[erlang-questions] [erlang-question] How to comprehend two lists synchronously?

Dmitry Demeshchuk <>
Fri Nov 18 08:49:40 CET 2011


I think it's meant that lists:reverse/1 is called at the end of the
_optimized_ code.

Here's the module code:

-module(test).
-export([a/0, b/0]).

a() ->
    L = lists:seq(1, 10000000),    map2(fun (I, J) -> I + J end, L, L).

b() ->
    L = lists:seq(1, 10000000),
    map3(fun (I, J) -> I + J end, L, L).


map2(Fun, [H1 | T1], [H2 | T2]) ->
    [Fun(H1, H2) | map2(Fun, T1, T2)];
map2(_, [], []) ->
    [].

map3(_Fun, [], []) ->
   [];
map3(Fun, L1, L2) ->
   map3(Fun, L1, L2, []).

map3(_Fun, [], [], L) ->
   lists:reverse(L);
map3(Fun, [H1 | T1], [H2 | T2], L) ->
   map3(Fun, T1, T2, [Fun(H1, H2) | L])

Try to call timer:tc/3 yourself.

On Fri, Nov 18, 2011 at 11:48 AM, Barco You <> wrote:
> According to the instruction attached by Ulf, the body-recursive and
> tail-recursive list function will be the same in consuming memory only when
> they call lists:reverse/1 at the end.
> So, I don't know how did you do the benchmarks. Did you compare these two
> methods with big enough lists?
> Or, I misunderstand the optimization instructions?
>
> BR,
> Barco
>
> On Fri, Nov 18, 2011 at 3:37 PM, Dmitry Demeshchuk <>
> wrote:
>>
>> Okay, I admit, this isn't an "honest" tail-recursed function, since a
>> list concatenation operator is going to be called at the end. However,
>> Erlang compiler optimizes such cases and converts them to
>> tail-recursive:
>> http://www.erlang.org/doc/efficiency_guide/myths.html#tail_recursive
>>
>> Also, I've ran benchmarks with both implementations: mine and yours.
>> And they result in the same performance.
>>
>> On Fri, Nov 18, 2011 at 11:06 AM, Barco You <> wrote:
>> > Yes, Ryan's suggestion is a good generic solution for n lists and it's
>> > tail-recursed.
>> > Hi Dmitry,
>> > Your version is just recursed but not tail-recursed, because your
>> > function
>> > needs a piece of memory to stack the intermediate result for every round
>> > of
>> > recursive calls. To be tail-recursed, the recursive calls should
>> > eliminate
>> > the linearly-increased memory consumption by adding an extra variable
>> > (accumulator) and let the recursive function call it alone for every
>> > round.
>> >
>> > On Fri, Nov 18, 2011 at 2:53 PM, Dmitry Demeshchuk
>> > <>
>> > wrote:
>> >>
>> >> Hi, Barco.
>> >>
>> >> Why do you think my version isn't tail-recursed? :) Take a look at
>> >> lists:map/2 implementation, for example. It's just the same.
>> >>
>> >> List comprehensions just serve different purpose: for combinations
>> >> from multiple list sources. My guess is that people need this
>> >> operation more often than mapping over multiple list. Another problem
>> >> is that you should be sure that all those lists have the same length.
>> >>
>> >> On Fri, Nov 18, 2011 at 10:38 AM, Barco You <> wrote:
>> >> > Hi Dmitry,
>> >> > What your suggested can really solve my problem, but it's not
>> >> > Tail-Recursion. The tail-recursed solution should look like this;
>> >> > map2(_Fun, [], []) ->
>> >> >    [];
>> >> > map2(Fun, L1, L2) ->
>> >> >    map2(Fun, L1, L2, []).
>> >> > map2(_Fun, [], [], L) ->
>> >> >    lists:reverse(L);
>> >> > map2(Fun, [H1 | T1], [H2 | T2], L) ->
>> >> >    map2(Fun, T1, T2, [Fun(H1, H2) | L]).
>> >> >
>> >> > However, I'm still disappointed with the list comprehension which is
>> >> > different from what I intuitively imagine about it.
>> >> >
>> >> > Regards,
>> >> > Barco
>> >> > On Fri, Nov 18, 2011 at 1:49 PM, Dmitry Demeshchuk
>> >> > <>
>> >> > wrote:
>> >> >>
>> >> >> My guess is you have to zip them together, or just write a
>> >> >> tail-recursed function:
>> >> >>
>> >> >> map2(Fun, [H1 | T1], [H2 | T2]) ->
>> >> >>    [Fun(H1, H2) | map2(Fun, T1, T2)];
>> >> >> map2(Fun, [], []) ->
>> >> >>    [].
>> >> >>
>> >> >> The second option definitely isn't a list comprehension, but it
>> >> >> requires less memory and has lesser complexity.
>> >> >>
>> >> >> On Fri, Nov 18, 2011 at 9:45 AM, Barco You <>
>> >> >> wrote:
>> >> >> > Dear Erlangers,
>> >> >> >
>> >> >> > I hope to get a list from two lists like this:
>> >> >> > [{a1,b1}, {a2,b2}, {a3,b3}]      <-     [a1, a2 a3],  [b1, b2,
>> >> >> > b3].
>> >> >> > But if I use list comprehension, I got:
>> >> >> > 10>  [{D1,D2} ||  D1 <- [a1,a2,a3], D2 <- [b1,b2,b3]].
>> >> >> > [{a1,b1},
>> >> >> >  {a1,b2},
>> >> >> >  {a1,b3},
>> >> >> >  {a2,b1},
>> >> >> >  {a2,b2},
>> >> >> >  {a2,b3},
>> >> >> >  {a3,b1},
>> >> >> >  {a3,b2},
>> >> >> >  {a3,b3}]
>> >> >> >
>> >> >> > So, my questions is how to comprehend list in synchronous way in
>> >> >> > order
>> >> >> > to
>> >> >> > get what I want, rather than to compose the elements from two
>> >> >> > lists
>> >> >> > in
>> >> >> > all
>> >> >> > possible situations.
>> >> >> >
>> >> >> > Thank you,
>> >> >> > Barco
>> >> >> > _______________________________________________
>> >> >> > erlang-questions mailing list
>> >> >> > 
>> >> >> > http://erlang.org/mailman/listinfo/erlang-questions
>> >> >> >
>> >> >> >
>> >> >>
>> >> >>
>> >> >>
>> >> >> --
>> >> >> Best regards,
>> >> >> Dmitry Demeshchuk
>> >> >
>> >> >
>> >>
>> >>
>> >>
>> >> --
>> >> Best regards,
>> >> Dmitry Demeshchuk
>> >
>> >
>>
>>
>>
>> --
>> Best regards,
>> Dmitry Demeshchuk
>
>



-- 
Best regards,
Dmitry Demeshchuk



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