[erlang-questions] Update with complicated data structure

[Jesper Louis Andersen]

On Mon, Nov 23, 2015 at 11:06 AM, YuanZhiqian <on-your-mark@REDACTED>
wrote:

> Sorry to bother you again... I am totally a green hand in Erlang :(


Suppose for the moment you are willing to use a map to store all your
records. Then we can simplify the code quite a lot. List-representations
are best used if you often process all of the the elements in the list, or
you can use any element of the list, particularly the first one. If you are
pin-pointing a specific entry, the map is a far better data structure.
First a module definition:

-module(z).
-export([cal_win_notice/2]).

Next, lets hack up a couple of records to match what is going on in your
code:

-record(state, {
        '...',
        companies,
        campaigns
}).

-record(company_info, {
        id,
        budget,
        consumption,
        campaigns
}).

The mapply/3 function here tries to apply F to the map value given by the
key ID. If no such ID exists, it leaves the map alone.

mapply(F, ID, Map) ->
    case maps:get(ID, Map, undefined) of
        undefined -> Map;
        T -> Map#{ ID := F(T) }
    end.

Now, we can use this function. Make an F which does what we want. Then
apply it. We can check if any change were made to the map by matching on
the unchanged map. Otherwise, the map has been changed, and we can update
the map with the new one.

cal_win_notice({CoID, _Ca, _AdGrp, Price, _Cur}, #state { companies = Cos }
= State) ->
    F = fun(#company_info{ consumption = Cons } = X) -> X#company_info{
consumption = Cons + Price } end,
    case mapply(F, CoID, Cos) of
        Cos -> {not_found, State}; %% Unchanged map
        Changed -> {ok, State#state { companies = Changed }}
    end.

Want to make two changes?

G = fun(#company_info { consumption = Cons } = X) -> X#company_info {
consumption = Cons - Price } end,

compose(F, G) ->
    fun(X) -> G(F(X)) end.

then mapply(compose(F, G), CoID, Cos) ...




-- 
J.
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