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

Barco You <>
Fri Nov 18 10:32:04 CET 2011


Very strange! By using another measuring method as below I got that your
function is far better than mine in memory consumption.


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

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

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


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]).


Eshell V5.8.4  (abort with ^G)
1> A = spawn(test,a,[]).
<0.32.0>
2> B = spawn(test,b,[]).
<0.34.0>
3> erlang:process_info(A,memory).
{memory,176316700}
4> erlang:process_info(B,memory).
{memory,306105040}




On Fri, Nov 18, 2011 at 4:53 PM, Dmitry Demeshchuk <>wrote:

> I'm not sure it is so. Try running this function several times:
>
> memtest() ->
>    erlang:garbage_collect(),
>    M1 = proplists:get_value(total, erlang:memory()),
>    a(),
>    M2 = proplists:get_value(total, erlang:memory()),
>    b(),
>    M3 = proplists:get_value(total, erlang:memory()),
>    {M2 - M1, M3 - M2}.
>
> The results are pretty the same.
>
> On Fri, Nov 18, 2011 at 12:36 PM, Barco You <> wrote:
> > Umm! I don't think there would be some difference between times consumed
> by
> > these two functions, but I will assume there are difference in memory
> > consumption.
> >
> >
> > On Fri, Nov 18, 2011 at 3:49 PM, Dmitry Demeshchuk <
> >
> > wrote:
> >>
> >> 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
> >
> >
>
>
>
> --
> Best regards,
> Dmitry Demeshchuk
>
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