[erlang-questions] Filtering hierarchical data structure

[Stanislav Ledenev]

Yes, I've skipped this one. Thank you!


On 14.07.2018 16:35, Dmitry Belyaev wrote:
> You can improve it by removing the last clause
> F({section, _, []}) -> false
>
>
> On 14 July 2018 22:41:34 GMT+10:00, Stanislav Ledenev 
> <s.ledenev@REDACTED> wrote:
>
>     Mikael, thank you very much for a hint!
>
>     After some reasoning I came up to this solution:
>
>     filter(List, Search) ->
>           F = fun F({policy, I, _}) -> lists:member(I, Search);
>                   F({section, SI, C}) ->
>                       R = lists:filtermap(F, C),
>                       case R of
>                           [] -> false;
>                           _ -> {true, {section, SI, R}}
>                       end;
>                   F({section, _, []}) -> false
>               end,
>           lists:filtermap(F, List).
>
>     It doesn't seem that it could be improved more.
>
>     On 14.07.2018 11:54, Mikael Pettersson wrote:
>
>         On Fri, Jul 13, 2018 at 3:25 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. 
>
>         Just some high-level comments. - Your data is clearly layered
>         (there are items which can be policies or sections, and
>         sections in turn contain lists of items), but your filtering
>         code conflates these and in turn gets confused as to what it's
>         processing, hence the large number of cases in filter/3. Also,
>         your code would crash if the top-level lists starts with a
>         policy, but accepts it elsewhere, which smells like a bug. -
>         You roll your own logic for reassembling the filtered-out
>         fragments, which gets unnecessarily fiddly, and in this case I
>         think wrong (you reorder items). - Using lists:filtermap/2
>         would greatly simplify your code. 
>
>
>     ------------------------------------------------------------------------
>
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>
>
> -- 
> Kind regards,
> Dmitry Belyaev
>
>
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-- 
-- Stanislav Ledenev

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