Prolog中的聚合和

我在Prolog中有这样一个表：

invoice(number, locality, value)

invoice(1, madrid, 100)
invoice(2, lisbon, 200)
invoice(3, london, 300)
invoice(4, madrid, 300)
invoice(5, lisbon, 200)
invoice(6, paris, 100)

我怎样才能得到当地的总金额

答案应该类似于在Excel中执行数据透视表

里斯本总数：400 马德里总数：400 巴黎总数：100

---

伦敦总数：300

有各种方法。 以下程序就是其中之一

/* agreggate_sum. */

agreggate_sum :-
        sum_calc( Data_list ),

        write( 'locality, total' ), nl,

        write_data( Data_list ).


sum_calc( Data_list ) :-
        findall( (Locality, Value), invoice(_, Locality, Value ), Invoice_list ),
        data_reduction( Invoice_list, [], Data_list ).


data_reduction( [], Data_list, Data_list ) :- !.

data_reduction( [(Locality, Value) | Result], Tmp_list, Data_list ) :-
        data_reduction2( Tmp_list, Locality, Value, Tmp_list2  ),
        data_reduction( Result, Tmp_list2, Data_list ).


data_reduction2( [], Locality, Value, [(Locality, Value)] ) :- !.

data_reduction2( [(Locality, Total_Value) | Result], Locality, Value, [(Locality, Total_Value2) | Result] ) :-
        !,
        Total_Value2 is Total_Value + Value.

data_reduction2( [(Locality2, Total_Value) | Result], Locality, Value, [(Locality2, Total_Value) | Result2] ) :-
        data_reduction2( Result, Locality, Value, Result2 ).


write_data( [] ) :- !.

write_data( [(Locality, Total_Value) | Result] ) :-
        write( Locality ), write( ', ' ), write( Total_Value ), nl,
        write_data( Result ).


invoice(1, madrid, 100).
invoice(2, lisbon, 200).
invoice(3, london, 300).
invoice(4, madrid, 300).
invoice(5, lisbon, 200).
invoice(6, paris, 100).

结果

9 ?- agreggate_sum.

locality, total

madrid, 400

lisbon, 400

london, 300

paris, 100

---

有多种方法。 以下程序就是其中之一

/* agreggate_sum. */

agreggate_sum :-
        sum_calc( Data_list ),

        write( 'locality, total' ), nl,

        write_data( Data_list ).


sum_calc( Data_list ) :-
        findall( (Locality, Value), invoice(_, Locality, Value ), Invoice_list ),
        data_reduction( Invoice_list, [], Data_list ).


data_reduction( [], Data_list, Data_list ) :- !.

data_reduction( [(Locality, Value) | Result], Tmp_list, Data_list ) :-
        data_reduction2( Tmp_list, Locality, Value, Tmp_list2  ),
        data_reduction( Result, Tmp_list2, Data_list ).


data_reduction2( [], Locality, Value, [(Locality, Value)] ) :- !.

data_reduction2( [(Locality, Total_Value) | Result], Locality, Value, [(Locality, Total_Value2) | Result] ) :-
        !,
        Total_Value2 is Total_Value + Value.

data_reduction2( [(Locality2, Total_Value) | Result], Locality, Value, [(Locality2, Total_Value) | Result2] ) :-
        data_reduction2( Result, Locality, Value, Result2 ).


write_data( [] ) :- !.

write_data( [(Locality, Total_Value) | Result] ) :-
        write( Locality ), write( ', ' ), write( Total_Value ), nl,
        write_data( Result ).


invoice(1, madrid, 100).
invoice(2, lisbon, 200).
invoice(3, london, 300).
invoice(4, madrid, 300).
invoice(5, lisbon, 200).
invoice(6, paris, 100).

结果

9 ?- agreggate_sum.

locality, total

madrid, 400

lisbon, 400

london, 300

paris, 100

---

为了避免偶尔的访问者认为完成这些基本任务需要冗长的代码，以下是library（）方法：

也就是说，我们可以选择“透视”列。要获得完整结果，请包装到另一个聚合中：

?- aggregate(set(Locality=Sum), aggregate(sum(Value), Number^invoice(Number,Locality,Value), Sum), All).
All = [lisbon=400, london=300, madrid=400, paris=100].

---

为了避免偶尔的访问者认为完成这些基本任务需要冗长的代码，以下是library（）方法：

也就是说，我们可以选择“透视”列。要获得完整结果，请包装到另一个聚合中：

?- aggregate(set(Locality=Sum), aggregate(sum(Value), Number^invoice(Number,Locality,Value), Sum), All).
All = [lisbon=400, london=300, madrid=400, paris=100].

---

另一种解决方案是制作城镇价值对，按城镇分组，并计算城镇的总和：

aggregate_sum_by_town(Result) :-
    findall(T-V, invoice(_, T, V), TVs),
    keysort(TVs, Ps),
    group_pairs_by_key(Ps, G),
    pairs_keys_values(G, Ks, Vs),
    maplist(sumlist, Vs, Ss),
    pairs_keys_values(Result, Ks, Ss).

?- aggregate_sum_by_town(Result).
Result = [lisbon-400, london-300, madrid-400, paris-100].

---

另一种解决方案是制作城镇价值对，按城镇分组，并计算城镇的总和：

aggregate_sum_by_town(Result) :-
    findall(T-V, invoice(_, T, V), TVs),
    keysort(TVs, Ps),
    group_pairs_by_key(Ps, G),
    pairs_keys_values(G, Ks, Vs),
    maplist(sumlist, Vs, Ss),
    pairs_keys_values(Result, Ks, Ss).

?- aggregate_sum_by_town(Result).
Result = [lisbon-400, london-300, madrid-400, paris-100].

---

切得太多了！事实上，OP可能会给人留下这样的印象，他们是必要的，而不是…太多的削减！事实上，OP可能会觉得它们是必要的，而不是。。。