Sequence splitting

tsangpo tsangpo.newsgroup at gmail.com
Fri Jul 3 11:03:51 CEST 2009


Just a shorter implementation:

from itertools import groupby
def split(lst, func):
    gs = groupby(lst, func)
    return list(gs[True]), list(gs[False])


"Lie Ryan" <lie.1296 at gmail.com> дÈëÏûÏ¢ÐÂÎÅ:nfi3m.2341$ze1.1151 at news-server.bigpond.net.au...
> Brad wrote:
>> On Jul 2, 9:40 pm, "Pablo Torres N." <tn.pa... at gmail.com> wrote:
>>> If it is speed that we are after, it's my understanding that map and
>>> filter are faster than iterating with the for statement (and also
>>> faster than list comprehensions).  So here is a rewrite:
>>>
>>> def split(seq, func=bool):
>>>         t = filter(func, seq)
>>>         f = filter(lambda x: not func(x), seq)
>>>         return list(t), list(f)
>>>
>>
>> In my simple tests, that takes 1.8x as long as the original solution.
>> Better than the itertools solution, when "func" is short and fast. I
>> think the solution here would worse if func was more complex.
>>
>> Either way, what I am still wondering is if people would find a built-
>> in implementation useful?
>>
>> -Brad
>
> A built-in/itertools should always try to provide the general solution
> to be as useful as possible, something like this:
>
> def group(seq, func=bool):
>    ret = {}
>    for item in seq:
>        fitem = func(item)
>        try:
>            ret[fitem].append(item)
>        except KeyError:
>            ret[fitem] = [item]
>    return ret
>
> definitely won't be faster, but it is a much more general solution.
> Basically, the function allows you to group sequences based on the
> result of func(item). It is similar to itertools.groupby() except that
> this also group non-contiguous items. 





More information about the Python-list mailing list