[erlang-questions] Filtering hierarchical data structure

黃耀賢 (Yau-Hsien Huang) g9414002.pccu.edu.tw@REDACTED
Sun Jul 15 06:17:09 CEST 2018


 filter([H | T], Search) ->
    lists:flatten([filter(H, Search, []) | filter(T, Search, [])]).

and

filter([], _S, Acc) -> Acc.

may be rewrote to

+ filter(List, Search) -> filter(List, Search, []).

+ filter([], _, Acc) -> lists:reverse(Acc).




On Fri, Jul 13, 2018 at 9:25 PM, Stanislav Ledenev <s.ledenev@REDACTED>
wrote:

> Hello,
>
> I am new to Erlang and functional programming (1-1.5 months) and have task:
>
> 1. Suppose we have hierarchical tree-like data structure:
> L = [
> {section, "id1", [
> {section, "id1.1", [
> {policy, "p1", "v1"},
> {policy, "p2", "v2"}
> ]},
> {section, "id1.2", [
> ]}
> ]},
> {section, "id2", [
> {policy, "p3", "v3"}
> ] },
> {section, "id3", [
> ] }
> ]
>
> 2. List of some "section"'s with children elements any of which could be
> another "section" with children or "policy" with name and value.
>
> 3. Also we have another list which contains identifiers of policies:
> F = ["p1", "p3"].
> 4. Now I need to write a function which takes lists L, F and returns list
> L2
> such as L2 contains only those "section"'s which has "policies" wich Id
> equals to those in F. Empty sections must be removed.
>
> For lists L and F result should be:
> L2 = [
> {section, "id1", [
> {section, "id1.1", [
> {policy, "p1", "v1"},
> ]},
> ]},
> {section, "id2", [
> {policy, "p3", "v3"}
> ] },
> ]
>
> 5. I have solution (please see below) but it seems to me not as good as
> it can be and even little weird. I am sure that it could be improved
> and done in proper way.
> If anyone could give me any glimpse or advice I would appreciate.
>
> Source:
> ------------------------------------------------------------
> ---------------
> main() ->
>     L = [
>             {section, "id1", [
>                 {section, "id1.1", [
>                     {policy, "p1", "v1"},
>                     {policy, "p2", "v2"}
>                 ]},
>                 {section, "id1.2", [
>                 ]}
>             ]},
>             {section, "id2", [
>                 {policy, "p3", "v3"}
>             ] },
>             {section, "id3", [
>             ] }
>         ],
>     %filter(L, ["p1", "p3"]).
>     filter(L, ["p3"]).
>
> filter([H | T], Search) ->
>     lists:flatten([filter(H, Search, []) | filter(T, Search, [])]).
>
> filter({section, I, C}, Search, _Acc) ->
>     NewC = filter(C, Search, []),
>     case NewC of
>         [] -> [];
>         _ -> {section, I, NewC}
>     end;
>
> filter([{section, I, C} | T], Search, Acc) ->
>     NewC = filter(C, Search, []),
>     case NewC of
>         [] -> filter(T, Search, Acc);
>         _ -> filter(T, Search, [{section, I, NewC} | Acc])
>     end;
>
> filter([{policy, I, V} | T], Search, Acc) ->
>     case lists:member(I, Search) of
>         true -> filter(T, Search, [{policy, I, V} | Acc]);
>         false -> filter(T, Search, Acc)
>     end;
> filter([], _S, Acc) -> Acc.
>
> -- Stanislav Ledenev
>
> _______________________________________________
> erlang-questions mailing list
> erlang-questions@REDACTED
> http://erlang.org/mailman/listinfo/erlang-questions
>
>


-- 

Best Regards.

--- Y-H. H.
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