Erlang, replacing an atom with another one in a list - erlang

I want to write a function to replace a specific atom with the given atom in an input list. But I want to do it using pattern matching and not using conditional statements. Any idea?
And also I want to write a function to return unique atoms in an expression.
e.g.
Input:
[a, b, c, a, b]
Output:
c
Input:
[b, b, b, r, t, y, y]
Output:
[t, r]

Assuming you want to replace all instances and keep the order of the list (works with all terms):
replace(Old, New, List) -> replace(Old, New, List, []).
replace(_Old, _New, [], Acc) -> lists:reverse(Acc);
replace(Old, New, [Old|List], Acc) -> replace(Old, New, List, [New|Acc]);
replace(Old, New, [Other|List], Acc) -> replace(Old, New, List, [Other|Acc]).
For the unique elements filter, you need to keep a state of which elements you have looked at already.
It would be really awkward to implement such a function using only pattern matching in the function headers and you would not really gain anything (performance) from it. The awkwardness would come from having to loop through both the list in question and the list(s) keeping your state of already parsed elements. You would also loose a lot of readability.
I would recommend going for something simpler (works with all terms, not just atoms):
unique(List) -> unique(List, []).
unique([], Counts) ->
lists:foldl(fun({E, 1}, Acc) -> [E|Acc];
(_, Acc) -> Acc
end, [], Counts);
unique([E|List], Counts) ->
unique(List, count(E, Counts).
count(E, []) -> [{E, 1}];
count(E, [{E, N}|Rest]) -> [{E, N + 1}|Rest];
count(E, [{X, N}|Rest]) -> [{X, N}|count(E, Rest)].

One way I'm looking for solving your first question would be to use guards, instead of if statements. Using only pattern matching doesn't seem possible (or desirable, even if you can do it).
So, for instance, you could do something like:
my_replace([H|T], ToReplace, Replacement, Accum) when H == ToReplace ->
my_replace(T, ToReplace, Replacement, [Replacement|Accum]);
my_replace([H|T], ToReplace, Replacement, Accum) ->
my_replace(T, ToReplace, Replacement, [H|Accum]);
my_replace([], ToReplace, Replacement, Accum) ->
lists:reverse(Accum).
EDIT: Edited for simplicity and style, thanks for the comments. :)
For the second part of your question, what do you consider an "expression"?
EDIT: Nevermind that, usort doesn't completely remove duplicates, sorry.

Related

Erlang: serial implementation of accumulator

I am trying to create a method that takes an associative and commutative operator, as well a list of values, and then returns the answer by applying an operator to the values in the list.
The following two examples represent what the input/output are supposed to look like.
Example 1
Input: sum(fun(A,B) -> A+B end, [2,6,7,10,12]).
Output: 37
Example 2
Input: sum(fun (A,B) -> A++B end , ["C", "D", "E"]).
Output: "CDE"
This is the code I am working with so far.
-module(tester).
-compile(export_all).
sum(Func, Data, Acc) ->
lists:foldr(Func, Acc, Data).
This code produces the correct result, however, there are two problems I am trying to figure out how to approach answering.
(1) In order for this code to work, it requires an empty list to be included at the end of the command line statements. In other words, if I enter the input above (as in the examples), it will err out, because I did not write it in the following way:
12> tester:sum(fun(X, Acc) -> X+Acc end, [2,6,7,10,12], 0).
How would I implement this without an empty list as in the examples above and get the same result?
(2) Also, how would the code be implemented without the list function, or in an even more serial way?
How would I implement this without an empty list as in the examples above and get the same result?
Assuming the list always has one element (you can't really do it without this assumption), you can extract the first element from the list and pass that as the initial accumulator. You'll need to switch to foldl to do this efficiently. (With foldr you'll essentially need to make a copy of the list to drop the last element.)
sum(Func, [X | Xs]) ->
lists:foldl(fun (A, B) -> Func(B, A) end, X, Xs).
1> a:sum(fun(A,B) -> A+B end, [2,6,7,10,12]).
37
2> a:sum(fun (A,B) -> A++B end , ["C", "D", "E"]).
"CDE"
Also, how would the code be implemented without the list function, or in an even more serial way?
Here's a simple implementation using recursion and pattern matching:
sum2(Func, [X | Xs]) ->
sum2(Func, Xs, X).
sum2(Func, [], Acc) ->
Acc;
sum2(Func, [X | Xs], Acc) ->
sum2(Func, Xs, Func(Acc, X)).
We define two versions of the function. The first one extracts the head and uses that as the initial accumulator. The second one, with arity 3, does essentially what the fold functions in lists do.
After working on this for a while, this was my solution. I've left some comments about the general idea of what I did, but there's a lot more to be said.
-module(erlang2).
-compile(export_all).
-export([reduce/2]).
reduce(Func, List) ->
reduce(root, Func, List).
%When done send results to Parent
reduce(Parent, _, [A]) ->
%send to parent
Parent ! { self(), A};
%I tried this at first to take care of one el in list, but it didn't work
%length ([]) ->
% Parent ! {self(), A};
%get contents of list, apply function and store in Parent
reduce(Parent, Func, List) ->
{ Left, Right } = lists:split(trunc(length(List)/2), List),
Me = self(),
%io:format("Splitting in two~n"),
Pl = spawn(fun() -> reduce(Me, Func, Left) end),
Pr = spawn(fun() -> reduce(Me, Func, Right) end),
%merge results in parent and call Func on final left and right halves
combine(Parent, Func,[Pl, Pr]).
%merge pl and pl and combine in parent
combine(Parent, Func, [Pl, Pr]) ->
%wait for processes to complete (using receive) and then send to Parent
receive
{ Pl, Sorted } -> combine(Parent, Func, Pr, Sorted);
{ Pr, Sorted } -> combine(Parent, Func, Pl, Sorted)
end.
combine(Parent, Func, P, List) ->
%wait and store in results and then call ! to send
receive
{ P, Sorted } ->
Results = Func(Sorted, List),
case Parent of
root ->
Results;
%send results to parent
_ -> Parent ! {self(), Results}
end
end.

Copy one list to another list

I am stuck at some basic recursion since this language is completely new for me, I want to copy integers from one list to another.
This is what I have:
cpy_list(L) -> cpy_list(L, []).
cpy_list([], Acc) -> Acc;
could somebody show me how a working solution can look like?
You need one more clause for thecpy_list/2
cpy_list(L) -> cpy_list(L, []). % Starting condition
cpy_list([H|T], Acc) -> cpy_list(T, Acc ++ [H]); % take the first element and add to Acc
cpy_list([], Acc) -> Acc.
Of course, it is not an ideal solution because it is not efficient. See note here: http://www.erlang.org/doc/efficiency_guide/listHandling.html
And since variables are immutable in erlang, I doubt there is any point in copying a list.
In Erlang, assigning a variable into a new one is "copying" the content from the on to the other. (no pointers!)
So you don't need a special function for it.
TheList = [1,2,3,4,5].
CpyList = TheList.
cpy_list(L) -> cpy_list(L, []).
cpy_list([], Acc) -> lists:reverse(Acc);
cpy_list([H|T], Acc) ->
cpy_list(T, [H|Acc]).
Here is the code, you can use Tail-recursive
In Erlang since all data is immutiable there is no point in copying a list. Anything you do to manipulate the list will produce an altered copy of the original list. If you did want to copy a list it would be as simple as running:
List = [1,2,3],
NewList = lists:map(fun(X) -> X end, List).
Because lists:map/2 will always returns a new list this will work! You could just have easily have performed your alterations on the original list with map/2 and it would have returned a completely new list.
Once can also use a list comprehension:
ListCopy = [Entry || Entry <- ListOriginal]

Erlang un-zip-flatten

I have a list of items that I would like to "un-zip-flatten". Basically what that means is that if I have a list of items:
[a, b, c, d, e, f, g]
I want to turn it into a list of lists like the following:
[[a, d, g], [b, e], [c, f]]
So far my solution looks like this:
unzipflatten(NumberOfLists, List) ->
lists:map(fun(Start) ->
lists:map(fun(N) ->
lists:nth(N, List)
end,
lists:seq(Start, length(List), NumberOfLists))
end,
lists:seq(1, NumberOfLists)).
I'm pretty new to Erlang so I'm wondering if I've missed some standard library function that would do what I want, or if there's a more "Erlangish" way to do this, or if the performance of my above solution will stink.
I think this would be a more "Erlangish" method to do this. Basically you would create the list of lists that will be your result, and use two lists to manage those lists like a queue. The "Heads" list contains the lists that you will prepend to next, and the "Tails" list are the ones most recently prepended to. When Heads is empty you simply reverse Tails and use that as the new Heads. Before returning the result you will need to reverse all of the lists inside Tails and Heads and then append Heads as-is to the reversed Tails. Excuse the confusing variable names, I think coming up with several good names for breaking up lists in an Erlang program is the hardest part ;)
unzipflatten(NumberOfLists, List) when NumberOfLists > 0 ->
unzipflatten(List, lists:duplicate(NumberOfLists, []), []).
unzipflatten([], Heads, Tails) ->
[lists:reverse(L) || L <- lists:reverse(Tails, Heads)];
unzipflatten(L, [], Tails) ->
unzipflatten(L, lists:reverse(Tails), []);
unzipflatten([Elem | Rest], [Head | Tail], Tails) ->
unzipflatten(Rest, Tail, [[Elem | Head] | Tails]).
It's also possible to do the "unzip" phase in a non tail-recursive way to avoid the lists:reverse step, but that is a more complicated solution. Something like this:
unzipflatten(NumberOfLists, List) when NumberOfLists > 0 ->
unzipflatten({List, lists:duplicate(NumberOfLists, [])}).
unzipflatten({[], Heads}) ->
[lists:reverse(L) || L <- Heads];
unzipflatten({L, Heads}) ->
unzipflatten(unzipper({L, Heads})).
unzipper({[], Heads}) ->
{[], Heads};
unzipper({L, []}) ->
{L, []};
unzipper({[H | T], [Head | Tail]}) ->
{T1, Tail1} = unzipper({T, Tail}),
{T1, [[H | Head] | Tail1]}.
Yes, performance will stink (basic advice for using lists:nth: never call it several times with growing N!). Something like this should be better (not tested):
unzipflatten(NumberOfLists, List) ->
unzipflatten(NumberOfLists, List, array:new(NumberOfLists, {default, []}), 0).
unzipflatten(_, [], Lists, _) ->
lists:map(fun lists:reverse/1, array:to_list(Lists));
unzipflatten(NumberOfLists, [H | T], Lists, CurrentIndex) ->
NewLists = array:set(CurrentIndex, [H | array:get(CurrentIndex, Lists)], Lists),
unzipflatten(NumberOfLists, T, NewLists, (CurrentIndex + 1) rem NumberOfLists).

Overuse of guards in Erlang?

I have the following function that takes a number like 5 and creates a list of all the numbers from 1 to that number so create(5). returns [1,2,3,4,5].
I have over used guards I think and was wondering if there is a better way to write the following:
create(N) ->
create(1, N).
create(N,M) when N =:= M ->
[N];
create(N,M) when N < M ->
[N] ++ create(N + 1, M).
The guard for N < M can be useful. In general, you don't need a guard for equality; you can use pattern-matching.
create(N) -> create(1, N).
create(M, M) -> [M];
create(N, M) when N < M -> [N | create(N + 1, M)].
You also generally want to write functions so they are tail-recursive, in which the general idiom is to write to the head and then reverse at the end.
create(N) -> create(1, N, []).
create(M, M, Acc) -> lists:reverse([M | Acc]);
create(N, M, Acc) when N < M -> create(N + 1, M, [N | Acc]).
(Of course, with this specific example, you can alternatively build the results in the reverse order going down to 1 instead of up to M, which would make the lists:reverse call unnecessary.)
If create/2 (or create/3) is not exported and you put an appropriate guard on create/1, the extra N < M guard might be overkill. I generally only check on the exported functions and trust my own internal functions.
create(N,N) -> [N];
create(N,M) -> [N|create(N + 1, M)]. % Don't use ++ to prefix a single element.
This isn't quite the same (you could supply -5), but it behaves the same if you supply meaningful inputs. I wouldn't bother with the extra check anyway, since the process will crash very quickly either way.
BTW, you have a recursion depth problem with the code as-is. This will fix it:
create(N) ->
create(1, N, []).
create(N, N, Acc) -> [N|Acc];
create(N, M, Acc) -> create(N, M - 1, [M|Acc]).
I don't really think you have over used guards. There are two cases:
The first is the explicit equality test in the first clause of create/2
create(N, M) when N =:= M -> [M];
Some have suggested transforming this to use pattern matching like
create(N, N) -> [N];
In this case it makes no difference as the compiler internally transforms the pattern matching version to what you have written. You can safely pick which version you think feels best in each case.
In the second case you need some form of sanity check that the value of the argument in the range you expect it to be. Doing in every loop is unnecessary and I would move it to an equivalent test in create/1:
create(M) when M > 1 -> create(1, M).
If you want to use an accumulator I would personally use the count version as it saves reversing the list at the end. If the list is not long I think the difference is very small and you can pick the version which feels most clear to you. Anyway, it is very easy to change later if you find it to be critical.

right rotate a List in Erlang

I am getting myself familiar to Sequential Erlang (and the functional programming thinking) now. So I want to implement the following two functionality without the help of BIF. One is left_rotate (which I have come up with the solution) and the other is right_rotate (which I am asking here)
-export(leftrotate/1, rightrotate/1).
%%(1) left rotate a lits
leftrotate(List, 0) ->
List;
leftrotate([Head | Tail], Times) ->
List = append(Tail, Head),
leftrotate(List, Times -1).
append([], Elem)->
[Elem];
append([H|T], Elem) ->
[H | append(T, Elem)].
%%right rotate a list, how?
%%
I don't want to use BIF in this exercise. How can I achieve the right rotation?
A related question and slightly more important question. How can I know one of my implementation is efficient or not (i.e., avoid unnecessary recursion if I implement the same thing with the help of a BIF, and etc.)
I think BIF is built to provide some functions to improve efficiency that functional programming is not good at (or if we do them in a 'functional way', the performance is not optimal).
The efficiency problem you mention has nothing to do with excessive recursion (function calls are cheap), and everything to do with walking and rebuilding the list. Every time you add something to the end of a list you have to walk and copy the entire list, as is obvious from your implementation of append. So, to rotate a list N steps requires us to copy the entire list out N times. We can use lists:split (as seen in one of the other answers) to do the entire rotate in one step, but what if we don't know in advance how many steps we need to rotate?
A list really isn't the ideal data structure for this task. Lets say that instead we use a pair of lists, one for the head and one for the tail, then we can rotate easily by moving elements from one list to the other.
So, carefully avoiding calling anything from the standard library, we have:
rotate_right(List, N) ->
to_list(n_times(N, fun rotate_right/1, from_list(List))).
rotate_left(List, N) ->
to_list(n_times(N, fun rotate_left/1, from_list(List))).
from_list(Lst) ->
{Lst, []}.
to_list({Left, Right}) ->
Left ++ reverse(Right).
n_times(0, _, X) -> X;
n_times(N, F, X) -> n_times(N - 1, F, F(X)).
rotate_right({[], []}) ->
{[], []};
rotate_right({[H|T], Right}) ->
{T, [H|Right]};
rotate_right({[], Right}) ->
rotate_right({reverse(Right), []}).
rotate_left({[], []}) ->
{[], []};
rotate_left({Left, [H|T]}) ->
{[H|Left], T};
rotate_left({Left, []}) ->
rotate_left({[], reverse(Left)}).
reverse(Lst) ->
reverse(Lst, []).
reverse([], Acc) ->
Acc;
reverse([H|T], Acc) ->
reverse(T, [H|Acc]).
The module queue provides a data structure something like this. I've written this without reference to that though, so theirs is probably more clever.
First, your implementation is a bit buggy (try it with the empty list...)
Second, I would suggest you something like:
-module(foo).
-export([left/2, right/2]).
left(List, Times) ->
left(List, Times, []).
left([], Times, Acc) when Times > 0 ->
left(reverse(Acc), Times, []);
left(List, 0, Acc) ->
List ++ reverse(Acc);
left([H|T], Times, Acc) ->
left(T, Times-1, [H|Acc]).
right(List, Times) ->
reverse(foo:left(reverse(List), Times)).
reverse(List) ->
reverse(List, []).
reverse([], Acc) ->
Acc;
reverse([H|T], Acc) ->
reverse(T, [H|Acc]).
Third, for benchmarking your functions, you can do something like:
test(Params) ->
{Time1, _} = timer:tc(?MODULE, function1, Params),
{Time2, _} = timer:tc(?MODULE, function2, Params),
{{solution1, Time1}, {solution2, Time2}}.
I didn't test the code, so look at it critically, just get the idea.
Moreover, you might want to implement your own "reverse" function. It will be trivial by using tail recursion. Why not to try?
If you're trying to think in functional terms then perhaps consider implementing right rotate in terms of your left rotate:
rightrotate( List, 0 ) ->
List;
rightrotate( List, Times ) ->
lists:reverse( leftrotate( lists:reverse( List ), Times ) ).
Not saying this is the best idea or anything :)
Your implementation will not be efficient since the list is not the correct representation to use if you need to change item order, as in a rotation. (Imagine a round-robin scheduler with many thousands of jobs, taking the front job and placing it at the end when done.)
So we're actually just asking ourself what would be the way with least overhead to do this on lists anyway. But then what qualifies as overhead that we want to get rid of? One can often save a bit of computation by consing (allocating) more objects, or the other way around. One can also often have a larger than needed live-set during the computation and save allocation that way.
first_last([First|Tail]) ->
put_last(First, Tail).
put_last(Item, []) ->
[Item];
put_last(Item, [H|Tl]) ->
[H|put_last(Item,Tl)].
Ignoring corner cases with empty lists and such; The above code would cons the final resulting list directly. Very little garbage allocated. The final list is built as the stack unwinds. The cost is that we need more memory for the entire input list and the list in construction during this operation, but it is a short transient thing. My damage from Java and Lisp makes me reach for optimizing down excess consing, but in Erlang you dont risk that global full GC that kills every dream of real time properties. Anyway, I like the above approach generally.
last_first(List) ->
last_first(List, []).
last_first([Last], Rev) ->
[Last|lists:reverse(Rev)];
last_first([H|Tl], Rev) ->
last_first(Tl, [H|Rev]).
This approach uses a temporary list called Rev that is disposed of after we have passed it to lists:reverse/1 (it calls the BIF lists:reverse/2, but it is not doing anything interesting). By creating this temporary reversed list, we avoid having to traverse the list two times. Once for building a list containing everything but the last item, and one more time to get the last item.
One quick comment to your code. I would change the name of the function you call append. In a functional context append usually means adding a new list to the end of a list, not just one element. No sense in adding confusion.
As mentioned lists:split is not a BIF, it is a library function written in erlang. What a BIF really is is not properly defined.
The split or split like solutions look quite nice. As someone has already pointed out a list is not really the best data structure for this type of operation. Depends of course on what you are using it for.
Left:
lrl([], _N) ->
[];
lrl(List, N) ->
lrl2(List, List, [], 0, N).
% no more rotation needed, return head + rotated list reversed
lrl2(_List, Head, Tail, _Len, 0) ->
Head ++ lists:reverse(Tail);
% list is apparenly shorter than N, start again with N rem Len
lrl2(List, [], _Tail, Len, N) ->
lrl2(List, List, [], 0, N rem Len);
% rotate one
lrl2(List, [H|Head], Tail, Len, N) ->
lrl2(List, Head, [H|Tail], Len+1, N-1).
Right:
lrr([], _N) ->
[];
lrr(List, N) ->
L = erlang:length(List),
R = N rem L, % check if rotation is more than length
{H, T} = lists:split(L - R, List), % cut off the tail of the list
T ++ H. % swap tail and head

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