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

Barco You <>
Fri Nov 18 16:08:33 CET 2011


Seemingly our methods are all not correct for measuring memory consumption.
If measuring in the middle as I did, we just get the instant value at a
moment, whereas if measuring after the function returns by waiting for
several seconds as you did, all the memory ever used have been garbage
collected. So I think what we need to do for profiling the functions is to
draw a curve with memory consumption against time elapsed. But I dont know
how to do that.
On Nov 18, 2011 5:41 PM, "Dmitry Demeshchuk" <> wrote:

> That's not the correct way. You were measuring the processes memory in
> the middle of calculations.
>
> 1> A = spawn(fun () -> test:a(), receive _ -> ok end end).
> <0.46.0>
> 2> B = spawn(fun () -> test:b(), receive _ -> ok end end).
> <0.48.0>
> 3> process_info(A, memory).
> {memory,391814792}
> 4> process_info(B, memory).
> {memory,401610112}
>
> Note that I waited for several seconds before entering commands 3 and 4.
>
> On Fri, Nov 18, 2011 at 1:32 PM, Barco You <> wrote:
> > 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
> >
> >
>
>
>
> --
> Best regards,
> Dmitry Demeshchuk
>
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