

edited Feb 03 '14 at 12:29

answered Feb 03 '14 at 12:17

Nigel Tufnel

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Well that answers my question! I had a suspicion it would involve itertools. – ptr Feb 03 '14 at 12:19
This second method proved to be the fastest way to do this. – soroosh.strife Feb 03 '14 at 12:41

score 3 · Answer 4 · answered Feb 03 '14 at 12:22

3

Without any explicit "for" :)

>>> list(chain(*zip(*([range(5)] * 5))))
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4]

answered Feb 03 '14 at 12:22

UltraInstinct

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1

Yeah, no obvious loop ;-) – Alfe Feb 03 '14 at 12:23
This one is very fast as well – soroosh.strife Feb 03 '14 at 12:42

score 1 · Answer 5 · answered Feb 03 '14 at 12:13

1

What about this:

>>> sum([ [x]*4 for x in range(5)],[])
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]
>>>

or

>>> reduce(lambda x,y: x+y, [ [x]*4 for x in range(5)])
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]
>>>

answered Feb 03 '14 at 12:13

James Sapam

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score 0 · Answer 6 · edited Feb 03 '14 at 12:21

0

If you can't use a loop in your current method, create one in an other?

range(0,1)*4 + range(1,2)*4 + range(2,3)*4

edited Feb 03 '14 at 12:21

dersvenhesse

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answered Feb 03 '14 at 12:19

AlanF

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That doesn't scale ;-) – Alfe Feb 03 '14 at 12:22

python - make a list with repetitive elements

6 Answers6