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How to efficiently read thousand of lines from STDIN in Erlang?
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Closed 2 years ago.
A common programming problem which I've done in Python, Java, etc. with no problem. The following in Erlang (which I'm just learning) runs very slowly (~44s user time for 10^5 operations), and I am lost as to why.
As written on HackerRank, the program takes one line from stdin with an integer representing the number of operations that follow. Each subsequent line should be 1 X (enqueue X), 2 (dequeue and discard), or 3 (peek and print the next value on the queue).
Am I using lists:reverse/1 incorrectly?
-module(two_stacks).
%% API exports
-export([main/1]).
enqueue(Num, F, B) ->
{[Num | F], B}.
dequeue(F, []) ->
[_|B] = lists:reverse(F),
{[], B};
dequeue(F, [_|B]) ->
{F, B}.
peek(F, []) ->
[H|T] = lists:reverse(F),
io:format(H),
{[], [H|T]};
peek(F, [H|T]) ->
io:format(H),
{F, [H|T]}.
dispatchOperation(_, {F, B}) ->
[Code|Line] = io:get_line(""),
case Code of
49 ->
[_|Num] = Line,
enqueue(Num, F, B);
50 -> dequeue(F, B);
51 -> peek(F, B)
end.
main(_) ->
{Count, _} = string:to_integer(io:get_line("")),
_ = lists:foldl(fun dispatchOperation/2, {[], []}, lists:seq(1, Count)),
erlang:halt(0).
https://www.hackerrank.com/challenges/queue-using-two-stacks/problem
Are you running an escript? If that's the case, you should add a -mode(compile). there, because otherwise it runs the script in interpreted mode.
Also, you can compare the times against using the queue module (which is implemented using two stacks)
The issue is with the way I'm parsing input. See 46493207 for discussion.
Since all the inputs are integers I was able to make use of the same technique used there. The completed code is:
-module(solution).
-export([main/0]).
enqueue(Num, F, B) ->
{[Num | F], B}.
dequeue(F, []) ->
[_|B] = lists:reverse(F),
{[], B};
dequeue(F, [_|B]) ->
{F, B}.
peek(F, []) ->
[H|T] = lists:reverse(F),
io:format("~B~n", [H]),
{[], [H|T]};
peek(F, [H|T]) ->
io:format("~B~n", [H]),
{F, [H|T]}.
run(_, {F, B}) ->
Line = io:get_line(""),
[X| Y] = binary:split(Line, [<<$\s>>, <<$\n>>], [global]),
Code = binary_to_integer(X),
case Code of
1 ->
Num = binary_to_integer(lists:nth(1, Y)),
enqueue(Num, F, B);
2 -> dequeue(F, B);
3 -> peek(F, B)
end.
main() ->
ok = io:setopts(standard_io, [binary]),
{Count, _} = string:to_integer(io:get_line("")),
_ = lists:foldl(fun run/2, {[], []}, lists:seq(1, Count)),
erlang:halt(0).
I'm trying to make a sumif function in Erlang that would return a sum of all elements in a list if the predicate function evaluates to true. Here is what I have:
sumif(_, []) -> undefined;
sumif(Fun, [H|T]) -> case Fun(H) of
true -> H + sumif(Fun, T);
false -> sumif(Fun, T)
end.
I also implemented my own pos function which returns true if a number is greater than 0 and false otherwise:
pos(A) -> A > 0.
I tried using pos with sumif but I'm getting this error:
exception error: bad function pos
Why is this happening? Is it because of my sumif function or pos? I have tested pos on its own and it seems to work just fine.
Edit: It might be because how I'm calling the function. This is how I'm currently calling it: hi:sumif(pos,[-1,1,2,-3]). Where hi is my module name.
Is it because of my sumif function or pos?
It's because of sumif. You should return 0 when an empty list is passed, as it'll be called from the 2nd clause when T is []:
-module(a).
-compile(export_all).
sumif(_, []) -> 0;
sumif(Fun, [H|T]) -> case Fun(H) of
true -> H + sumif(Fun, T);
false -> sumif(Fun, T)
end.
pos(A) -> A > 0.
Test:
1> c(a).
{ok,a}
2> a:sumif(fun a:pos/1, [-4, -2, 0, 2, 4]).
6
List comprehensions make things far simpler:
sumif(F, L) ->
lists:sum([X || X <- L, F(X)]).
Dobert's answer is of cousrse right, problem is your sum for empty list.
If your concern is performance a little bit you should stick to tail recursive solution (in this case it matter because there is not lists:reverse/1 involved).
sumif(F, L) ->
sumif(F, L, 0).
sumif(F, [], Acc) when is_function(F, 1) -> Acc;
sumif(F, [H|T], Acc) ->
New = case F(H) of
true -> H+Acc;
false -> Acc
end,
sumif(F, T, New).
Ways how to make correct function for first parameter:
F1 = fun pos/1, % inside module where pos/1 defined
F2 = fun xyz:pos/1, % exported function from module xyz (hot code swap works)
N = 0,
F3 = fun(X) -> X > N end, % closure
% test it
true = lists:all(fun(F) -> is_function(F, 1) end, [F1, F2, F3]).
There has tow error in your code:
1. sumif(_, []) -> undefined; should return 0, not undefined.
2. when you pass pos(A) -> A > 0. to sumif/2,you should use fun pos/1, please read http://erlang.org/doc/programming_examples/funs.html#id59138
sumif(F, L) ->
lists:foldl(fun(X, Sum) when F(X) -> Sum+X; (_) -> Sum end, 0, L).
You can use lists:foldl.
The following is my solution to Project Euler 14, which works (in 18 s):
%Which starting number, under one million, produces the longest Collartz chain?
-module(soln14).
-export([solve/0]).
collatz(L) ->
[H|T] = L,
F = erlang:get({'collatz', H}),
case is_list(F) of
true ->
R = lists:append(F, T);
false ->
if H == 1 ->
R = L;
true ->
if H rem 2 == 0 ->
R = collatz([H div 2 | L]);
true ->
R = collatz([3*H+1 | L])
end
end,
erlang:put({'collatz', lists:last(L)}, R),
R
end.
dosolve(N, Max, MaxN, TheList) ->
if N == 1000000 -> MaxN;
true ->
L = collatz([N]),
M = length(L),
if M > Max -> dosolve(N+1, M, N, L);
true ->
dosolve(N+1, Max, MaxN, TheList)
end
end.
solve() ->
{Megass, Ss, Micros} = erlang:timestamp(),
S = dosolve(1, -1, 1, []),
{Megase, Se, Microe} = erlang:timestamp(),
{Megase-Megass, Se-Ss, Microe-Micros, S}.
However, the compiler complains:
8> c(soln14).
soln14.erl:20: Warning: variable 'R' is unused
{ok,soln14}
9> soln14:solve().
{0,18,-386776,837799}
Is this a compiler scoping error, or do I have a legit bug?
It's not a compiler error, just a warning that in the true case of "case is_list(F) of", the bindning of R to the result of lists:append() is pointless, since this value of R will not be used after that point, just returned immediately. I'll leave it to you to figure out if that's a bug or not. It may be that you are fooled by your indentation. The lines "erlang:put(...)," and "R" are both still within the "false" case of "case is_list(F) of", and should be deeper indented to reflect this.
The error message and the code are not "synchronized". with the version you give, the warning is on line 10: R = lists:append(F, T);.
What it means is that you bind the result of the lists:append/2 call to R and that you don't use it later in the true statement.
this is not the case in the false statement since you use R in the function erlang:put/2.
You could write the code this way:
%Which starting number, under one million, produces the longest Collartz chain?
-module(soln14).
-export([solve/0,dosolve/4]).
collatz(L) ->
[H|T] = L,
F = erlang:get({'collatz', H}),
case is_list(F) of
true ->
lists:append(F, T);
false ->
R = if H == 1 ->
L;
true ->
if H rem 2 == 0 ->
collatz([H div 2 | L]);
true ->
collatz([3*H+1 | L])
end
end,
erlang:put({'collatz', lists:last(L)}, R),
R
end.
dosolve(N, Max, MaxN, TheList) ->
if N == 1000000 -> MaxN;
true ->
L = collatz([N]),
M = length(L),
if M > Max -> dosolve(N+1, M, N, L);
true ->
dosolve(N+1, Max, MaxN, TheList)
end
end.
solve() ->
timer:tc(?MODULE,dosolve,[1, -1, 1, []]).
Warning the code uses a huge amount of memory, collatz is not tail recursive, and it seems that there is some garbage collecting witch is not done.
What's the Erlang equivalent to the following Python code:
for x in range(9):
for y in range(9):
for z in range(9):
foo(x, y, z)
I know I can generate the product first with C = [{X,Y,Z} || X<- lists:seq(1,9), Y<- lists:seq(1,9), Z<- lists:seq(1,9)] then foo([])->done; foo([H|T])->blah blah.
How do I do it without an auxiliary list, using recursion only?
You could do it with three recursive functions.
You might be able to do it with some complex pattern-matching in function head.
But easiest way to skip creation of auxiliary list is to call your function inside list comprehension
C = [foo(X, Y, Z) || X<- lists:seq(1,9),
Y<- lists:seq(1,9),
Z<- lists:seq(1,9)]
Where foo/3 process one element.
List comprehension still forces you to create auxiliary lists in memory.
In case of dealing with huge data sets you should avoid it. Writing recursive functions every time is also awkward so i came up with my own generic for function. It's a little bit slower in traversing than direct recursion or list comprehension but it's memory stable, generic and easy to use.
Usage:
(for({10}))(
fun (X) -> io:format("~p ",[X]) end).
> 1 2 3 4 5 6 7 8 9 10
(for({10, -10, -2}))(
fun (X) -> io:format("~p ",[X]) end).
> 10 8 6 4 2 0 -2 -4 -6 -8 -10
Works with lists too:
(for(lists:seq(10, -10, -2)))(
fun (X) -> io:format("~p ",[X]) end).
> 10 8 6 4 2 0 -2 -4 -6 -8 -10
It's also possible to define step or guard as a function:
(for({256, 1.1, fun (X) -> math:sqrt(X) end, fun (X, Range) -> X > Range end}))(
fun (X) -> io:format("~p ",[X]) end).
> 256 16.0 4.0 2.0 1.4142135623730951 1.189207115002721
If you pass to for a two parameter function, then you can use accumulator feature just like with lists:foldl/3. You also need to pass initial accumulator to for:
Fact = (for(1, {1, 5}))(
fun(X, Acc) ->
X * Acc
end),
io:format("~p", [Fact]).
> 120
e_fact(N) ->
{_, E} = (for({1, 1}, {1, N}))( % i assumed 1/0! equals 1
fun(X, {LastFact, Sum}) ->
Fact = LastFact * X,
{Fact, Sum + 1 / Fact}
end),
E.
io:format("e=~p", [e_fact(10)]).
> e=2.7182818011463845
Also step and guard functions can be dependent on accumulator. Just pass function with one more parameter.
Nested loops finding Pythagorean triples. Easy with closures:
pyth_lists(N) ->
[io:format("~p ", [{A, B, C}]) ||
A <- lists:seq(1, N),
B <- lists:seq(A + 1, N),
C <- lists:seq(B + 1, N),
A * A + B * B == C * C].
pyth_for(N) ->
(for({1, N}))(
fun(A) ->
(for({A + 1, N}))(
fun(B) ->
(for({B + 1, N}))(
fun(C) ->
case A * A + B * B == C * C of
true -> io:format("~p ", [{A, B, C}]);
false -> ok
end
end)
end)
end).
It's too small for external repository. I keep it in my utilities module.
If you find it helpful, here is code:
-export([for/1, for/2]).
for(Through) ->
for([], Through).
for(InitAcc, Opts) when is_tuple(Opts) ->
{Init, Range, Step, Guard} = for_apply_default_opts(Opts),
fun(Fun) ->
UpdFun = if
is_function(Fun, 1) ->
fun(I, _FAcc) -> Fun(I) end;
is_function(Fun, 2) ->
Fun
end,
for_iter(UpdFun, InitAcc, Init, Range, Step, Guard) end;
for(InitAcc, List) when is_list(List) ->
fun(Fun) -> for_list_eval(Fun, InitAcc, List) end.
for_iter(Fun, Acc, I, Range, Step, Guard) ->
case Guard(I, Range, Acc) of
false ->
Acc;
true ->
NewAcc = Fun(I, Acc),
for_iter(Fun, NewAcc, Step(I, NewAcc), Range, Step, Guard)
end.
for_list_eval(Fun, Acc, List) ->
if
is_function(Fun, 1) ->
lists:foreach(Fun, List);
is_function(Fun, 2) ->
lists:foldl(Fun, Acc, List)
end.
for_apply_default_opts({Range}) ->
DefaultInit = 1,
for_apply_default_opts({DefaultInit, Range});
for_apply_default_opts({Init, Range}) ->
DefaultStep = 1,
for_apply_default_opts({Init, Range, DefaultStep});
for_apply_default_opts({Init, Range, Step}) ->
DefaultGuard = case (Step > 0) or is_function(Step) of
true -> fun(I, IterRange, _Acc) -> I =< IterRange end;
false -> fun(I, IterRange, _Acc) -> I >= IterRange end
end,
for_apply_default_opts({Init, Range, Step, DefaultGuard});
for_apply_default_opts({Init, Range, Step, Guard}) when is_function(Guard, 2) ->
for_apply_default_opts({Init, Range, Step, fun(I, IterRange, _Acc) -> Guard(I, IterRange) end});
for_apply_default_opts({Init, Range, Step, DefaultGuard}) when is_number(Step) ->
for_apply_default_opts({Init, Range, fun(I, _Acc) -> I + Step end, DefaultGuard});
for_apply_default_opts({Init, Range, Step, DefaultGuard}) when is_function(Step, 1) ->
for_apply_default_opts({Init, Range, fun(I, _Acc) -> Step(I) end, DefaultGuard});
for_apply_default_opts({_Init, _Range, _Step, _DefaultGuard} = Opts) ->
Opts.
I have a string:
"abc abc abc abc"
How do I calculate the number of "abc" repetitions?
If you are looking for practical and efficient implementation which will scale well for even longer substrings you can use binary:matches/2,3 which is using Boyer–Moore string search algorithm (and Aho-Corasic for multiple substrings). It obviously works only for ASCII or Latin1 strings.
repeats(L, S) -> length(binary:matches(list_to_binary(L), list_to_binary(S))).
If it is for education purposes, you can write your own less efficient version for lists of any kind. If you know substring in compile time you can use very simple and not so much bad in performance:
-define(SUBSTR, "abc").
repeats(L) -> repeats(L, 0).
repeats(?SUBSTR ++ L, N) -> repeats(L, N+1);
repeats([_|L] , N) -> repeats(L, N);
repeats([] , N) -> N.
If you don't know substring you can write a little bit more complicated and less efficient
repeats(L, S) -> repeats(L, S, 0).
repeats([], _, N) -> N;
repeats(L, S, N) ->
case prefix(L, S) of
{found, L2} -> repeats( L2, S, N+1);
nope -> repeats(tl(L), S, N)
end.
prefix([H|T], [H|S]) -> prefix(T, S);
prefix( L, [ ]) -> {found, L};
prefix( _, _ ) -> nope.
And you, of course, can try write some more sophisticated variant as simplified Boyer–Moore for lists.
1> F = fun
F([],_,_,N) -> N;
F(L,P,S,N) ->
case string:sub_string(L,1,S) == P of
true -> F(tl(string:sub_string(L,S,length(L))),P,S,N+1);
_ -> F(tl(L),P,S,N)
end
end.
#Fun<erl_eval.28.106461118>
2> Find = fun(L,P) -> F(L,P,length(P),0) end.
#Fun<erl_eval.12.106461118>
3> Find("abc abc abc abc","abc").
4
4>
this works if defined in a module, or in the shell but only with the R17.
length(lists:filter(fun(X) -> X=="abc" end, string:tokens("abc abc abc abc", " "))).