faster than list.extend()

[Hyunchul Kim]

Thank Tim, Raymond, and all.

In my test, Tim's t1 was fastest and Raymond's triple3 is very near to Tim's
t1.
Anyway both of them took only 6~7% time of my original .extend() version.

I think that following n_ple() is a good generalized function that was 1~3%
slower than t1() and triple3() for "triple".

*********
from itertools import *
def n_ple(inputlist, n):
    chain_from_iterable = chain.from_iterable; izip = izip
    return chain_from_iterable(izip(*[inputlist]*n))
*********

 def t2(i):
   r = range(3)
   return (x for x in i for _ in r)
 def t2_1(i):
   r = range(3)
   return [x for x in i for _ in r]

I didn't know that list(t2(inputlist)) is much faster than t2_1(inputlist),
it's gonna be a helpful tip for me.

Thank you all.

Hyunchul


On Tue, Nov 17, 2009 at 7:55 AM, Tim Chase <python.list at tim.thechases.com>wrote:

> Hyunchul Kim wrote:
>
>> Hi, all.
>>
>> I want to improve speed of following simple function.
>> Any suggestion?
>>
>> **********
>> def triple(inputlist):
>>  results = []
>>  for x in inputlist:
>>    results.extend([x,x,x])
>>  return results
>> **********
>>
>
> Several ways occur to me:
>
>  def t1(i):
>    for x in i:
>      yield x
>      yield x
>      yield x
>
>  def t2(i):
>    r = range(3)
>    return (x for x in i for _ in r)
>
>
> I prefer to return generators in case the input is large, but in both
> cases, you can just wrap it in list() like
>
>  list(t1(inputlist))
>
> or in t2(), you can just change it to use
>
>    return [x for x in i for _ in r]
>
> -tkc
>
>
>
>
>
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