合并/合并两个Erlang列表

Question

如何在erlang中组合元组列表？我有清单:

L1 = [{k1, 10}, {k2, 20}, {k3, 30}, {k4, 20.9}, {k6, "Hello world"}],

和

L2 = [{k1, 90}, {k2, 210}, {k3, 60}, {k4, 66.9}, {k6, "Hello universe"}],

现在我想要一个组合列表:

L3 = [
       {k1, [10, 90]},
       {k2, [20, 210]},
       {K3, [30, 60]},
       {k4, [20.9, 66.9]},
       {K6, ["Hello world", "Hello universe"]}
     ].

Answer 1

这种技术称为合并连接。这在数据库设计中是众所周知的。

merge(L1, L2) ->
    merge_(lists:sort(L1), lists:sort(L2)).

merge_([{K, V1}|T1], [{K, V2}|T2]) -> [{K, [V1, V2]}|merge_(T1, T2)];
merge_([], []) -> [].

如果两个列表中可以有不同的键集并且您愿意删除可以使用的这些值

merge_([{K, V1}|T1], [{K, V2}|T2]) -> [{K, [V1, V2]}|merge_(T1, T2)];
merge_([{K1, _}|T1], [{K2, _}|_]=L2) when K1 < K2 -> merge_(T1, L2);
merge_(L1, [{_, _}|T2]) -> merge_(L1, T2);`
merge_(_, []) -> [].

或者如果您想将这些值存储在列表中

merge_([{K, V1}|T1], [{K, V2}|T2]) -> [{K, [V1, V2]}|merge_(T1, T2)];
merge_([{K1, V1}|T1], [{K2, _}|_]=L2) when K1 < K2 -> [{K1, [V1]}|merge_(T1, L2)];
merge_(L1, [{K2, V2}|T2]) -> [{K2, [V2]}|merge_(L1, T2)];
merge_(L1, []) -> [{K, [V]} || {K, V} <- L1].

如果您不介意结果相反，您当然可以使用尾递归版本，或者您始终可以使用lists:reverse/1

merge(L1, L2) ->
    merge(lists:sort(L1), lists:sort(L2), []).

merge([{K, V1}|T1], [{K, V2}|T2], Acc) -> merge(T1, T2, [{K, [V1, V2]}|Acc]);
merge([], [], Acc) -> Acc. % or lists:reverse(Acc).

或者

merge([{K, V1}|T1], [{K, V2}|T2], Acc) -> merge(T1, T2, [{K, [V1, V2]}|Acc]);
merge([{K1, _}|T1], [{K2, _}|_]=L2, Acc) when K1 < K2 -> merge(T1, L2, Acc);
merge(L1, [{_, _}|T2], Acc) -> merge(L1, T2, Acc);`
merge(_, [], Acc) -> Acc. % or lists:reverse(Acc).

或者

merge([{K, V1}|T1], [{K, V2}|T2], Acc) -> merge(T1, T2, [{K, [V1, V2]}|Acc]);
merge([{K1, V1}|T1], [{K2, _}|_]=L2, Acc) when K1 < K2 -> merge(T1, L2, [{K1, [V1]}|Acc]);
merge(L1, [{K2, V2}|T2], Acc) -> merge(L1, T2, [{K2, [V2]}|Acc]);`
merge([{K1, V1}|T1], [], Acc) -> merge(T1, [], [{K1, [V1]} | Acc]);
merge([], [], Acc) -> Acc. % or lists:reverse(Acc).
% or merge(L1, [], Acc) -> lists:reverse(Acc, [{K, [V]} || {K, V} <- L1]).
% instead of two last clauses.

如果其中一个列表可能包含相同的键并且您愿意收集所有值，您可以考虑这样做

merge(L1, L2) ->
    merge(lists:sort(L1), lists:sort(L2), []).

merge([{K1, _}|_]=L1, {K2, _}|_]=L2, Acc) ->
    K = min(K1, K2),
    {Vs1, T1} = collect(K, L1, []),
    {Vs2, T2} = collect(K, L2, Vs1),
    merge(T1, T2, [{K, Vs2}|Acc]);
merge([{K, _}|_]=L1, [], Acc) ->
    {Vs, T1} = collect(K, L1, []),
    merge(T1, [], [{K, Vs}|Acc]);
merge([], [{K, _}|_]=L2, Acc) ->
    {Vs, T2} = collect(K, L2, []),
    merge([], T2, [{K, Vs}|Acc]);
merge([], [], Acc) -> lists:reverse(Acc).

collect(K, [{K, V}|T], Acc) -> collect(K, T, [V|Acc]);
collect(_, T, Acc) -> {Acc, T}.

Answer 2

通过使用sofsErlang标准库中的模块,可以很好地解决这个问题.该sofs模块描述了用于处理数学集的DSL.这是其中一种情况,您可以通过将数据转换为SOFS世界来利用它,在这个世界中操纵它们,然后再将它们转换回来.

请注意,我确实更改了L3,因为sofs不保留字符串顺序.

-module(z).

-compile(export_all). % Don't do this normally :)

x() ->
    L1 = [{k1, 10}, {k2, 20}, {k3, 30}, {k4, 20.9}, {k6, "Hello world"}],
    L2 = [{k1, 90}, {k2, 210}, {k3, 60}, {k4, 66.9}, {k6, "Hello universe"}],
    L3 = [{k1, [10, 90]},{k2, [20, 210]},{k3, [30, 60]},{k4, [20.9, 66.9]},{k6, ["Hello universe", "Hello world"]}],
    R = sofs:relation(L1 ++ L2),
    F = sofs:relation_to_family(R),
    L3 = sofs:to_external(F),
    ok.

Answer 3

更短的东西，列表甚至不必拥有相同的键，并且可以是无序的：

merge(In1,In2) ->
    Combined = In1 ++ In2,
    Fun      = fun(Key) -> {Key,proplists:get_all_values(Key,Combined)} end,
    lists:map(Fun,proplists:get_keys(Combined)).

Fun 可以直接写在lists:map/2函数中，但这使得它可读。

输出，包含示例中的数据：

1> test:merge(L1,L2).
[{k1,"\nZ"},
 {k2,[20,210]},
 {k3,[30,60]},
 {k4,[20.9,66.9]},
 {k6,["Hello world","Hello universe"]}]

"\nZ"是因为 erlang 将 [10,90] 解释为字符串（实际上是列表）。别打扰。