[erlang-questions] zip for N lists?

Daniel Goertzen daniel.goertzen@REDACTED
Fri Oct 19 14:30:50 CEST 2012


Here is one that I use.  My comments say that it comes from Haskell, but I
seem to remember that I got the Erlang translation from this mailing list.


% transpose a matrix represented as list of lists
% aka "zip" list of lists
% This implementation originates from Haskell
transpose([[X | Xs] | Xss]) ->
[[X | [H || [H | _] <- Xss]]
 | transpose([Xs | [T || [_ | T] <- Xss]])];
transpose([[] | Xss]) -> transpose(Xss);
transpose([]) -> [];

transpose(Tuple) when is_tuple(Tuple) ->  % wrapper to emulate zip
Xs = transpose(tuple_to_list(Tuple)),
[list_to_tuple(X) || X <- Xs].


Dan.




On Fri, Oct 19, 2012 at 4:17 AM, Ivan Uemlianin <ivan@REDACTED> wrote:

> Dear All
>
> Erlang provides lists:zip/2 and /3 for zipping together two or three
> lists.  Is there anything ready available for zipping together arbitrarily
> many lists?  e.g., taking a list of lists as its input,
>
>     zipN([[1,2,3],[4,5,6],[7,8,9],**[10,11,12]]) ->
>         [[1,4,7,10],[2,5,8,11],[3,6,9,**12]].
>
> If not, I cooked up the following:
>
>     zip_lol([[]|_]) ->
>         [];
>     zip_lol(Lols) ->
>         {Lohs, Lots} = lists:foldl(fun([H|T], {Hs,Ts}) ->
>                                        {[H|Hs],[T|Ts]}
>                                    end,
>                                    {[],[]},
>                                    Lols),
>         [lists:reverse(Lohs) | zip_lol(Lots)].
>
> This doesn't quite work, as only every other Lohs seems to need reversing:
>
>     zip_lol([[1,2,3],[4,5,6],[7,8,**9],[10,11,12]]) ->
>        [[1,4,7,10],[11,8,5,2],[3,6,9,**12]].
>
> Actually, it's enough for my purposes, as I only need the sum of each loh,
> so I can use:
>
>     zipwith_lol(_, [[]|_]) ->
>         [];
>     zipwith_lol(Fun, Lols) ->
>         {Lohs, Lots} = lists:foldl(fun([H|T], {Hs,Ts}) ->
>                                        {[H|Hs],[T|Ts]}
>                                    end,
>                                    {[],[]},
>                                    Lols),
>         [Fun(Lohs) | zipwith_lol(Fun, Lots)].
>
>     zip_lol([[1,2,3],[4,5,6],[7,8,**9],[10,11,12]]) ->
>         [22,26,30].
>
> What concerns me is performance.  My lols are likely to be quite large: 10
> to 20 lists, and each list could be 100k+ integers long (if it were Haskell
> each list would be one of those lazy infinite list stream things).
>
> Is the fold the right way to do this, or have I missed something that
> would do the job faster?
>
> With thanks and best wishes
>
> Ivan
>
>
> --
> ==============================**==============================
> Ivan A. Uemlianin PhD
> Llaisdy
> Speech Technology Research and Development
>
>                     ivan@REDACTED
>                      www.llaisdy.com
>                          llaisdy.wordpress.com
>               github.com/llaisdy
>                      www.linkedin.com/in/**ivanuemlianin<http://www.linkedin.com/in/ivanuemlianin>
>
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>                  (Spinoza, Ethica, IV, XLII)
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