我有一些包含数据的表单的大清单,这些表单需要与来自其他同名表单的其他数据进行比较。
列表格式非常复杂,如下所示:
[[{EVAL_SEQ_1, {FORMNAME, ListDataToConcat1}}], [{EVAL_SEQ_2, {FORMNAME, ListDataToConcat2}}], ...]
这是我想要的输出:
[[{EVAL_SEQ_1,{FORMNAME,ListDataToConcat1 + ListDataToConcat2}}]}}] ...]
其中:
EVAL_SEQ_1 = Form Sequence Number,
FORMNAME = Form Name
ListDataToConcat = List that Needs to concat
例如。这是我的样本数据:
[[{"eval_data_12",
{<<"prvl_mobable_asset_0000_h200401">>,
[{'F_01_0100',[1]},
{'F_01_0090',["3"]},
{'F_01_0080',[]},
{'F_01_0070',[9999]},
{'F_01_0060',[{era,0},{year,[]},{month,[]}]},
{'F_01_0050',[]},
{'F_01_0040',[]},
{'F_01_0030',[]},
{'F_01_0020',<<>>},
{'F_01_0010',<<"4 - 8">>}]}}],
[{"eval_data_11",
{<<"prvl_mobable_asset_0000_h200401">>,
[{'F_01_0100',[]},
{'F_01_0090',["2"]},
{'F_01_0080',[]},
{'F_01_0070',[22222]},
{'F_01_0060',[{era,0},{year,[]},{month,[]}]},
{'F_01_0050',[]},
{'F_01_0040',[]},
{'F_01_0030',[]},
{'F_01_0020',<<>>},
{'F_01_0010',<<"4 - 1">>}]}}], ...]
我希望结果输出如下:
[{"eval_data_11",
{<<"prvl_mobable_asset_0000_h200401">>,
[{'F_01_0100',[[], [1]]},
{'F_01_0090',[["2"], ["3"]]},
{'F_01_0080',[[], []]},
{'F_01_0070',[[22222], [9999]]},
{'F_01_0060',[[{era,0},{year,[]},{month,[]}], [{era,0},{year,[]},{month,[]}]]},
{'F_01_0050',[[], []]},
{'F_01_0040',[[], []]},
{'F_01_0030',[[], []]},
{'F_01_0020',[[<<>>], [<<>>]]},
{'F_01_0010',[[<<"4 - 1">>], [<<"4 - 8">>]}]}}]
答案 0 :(得分:1)
@trex:
FormList = [[{EVAL_SEQ_1, {FORMNAME, ListDataToConcat1}}], [{EVAL_SEQ_2, {FORMNAME, ListDataToConcat2}}], ....]
您需要单独列出具有<<"prvl_mobable_asset_0000_h200401">>
{MobableList, EvalDataList} = lists:partition(fun([{EVAL_SEQ, {FORMNAME, ListData}}]) ->
FORMNAME == <<"prvl_mobable_asset_0000_h200401">>
end, FormList),
然后,获取单独的表单序列和列表元组。将它们分开
{EvalSeq, MergingList} = lists:foldl(fun(X, {EvalNames, OutList}) ->
[{EVAL_SEQ, {FORMNAME, ListData}}] = X,
{[EVAL_SEQ|EvalNames], [ListData|OutList]}
end, {[], []}, MobableList),
因此,您将获得新的元组:
{[EVAL_SEQ_1, EVAL_SEQ_2, EVAL_SEQ_3, ...], [ListDataToConcat1, ListDataToConcat2, ListDataToConcat3,...]}
我不确定你想要哪个序列号,因为你没有清楚地提到它,这是你获得最小序列号的方法。 Evalsequence = lists:min(EvalSeq),
现在使用merge
功能合并您的代码,如下所示,或者您可以参考merging inner list Merge inner lists of a list erlang:
MergedList = merge(MergingList),
最后是一个单独的列表:
[{Evalsequence, {<<"prvl_mobable_asset_0000_h200401">>, MergedList}}].
merge(ListOfLists) ->
Combined = lists:append(ListOfLists),
Fun = fun(Key) -> {Key,proplists:get_all_values(Key,Combined)} end,
lists:map(Fun,proplists:get_keys(Combined)).
答案 1 :(得分:1)
我建议你这个解决方案:
<强> [编辑] 强>
我修改了代码来回答你的上一条评论,还有一些模糊的东西:
一句话,我错过了一些上下文,但如果我添加收集这些信息,我会将其存储在ets表中。如果需要,更新更快,易于遍历,并且易于转换为列表。
-module (t).
-compile([export_all]).
% rec = {atom,term}
% reclist = [rec,...]
% asset = {bin,reclist}
% eval_data = [{list,asset}]
% eval_set = [eval_data,...]
% recs = {atom,[term]}
% recslist = [recs,...]
addrec({Key,Val},Recslist) ->
Val_list = proplists:get_value(Key, Recslist, []),
[{Key,[Val|Val_list]}|proplists:delete(Key,Recslist)].
merge_rec(Reclist,Recslist) -> lists:foldl(fun(Rec,Acc) -> addrec(Rec,Acc) end,Recslist,Reclist).
merge_eval([{Eval,{Asset,Reclist}}],[]) ->
[{Eval,{Asset,[{Key,[Val]} || {Key,Val} <- Reclist]}}];
merge_eval([{Eval,{Asset,Reclist}}],[{Eval_low,{Asset_low,Recslist}}]) ->
[{min(Eval,Eval_low),{min(Asset,Asset_low),merge_rec(Reclist,Recslist)}}].
merge_set(Eval_set) -> lists:foldl(fun(Eval_data,Acc) -> merge_eval(Eval_data,Acc) end,[],Eval_set).
test() ->
Eval_set = [[{"eval_data_10",
{<<"prvl_mobable_asset_0000_h200401">>,
[{'F_01_0100',[1]},
{'F_01_0090',["3"]},
{'F_01_0080',[]},
{'F_01_0070',[9999]},
{'F_01_0060',[{era,0},{year,[]},{month,[]}]},
{'F_01_0050',[]},
{'F_01_0040',[]},
{'F_01_0030',[]},
{'F_01_0020',<<>>},
{'F_01_0010',<<"4 - 8">>}]}}],
[{"eval_data_11",
{<<"prvl_mobable_asset_0000_h200401">>,
[{'F_01_0100',[]},
{'F_01_0090',["2"]},
{'F_01_0080',[]},
{'F_01_0070',[22222]},
{'F_01_0060',[{era,0},{year,[]},{month,[]}]},
{'F_01_0050',[]},
{'F_01_0040',[]},
{'F_01_0030',[]},
{'F_01_0020',<<>>},
{'F_01_0010',<<"4 - 1">>}]}}]],
merge_set(Eval_set).