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]
Given a list and an integer, I want to split that list into the specified number of lists (inside a list).
For example:
Input:
[1,2,3,4,5,6,7,8,9], 3
Output:
[[1,2,3],[4,5,6],[7,8,9]]
What is a clean and efficient way to do this?
The solution written by Steve Vinoski calls length/1 in guard for each partition which makes it O(N^2). It simply bothers me because it can be done in O(N) and I am performance freak. It can be done in many ways so just for example there is one:
divide(L, N) when is_integer(N), N > 0 ->
divide(N, 0, L, []).
divide(_, _, [], Acc) ->
[lists:reverse(Acc)];
divide(N, N, L, Acc) ->
[lists:reverse(Acc) | divide(N, 0, L, [])];
divide(N, X, [H|T], Acc) ->
divide(N, X+1, T, [H|Acc]).
or as a modification of Steve's solution
divide(L, N) ->
divide(L, N, []).
divide([], _, Acc) ->
lists:reverse(Acc);
divide(L, N, Acc) ->
try lists:split(N, L) of
{H,T} -> divide(T, N, [H|Acc])
catch
error:badarg ->
lists:reverse([L|Acc])
end.
or even simpler:
divide([], _) -> [];
divide(L, N) ->
try lists:split(N, L) of
{H,T} -> [H|divide(T, N)]
catch
error:badarg -> [L]
end.
You can use lists:split/2 for this:
divide(L, N) ->
divide(L, N, []).
divide([], _, Acc) ->
lists:reverse(Acc);
divide(L, N, Acc) when length(L) < N ->
lists:reverse([L|Acc]);
divide(L, N, Acc) ->
{H,T} = lists:split(N, L),
divide(T, N, [H|Acc]).
The first function, divide/2, serves as the entry point. It merely calls the helper function divide/3 with an initial accumulator value of an empty list, and then divide/3 does all the work. The first clause of divide/3 matches when the list has been completely processed, so it just reverses the accumulator and returns that value. The second clause handles the case when the length of L is less than the requested N value; it creates a new accumulator by prepending Acc with L and then returning the reverse of that new accumulator. The third clause first calls lists:split/2 to split the incoming list into H, which is a list of N elements, and T, the remainder of the list. It then calls itself recursively, passing T as the new list value, the original N value, and a new accumulator consisting of H as the first element and the original accumulator, Acc, as the tail.
I want to split:
[1,2,3,4,5,6,7,8]
into:
[[1,2],[3,4],[5,6],[7,8]]
It generally works great with:
[ lists:sublist(List, X, 2) || X <- lists:seq(1,length(List),2) ] .
But it is really slow this way. 10000 Elements take amazing 2.5 seconds on my netbook. I have also written a really fast recursive function, but I am simply interested: Could this list comprehension also be written in a different way, so that it is faster?
Try this:
part(List) ->
part(List, []).
part([], Acc) ->
lists:reverse(Acc);
part([H], Acc) ->
lists:reverse([[H]|Acc]);
part([H1,H2|T], Acc) ->
part(T, [[H1,H2]|Acc]).
Test in erlang-shell (I've declared this function in module part):
2> part:part([1,2,3,4,5,6,7,8]).
[[1,2],[3,4],[5,6],[7,8]]
3>
3> timer:tc(part, part, [lists:seq(1,10000)]).
{774,
[[1,2],
[3,4],
[5,6],
[7,8],
"\t\n","\v\f",
[13,14],
[15,16],
[17,18],
[19,20],
[21,22],
[23,24],
[25,26],
[27,28],
[29,30],
[31,32],
"!\"","#$","%&","'(",")*","+,","-.","/0","12","34",
[...]|...]}
Just 774 microseconds (which is ~0,8 milliseconds)
Here are two quick solutions for you that are both flexible. One is easy to read, but only slightly faster than your proposed solution. The other is quite fast, but is a bit cryptic to read. And note that both of my proposed algorithms will work for lists of anything, not just numeric ordered lists.
Here is the "easy-to-read" one. Call by n_length_chunks(List,Chunksize). For example, to get a list of chunks 2 long, call n_length_chunks(List,2). This works for chunks of any size, ie, you could call n_length_chunks(List,4) to get [[1,2,3,4],[5,6,7,8],...]
n_length_chunks([],_) -> [];
n_length_chunks(List,Len) when Len > length(List) ->
[List];
n_length_chunks(List,Len) ->
{Head,Tail} = lists:split(Len,List),
[Head | n_length_chunks(Tail,Len)].
The much faster one is here, but is definitely harder to read, and is called in the same way: n_length_chunks_fast(List,2) (I've made one change to this compared with the one above, in that it pads the end of the list with undefined if the length of the list isn't cleanly divisible by the desired chunk length.
n_length_chunks_fast(List,Len) ->
LeaderLength = case length(List) rem Len of
0 -> 0;
N -> Len - N
end,
Leader = lists:duplicate(LeaderLength,undefined),
n_length_chunks_fast(Leader ++ lists:reverse(List),[],0,Len).
n_length_chunks_fast([],Acc,_,_) -> Acc;
n_length_chunks_fast([H|T],Acc,Pos,Max) when Pos==Max ->
n_length_chunks_fast(T,[[H] | Acc],1,Max);
n_length_chunks_fast([H|T],[HAcc | TAcc],Pos,Max) ->
n_length_chunks_fast(T,[[H | HAcc] | TAcc],Pos+1,Max);
n_length_chunks_fast([H|T],[],Pos,Max) ->
n_length_chunks_fast(T,[[H]],Pos+1,Max).
Tested on my (really old) laptop:
Your proposed solution took about 3 seconds.
My slow-but-readable one was slightly faster and takes about 1.5 seconds (still quite slow)
My fast version takes about 5 milliseconds.
For completeness, Isac's solution took about 180 milliseconds on my same machine.
Edit: wow, I need to read the complete question first. Oh well I'll keep here for posterity if it helps. As far as I can tell, there's not a good way to do this using list comprehensions. Your original version is slow because each iteration of sublist needs to traverse the list each time to get to each successive X, resulting in complexity just under O(N^2).
Or with a fold:
lists:foldr(fun(E, []) -> [[E]];
(E, [H|RAcc]) when length(H) < 2 -> [[E|H]|RAcc] ;
(E, [H|RAcc]) -> [[E],H|RAcc]
end, [], List).
I want to submit slightly complicated but more flexible (and mostly faster) solution of one proposed by #Tilman
split_list(List, Max) ->
element(1, lists:foldl(fun
(E, {[Buff|Acc], C}) when C < Max ->
{[[E|Buff]|Acc], C+1};
(E, {[Buff|Acc], _}) ->
{[[E],Buff|Acc], 1};
(E, {[], _}) ->
{[[E]], 1}
end, {[], 0}, List)).
so function part can be implemented as
part(List) ->
RevList = split_list(List, 2),
lists:foldl(fun(E, Acc) ->
[lists:reverse(E)|Acc]
end, [], RevList).
update
I've added reverse in case if you want to preserve order, but as I can see it adds no more than 20% of processing time.
You could do it like this:
1> {List1, List2} = lists:partition(fun(X) -> (X rem 2) == 1 end, List).
{[1,3,5|...],[2,4,6|...]}
2> lists:zipwith(fun(X, Y) -> [X, Y] end, List1, List2).
[[1,2],[3,4],[5,6]|...]
This takes ~73 milliseconds with a 10000 elements List on my computer. The original solution takes ~900 miliseconds.
But I would go with the recursive function anyway.
I was looking for a partition function which can split a large list to small amount of workers. With lkuty's partition you might get that one worker gets almost double work than all the others. If that's not what you want, here is a version which sublist lengths differ by at most 1.
Uses PropEr for testing.
%% #doc Split List into sub-lists so sub-lists lengths differ most by 1.
%% Does not preserve order.
-spec split_many(pos_integer(), [T]) -> [[T]] when T :: term().
split_many(N, List) ->
PieceLen = length(List) div N,
lists:reverse(split_many(PieceLen, N, List, [])).
-spec split_many(pos_integer(), pos_integer(), [T], [[T]]) ->
[[T]] when T :: term().
split_many(PieceLen, N, List, Acc) when length(Acc) < N ->
{Head, Tail} = lists:split(PieceLen, List),
split_many(PieceLen, N, Tail, [Head|Acc]);
split_many(_PieceLen, _N, List, Acc) ->
% Add an Elem to each list in Acc
{Appendable, LeaveAlone} = lists:split(length(List), Acc),
Appended = [[Elem|XS] || {Elem, XS} <- lists:zip(List, Appendable)],
lists:append(Appended, LeaveAlone).
Tests:
split_many_test_() ->
[
?_assertEqual([[1,2]], elibs_lists:split_many(1, [1,2])),
?_assertEqual([[1], [2]], elibs_lists:split_many(2, [1,2])),
?_assertEqual([[1], [3,2]], elibs_lists:split_many(2, [1,2,3])),
?_assertEqual([[1], [2], [4,3]], elibs_lists:split_many(3, [1,2,3,4])),
?_assertEqual([[1,2], [5,3,4]], elibs_lists:split_many(2, [1,2,3,4,5])),
?_assert(proper:quickcheck(split_many_proper1())),
?_assert(proper:quickcheck(split_many_proper2()))
].
%% #doc Verify all elements are preserved, number of groups is correct,
%% all groups have same number of elements (+-1)
split_many_proper1() ->
?FORALL({List, Groups},
{list(), pos_integer()},
begin
Split = elibs_lists:split_many(Groups, List),
% Lengths of sub-lists
Lengths = lists:usort(lists:map(fun erlang:length/1, Split)),
length(Split) =:= Groups andalso
lists:sort(lists:append(Split)) == lists:sort(List) andalso
length(Lengths) =< 2 andalso
case Lengths of
[Min, Max] -> Max == Min + 1;
[_] -> true
end
end
).
%% #doc If number of groups is divisable by number of elements, ordering must
%% stay the same
split_many_proper2() ->
?FORALL({Groups, List},
?LET({A, B},
{integer(1, 20), integer(1, 10)},
{A, vector(A*B, term())}),
List =:= lists:append(elibs_lists:split_many(Groups, List))
).
Here is a more general answer that works with any sublist size.
1> lists:foreach(fun(N) -> io:format("~2.10.0B -> ~w~n",[N, test:partition([1,2,3,4,5,6,7,8,9,10],N)] ) end, [1,2,3,4,5,6,7,8,9,10]).
01 -> [[1],[2],[3],[4],[5],[6],[7],[8],[9],[10]]
02 -> [[1,2],[3,4],[5,6],[7,8],[9,10]]
03 -> [[1,2,3],[4,5,6],[7,8,9],[10]]
04 -> [[1,2,3,4],[5,6,7,8],[10,9]]
05 -> [[1,2,3,4,5],[6,7,8,9,10]]
06 -> [[1,2,3,4,5,6],[10,9,8,7]]
07 -> [[1,2,3,4,5,6,7],[10,9,8]]
08 -> [[1,2,3,4,5,6,7,8],[10,9]]
09 -> [[1,2,3,4,5,6,7,8,9],[10]]
10 -> [[1,2,3,4,5,6,7,8,9,10]]
And the code to achieve this is stored inside a file called test.erl:
-module(test).
-compile(export_all).
partition(List, N) ->
partition(List, 1, N, []).
partition([], _C, _N, Acc) ->
lists:reverse(Acc) ;
partition([H|T], 1, N, Acc) ->
partition(T, 2, N, [[H]|Acc]) ;
partition([H|T], C, N, [HAcc|TAcc]) when C < N ->
partition(T, C+1, N, [[H|HAcc]|TAcc]) ;
partition([H|T], C, N, [HAcc|TAcc]) when C == N ->
partition(T, 1, N, [lists:reverse([H|HAcc])|TAcc]) ;
partition(L, C, N, Acc) when C > N ->
partition(L, 1, N, Acc).
It could probably be more elegant regarding the special case where C > N. Note that C is the size of the current sublist being constructed. At start, it is 1. And then it increments until it reaches the partition size of N.
We could also use a modified version of #chops code to let the last list contains the remaining items even if its size < N :
-module(n_length_chunks_fast).
-export([n_length_chunks_fast/2]).
n_length_chunks_fast(List,Len) ->
SkipLength = case length(List) rem Len of
0 -> 0;
N -> Len - N
end,
n_length_chunks_fast(lists:reverse(List),[],SkipLength,Len).
n_length_chunks_fast([],Acc,_Pos,_Max) -> Acc;
n_length_chunks_fast([H|T],Acc,Pos,Max) when Pos==Max ->
n_length_chunks_fast(T,[[H] | Acc],1,Max);
n_length_chunks_fast([H|T],[HAcc | TAcc],Pos,Max) ->
n_length_chunks_fast(T,[[H | HAcc] | TAcc],Pos+1,Max);
n_length_chunks_fast([H|T],[],Pos,Max) ->
n_length_chunks_fast(T,[[H]],Pos+1,Max).
I've slightly altered the implementation from #JLarky to remove the guard expression, which should be slightly faster:
split_list(List, Max) ->
element(1, lists:foldl(fun
(E, {[Buff|Acc], 1}) ->
{[[E],Buff|Acc], Max};
(E, {[Buff|Acc], C}) ->
{[[E|Buff]|Acc], C-1};
(E, {[], _}) ->
{[[E]], Max}
end, {[], Max}, List)).