[erlang-questions] Can I do the same with a fold easily

[PAILLEAU Eric]

Hi,
forget. By switching order of tests, result is contrary.
time is equivalent.


Le 19/10/2015 23:41, PAILLEAU Eric a écrit :
> Hi,
> by doing some testing, strangely using (X rem 2 ) =:= 1  is almost
> always better than X band 1 =:= 1. On my machine at least.
>
> ______________________________________________________________________
> -module(odd_even).
>
> -export([go/0]).
>
>
> go() ->   List = lists:seq(1, 1000) ,
>            T1 = erlang:timestamp(),
>            lists:partition(fun (X) -> (X rem 2 ) =:= 1 end, List),
>            io:format("~p
> microseconds~n",[timer:now_diff(erlang:timestamp(), T1)]),
>
>            T3 = erlang:timestamp(),
>            lists:partition(fun (X) -> X band 1 =:= 1 end, List),
>            io:format("~p
> microseconds~n",[timer:now_diff(erlang:timestamp(), T3)]).
> ______________________________________________________________________
>
>
> Le 19/10/2015 22:53, Éric Pailleau a écrit :
>> Hi,
>> I agree but like Joe says: let it work,  then optimize if needed.
>> My main remark was: Splitting odd and even does not need to check odd
>> then even, but check it is odd Else it is even (or contrary).
>> Regards
>>
>> Le 19 oct. 2015 4:54 AM, ok@REDACTED a écrit :
>>>
>>> "Éric Pailleau" <eric.pailleau@REDACTED>
>>>
>>>> A number cannot be both even and odd, so simply take odd numbers in a
>>>> variable O from list L , even numbers will be L -- O.
>>>
>>> Here we run into the question "what does BEST mean?"
>>>
>>> Now Xs -- Ys cannot assume anything about the contents of Xs or Ys,
>>> other than them both being well formed lists.  So while it *could*
>>> be implemented in O(|Xs|+|Ys|+|Xs|lg|Ys|) time by building some kind
>>> of balanced search tree from Ys, or even better by building some
>>> kind of hashed set, the cost of building the supporting data structure
>>> would likely be worse than the cost of traversing L twice or using
>>> lists:partition/2.
>>>
>>>
>>>
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>
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