Related
Im working on some erlang functions and im also not allowed to use library functions. I have to define a function that drops every other element from a list, starting with the first element.
I have worked on something similar before but i could use BIFs and now i am struggling.
For example, alternate([1,2,3,four,5,6]) is [2,four,6]. I am not sure how to implement it.
spec drop_word(string()) -> string().
drop_word([]) -> [];
drop_word([O|Op]) -> case wsp(O) of
true -> Op;
false -> drop_word(Op)
end.
alternate(List) ->
alternate(List, _Index=0).
alternate([_|T], Index) when Index rem 2 == 0 -> %even indexes
alternate(T, Index+1);
alternate([H|T], Index) when Index rem 2 == 1 -> %odd indexes
[H | alternate(T, Index+1)];
alternate([], _Index) ->
[].
In the shell:
12> a:alternate([1,2,3,four,5,6]).
[2,four,6]
13> a:alternate([1,2,3,four,5]).
[2,four]
But, that can be simplified to:
alternate(List) ->
evens(List).
evens([_|T]) ->
odds(T);
evens([]) -> [].
odds([H|T]) ->
[H | evens(T)];
odds([]) -> [].
In the shell:
6> a:alternate([1,2,3,four,5,6]).
[2,four,6]
7> a:alternate([1,2,3,four,5]).
[2,four]
Here's an accumulator version:
alternate(List) ->
evens(List, []).
evens([_|T], Acc) ->
odds(T, Acc);
evens([], Acc) ->
lists:reverse(Acc).
odds([H|T], Acc) ->
evens(T, [H|Acc]);
odds([], Acc) ->
lists:reverse(Acc).
In the shell:
20> a:alternate([1,2,3,four,5,6]).
[2,four,6]
21> a:alternate([1,2,3,four,5]).
[2,four]
Note that lists:reverse() is highly optimized, so you would never do List ++ [X] many times, which traverses the whole list every time you add an element to the end of the list. Rather, you should always choose to add an element to the head of a list, then call lists:reverse(). Oh yeah, no library functions...a reverse() function is easy to implement yourself, and although it won't be optimized like the erlang version, it will still be more efficient than doing List ++ [X] multiple times.
You can use two atoms drop and keep two match the alternating clauses of do_alternate. Details below in comments.
-module(so).
-export([alternate/1]).
% The exported function starts the actual function and tells it to match the `drop`
% clause. Kept elements of L will be collected in the third argument.
alternate(L) -> do_alternate(drop, L, []).
% The `drop` clause will call the `keep` clause and pass the tail T and the
% currently collected list Acc. The head H will be dropped.
do_alternate(drop, [_|T], Acc) -> do_alternate(keep, T, Acc);
% The `keep` clause will call the `drop` claues and pass the tail T and the
% currently collected list Acc with the head H prepented to it.
do_alternate(keep, [H|T], Acc) -> do_alternate(drop, T, Acc ++ [H]);
% If the arugment list is empty, return the accumulated list.
do_alternate(_, [], Acc) -> Acc.
Example usage:
> c(so).
{ok,so}
9> so:alternate([1,2,3,4,5,6]).
[2,4,6]
10> so:alternate([1,2,3,4,5,6,seven,eight,nine,ten,eleven]).
[2,4,6,eight,ten]
It seems that you want to drop any elements with position which is even. So you can do it like below:
-module(test).
-compile([export_all,debug_info]).
alternate(L) -> do_alternate(L, 0, length(L)).
do_alternate(_, L, L) -> [];
do_alternate([H|T], N, L) ->
case (N band 1) == 0 of
true -> do_alternate(T, N+1, L);
false -> [H] ++ do_alternate(T, N+1, L)
end.
Result in shell:
1> c(test).
test.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,test}
2> test:alternate([1,2,3,four,5,6]).
[2,four,6]
Moreover, if your List only has integer numbers, so you can use BIF like below:
3> lists:partition(fun(A) -> A rem 2 == 1 end, [1,2,3,4,5,6]).
{[1,3,5],[2,4,6]}
You could pattern match on the list, taking two elements at a time:
alternate([_Odd, Even | T]) ->
[Even] ++ alternate(T);
alternate([]) ->
[].
The specification doesn't say what happens if the list has an odd number of elements, so this function will just crash with a "function clause" error in that case. You could add a third clause to handle that - presumably dropping the last element would be a sensible thing to do:
alternate([_Last]) ->
[];
I have a list in erlang containing interger values.
I want to remove values that occur only one time.(Not Duplicates).
Input = [1,3,2,1,2,2]
Output = [1,2,1,2,2]
I am newbie to erlang. I have tried an approach to sorting them first using list:sort() and then removing a member if the member next to it is the same.
I am having trouble trying to iterate the list. It would be great help if you can show me how I can do it.
multiple(L) ->
M = L -- lists:usort(L),
[X || X <- L , lists:member(X,M)].
Use map to count values and then filter values which was not present just once.
-module(test).
-export([remove_unique/1]).
remove_unique(L) ->
Count = lists:foldl(fun count/2, #{}, L),
lists:filter(fun(X) -> maps:get(X, Count) =/= 1 end, L).
count(X, M) ->
maps:put(X, maps:get(X, M, 0) + 1, M).
And test:
1> c(test).
{ok,test}
2> test:remove_unique([1,2,3,3,3,5,5,6,7,7]).
[3,3,3,5,5,7,7]
3> test:remove_unique([1,2,3,3,3,5,5,6,7,8]).
[3,3,3,5,5]
4> test:remove_unique([1,3,2,1,2,2]).
[1,2,1,2,2]
Here's a solution I'd written when first seeing the question when posted, that uses the same logic as #A.Sarid's recursion/pattern matching answer, except that I use a "Last" parameter instead of the count.
-module(only_dupes).
-export([process/1]).
process([]) -> [];
process(L) when is_list(L) ->
[H|T] = lists:sort(L),
lists:sort(process(undefined, H, T, [])).
process(Last, Curr, [], Acc)
when Curr =/= Last ->
Acc;
process(_Last, Curr, [], Acc) ->
[Curr | Acc];
process(Last, Curr, [Next | Rest], Acc)
when Curr =/= Last, Curr =/= Next ->
process(Curr, Next, Rest, Acc);
process(_Last, Curr, [Next | Rest], Acc) ->
process(Curr, Next, Rest, [Curr | Acc]).
One way for iterating a list (that as a result will return a new list) is using recursion and pattern matching.
After you sort your list you want to iterate the list and to check not only that it is different from the next element, but that there was no other equal elements before it. Consider the list [3,3,3,5,5] if you will only check the next element, the last 3 will also be unique and that is incorrect.
Here is a working program, I used a counter to cover the above case as well. See the syntax for using [H|T] for iterating over the list. You may see more cases and read more about it here.
-module(test).
-export([remove_unique/1]).
remove_unique(Input) ->
Sorted = lists:sort(Input),
remove_unique(Sorted, [], 0).
% Base case - checks if element is unique
remove_unique([H|[]],Output,Count) ->
case Count of
0 -> Output;
_Other -> [H|Output]
end;
% Count is 0 - might be unique - check with next element
remove_unique([H1|[H2|T]],Output, 0)->
case (H1 =:= H2) of
true -> remove_unique([H2|T],[H1|Output],1);
false -> remove_unique([H2|T],Output,0)
end;
% Count is > 0 - not unique - proceed adding to list until next value
remove_unique([H1|[H2|T]],Output,Count) ->
case (H1 =:= H2) of
true -> remove_unique([H2|T],[H1|Output],Count+1);
false -> remove_unique([H2|T],[H1|Output],0)
end.
Test
7> test:remove_unique([1,2,3,3,3,5,5,6,7,7]).
[7,7,5,5,3,3,3]
8> test:remove_unique([1,2,3,3,3,5,5,6,7,8]).
[5,5,3,3,3]
I am doing somthing horrible but I don't know how to make it better.
I am forming all pairwise sums of the elements of a List called SomeList, but I don't want to see duplicates ( I guess I want "all possible pairwise sums" ):
sets:to_list(sets:from_list([A+B || A <- SomeList, B <- SomeList]))
SomeList does NOT contain duplicates.
This works, but is horribly inefficient, because the original list before the set conversion is GIGANTIC.
Is there a better way to do this?
You could simply use lists:usort/1
lists:usort([X+Y || X <- L, Y <- L]).
if the chance to have duplicates is very high, then you can generate the sum using 2 loops and store the sum in an ets set (or using map, I didn't check the performance of both).
7> Inloop = fun Inloop(_,[],_) -> ok; Inloop(Store,[H|T],X) -> ets:insert(Store,{X+H}), Inloop(Store,T,X) end.
#Fun<erl_eval.42.54118792>
8> Outloop = fun Outloop(Store,[],_) -> ok; Outloop(Store,[H|T],List) -> Inloop(Store,List,H), Outloop(Store,T,List) end.
#Fun<erl_eval.42.54118792>
9> Makesum = fun(L) -> S = ets:new(temp,[set]), Outloop(S,L,L), R =ets:foldl(fun({X},Acc) -> [X|Acc] end,[],S), ets:delete(S), R end.
#Fun<erl_eval.6.54118792>
10> Makesum(lists:seq(1,10)).
[15,13,8,11,20,14,16,12,7,3,10,9,19,18,4,17,6,2,5]
11> lists:sort(Makesum(lists:seq(1,10))).
[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
12>
This module will allow you to compare times of execution when using list comprehension, sets or ets. You can of course add additional functions to this comparison:
-module(pairwise).
-export([start/2]).
start(Type, X) ->
L = lists:seq(1, X),
timer:tc(fun do/2, [Type, L]).
do(compr, L) ->
sets:to_list(sets:from_list([A+B || A <- L, B <- L]));
do(set, L) ->
F = fun(Sum, Set) -> sets:add_element(Sum, Set) end,
R = fun(Set) -> sets:to_list(Set) end,
do(L, L, sets:new(), {F, R});
do(ets, L) ->
F = fun(Sum, Tab) -> ets:insert(Tab, {Sum}), Tab end,
R = fun(Tab) ->
Fun = fun({X}, Acc) -> [X|Acc] end,
Res = ets:foldl(Fun, [], Tab),
ets:delete(Tab),
Res
end,
do(L, L, ets:new(?MODULE, []), {F, R}).
do([A|AT], [B|BT], S, {F, _} = Funs) -> do([A|AT], BT, F(A+B, S), Funs);
do([_AT], [], S, {_, R}) -> R(S);
do([_A|AT], [], S, Funs) -> do(AT, AT, S, Funs).
Results:
36> {_, Res1} = pairwise:start(compr, 20).
{282,
[16,32,3,19,35,6,22,38,9,25,12,28,15,31,2,18,34,5,21,37,8,
24,40,11,27,14,30|...]}
37> {_, Res2} = pairwise:start(set, 20).
{155,
[16,32,3,19,35,6,22,38,9,25,12,28,15,31,2,18,34,5,21,37,8,
24,40,11,27,14,30|...]}
38> {_, Res3} = pairwise:start(ets, 20).
{96,
[15,25,13,8,21,24,40,11,26,20,14,28,23,16,12,39,34,36,7,32,
35,3,33,10,9,19,18|...]}
39> R1=lists:usort(Res1), R2=lists:usort(Res2), R3=lists:usort(Res3).
[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
24,25,26,27,28,29,30|...]
40> R1 = R2 = R3.
[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
24,25,26,27,28,29,30|...]
The last line is to compare that all functions return the same result but sorted differently.
First number in each resulted tuple is the time of execution as returned from timer:tc(fun do/2, [Type, L]).. In this example it's 282 for list comprehension, 155 for sets and 96 for ets.
An effective way is to use foldl instead of lists comprehension, because in this case you nedd a state on each step
sets:to_list(
lists:foldl(fun(A, S1) ->
lists:foldl(fun(B, S2) ->
sets:add_element(A+B, S2)
end, S1, SomeListA)
end, sets:new(), SomeListB)).
This solution keeps it relatively fast and makes use of as much pre-written library code as possible.
Note that I use lists:zip/2 here rather than numeric +, only to illustrate that this approach works for any kind of non-repeating permutation of a unique list. You may only care about arithmetic, but if you want more, this can do it.
-export([permute_unique/1]).
permute_unique([]) ->
[];
permute_unique([A|Ab]) ->
lists:zip(lists:duplicate(length(Ab)+1, A), [A|Ab])
++
permute_unique(Ab).
%to sum integers, replace the lists:zip... line with
% [B+C || {B,C} <- lists:zip(lists:duplicate(length(Ab)+1, A), [A|Ab])]
%to perform normal arithmetic and yield a numeric value for each element
I am not sure what you consider gigantic - you will end up with N*(N+1)/2 total elements in the permuted list for a unique list of N original elements, so this gets big really fast.
I did some basic performance testing of this, using an Intel (Haswell) Core i7 # 4GHz with 32GB of memory, running Erlang/OTP 17 64-bit.
5001 elements in the list took between 2 and 5 seconds according to timer:tc/1.
10001 elements in the list took between 15 and 17 seconds, and required about 9GB of memory. This generates a list of 50,015,001 elements.
15001 elements in the list took between 21 and 25 seconds, and required about 19GB of memory.
20001 elements in the list took 49 seconds in one run, and peaked at about 30GB of memory, with about 200 million elements in the result. That is the limit of what I can test.
I am supposed to collect frequencies of characters.
freq(Sample) -> freq(Sample,[]).
freq([],Freq) ->
Freq;
freq([Char|Rest],Freq)->
freq(Rest,[{Char,1}|Freq]).
This function does not work in the right way. If the input is "foo", then the output will be
[{f,1},{o,1},{o,1}].
But I wished to have the output like
[{f,1},{o,2}].
I can't manage to modify element in a tulpe. Can anyone help me out of this and show me how it can be fixed?
a one line solution :o)
% generate a random list
L = [random:uniform(26)+$a-1 || _ <- lists:seq(1,1000)].
% collect frequency
lists:foldl(fun(X,[{[X],I}|Q]) -> [{[X],I+1}|Q] ; (X,Acc) -> [{[X],1}|Acc] end , [], lists:sort(L)).
in action
1> lists:foldl(fun(X,[{[X],I}|Q]) -> [{[X],I+1}|Q] ; (X,Acc) -> [{[X],1}|Acc] end , [], lists:sort("foo")).
[{"o",2},{"f",1}]
quite fast with short list, but the execution time increase a lot with long list (on my PC, it needs 6.5s for a 1 000 000 character text) .
in comparison, with the same 1 000 000 character text Ricardo solution needs 5 sec
I will try another version using ets.
By far the easiest way is to use an orddict to store the value as it already comes with an update_counter function and returns the value in a (sorted) list.
freq(Text) ->
lists:foldl(fun (C, D) -> orddict:update_counter(C, 1, D) end, orddict:new(), Text).
Try with something like this:
freq(Text) ->
CharsDictionary = lists:foldl(fun(Char, Acc) -> dict:update_counter(Char, 1, Acc) end, dict:new(), Text),
dict:fold(fun(Char, Frequency, Acc) -> [{Char, Frequency} | Acc] end, [], CharsDictionary).
The first line creates a dictionary that uses the char as key and the frequency as value (dict:update_counter).
The second line converts the dictionary in the list that you need.
Using pattern matching and proplists.
-module(freq).
-export([char_freq/1]).
-spec char_freq(string()) -> [tuple()].
char_freq(L) -> char_freq(L, []).
char_freq([], PL) -> PL;
char_freq([H|T], PL) ->
case proplists:get_value([H], PL) of
undefined ->
char_freq(T, [{[H],1}|PL]);
N ->
L = proplists:delete([H], PL),
char_freq(T, [{[H],N+1}|L])
end.
Test
1> freq:char_freq("abacabz").
[{"z",1},{"b",2},{"a",3},{"c",1}]
L = [list_to_atom(X) || X <- Str].
D = lists:foldl(fun({Char, _}, Acc) -> dict:update_counter(Char, 1, Acc) end, dict:new(), L).
dict:to_list(D).
Part of my program requires me to be able to randomly shuffle list elements. I need a function such that when i give it a list, it will pseudo-randomly re-arrange the elements in the list. A change in arrangement Must be visible at each call with the same list. My implementation seems to work just fine but i feel that its rather long and is increasing my code base and also, i have a feeling that it ain't the best solution for doing this. So i need a much shorter implementation. Here is my implementation:
-module(shuffle).
-export([list/1]).
-define(RAND(X),random:uniform(X)).
-define(TUPLE(Y,Z,E),erlang:make_tuple(Y,Z,E)).
list(L)->
Len = length(L),
Nums = lists:seq(1,Len),
tuple_to_list(?TUPLE(Len,[],shuffle(Nums,L,[]))).
shuffle([],_,Buffer)-> Buffer;
shuffle(Nums,[Head|Items],Buffer)->
{Pos,NewNums} = pick_position(Nums),
shuffle(NewNums,Items,[{Pos,Head}|Buffer]).
pick_position([N])-> {N,[]};
pick_position(Nos)->
T = lists:max(Nos),
pick(Nos,T).
pick(From,Max)->
random:seed(begin
(case random:seed(now()) of
undefined ->
NN = element(3,now()),
{?RAND(NN),?RAND(NN),?RAND(NN)};
Any -> Any
end)
end
),
T2 = random:uniform(Max),
case lists:member(T2,From) of
false -> pick(From,Max);
true -> {T2,From -- [T2]}
end.
On running it in shell:
F:\> erl
Eshell V5.8.4 (abort with ^G)
1> c(shuffle).
{ok,shuffle}
2> shuffle:list([a,b,c,d,e]).
[c,b,a,e,d]
3> shuffle:list([a,b,c,d,e]).
[e,c,b,d,a]
4> shuffle:list([a,b,c,d,e]).
[a,b,c,e,d]
5> shuffle:list([a,b,c,d,e]).
[b,c,a,d,e]
6> shuffle:list([a,b,c,d,e]).
[c,e,d,b,a]
I am motivated by the fact that in the STDLIB there is no such function. Somewhere in my game, i need to shuffle things up and also i need to find the best efficient solution to the problem, not just one that works.
Could some one help build a shorter version of the solution ? probably even more efficient ? Thank you
1> L = lists:seq(1,10).
[1,2,3,4,5,6,7,8,9,10]
Associate a random number R with each element X in L by making a list of tuples {R, X}. Sort this list and unpack the tuples to get a shuffled version of L.
1> [X||{_,X} <- lists:sort([ {random:uniform(), N} || N <- L])].
[1,6,2,10,5,7,9,3,8,4]
2>
Please note that karl's answer is much more concise and simple.
Here's a fairly simple solution, although not necessarily the most efficient:
-module(shuffle).
-export([list/1]).
list([]) -> [];
list([Elem]) -> [Elem];
list(List) -> list(List, length(List), []).
list([], 0, Result) ->
Result;
list(List, Len, Result) ->
{Elem, Rest} = nth_rest(random:uniform(Len), List),
list(Rest, Len - 1, [Elem|Result]).
nth_rest(N, List) -> nth_rest(N, List, []).
nth_rest(1, [E|List], Prefix) -> {E, Prefix ++ List};
nth_rest(N, [E|List], Prefix) -> nth_rest(N - 1, List, [E|Prefix]).
For example, one could probably do away with the ++ operation in nth_rest/3. You don't need to seed the random algorithm in every call to random. Seed it initially when you start your program, like so: random:seed(now()). If you seed it for every call to uniform/1 your results become skewed (try with [shuffle:list([1,2,3]) || _ <- lists:seq(1, 100)]).
-module(shuffle).
-compile(export_all).
shuffle(L) ->
shuffle(list_to_tuple(L), length(L)).
shuffle(T, 0)->
tuple_to_list(T);
shuffle(T, Len)->
Rand = random:uniform(Len),
A = element(Len, T),
B = element(Rand, T),
T1 = setelement(Len, T, B),
T2 = setelement(Rand, T1, A),
shuffle(T2, Len - 1).
main()->
shuffle(lists:seq(1, 10)).
This will be a bit faster than the above solution, listed here as do2 for timing comparison.
-module(shuffle).
-export([
time/1,
time2/1,
do/1,
do2/1
]).
time(N) ->
L = lists:seq(1,N),
{Time, _} = timer:tc(shuffle, do, [L]),
Time.
time2(N) ->
L = lists:seq(1,N),
{Time, _} = timer:tc(shuffle, do2, [L]),
Time.
do2(List) ->
[X||{_,X} <- lists:sort([ {rand:uniform(), N} || N <- List])].
do(List) ->
List2 = cut(List),
AccInit = {[],[],[],[],[]},
{L1,L2,L3,L4,L5} = lists:foldl(fun(E, Acc) ->
P = rand:uniform(5),
L = element(P, Acc),
setelement(P, Acc, [E|L])
end, AccInit, List2),
lists:flatten([L1,L2,L3,L4,L5]).
cut(List) ->
Rand=rand:uniform(length(List)),
{A,B}=lists:split(Rand, List),
B++A.