CODEWITHSUNDEEP
erlang
Muhammad Saad
Muhammad Saad

Reputation: 23

Remove list element occur only once

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.

Upvotes: 2

Views: 422

Answers (4)

A. Sarid
A. Sarid

Reputation: 3996

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]

Upvotes: 0

Derek Brown
Derek Brown

Reputation: 521

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]).

Upvotes: 0

Pascal
Pascal

Reputation: 14042

multiple(L) ->
    M = L -- lists:usort(L),
    [X || X <- L , lists:member(X,M)].

Upvotes: 4

Hynek -Pichi- Vychodil
Hynek -Pichi- Vychodil

Reputation: 26141

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]

Upvotes: 2

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