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.1...@gmail.com>
дÈëÏûÏ¢ÐÂÎÅ:nfi3m.2341$ze1.1...@news-server.bigpond.net.au...
> Brad wrote:
>> On Jul 2, 9:40 pm, "Pablo Torres N." <tn.pa...@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.
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