Question

我的一个谓词给了我一个像这样的列表作为输出：

WITH T
     AS (SELECT *,
                prev = LAG(direction) OVER (ORDER BY ID),
                rn = ROW_NUMBER() OVER (ORDER BY ID)
         FROM   #data)
SELECT mytime,
       Distance,
       Direction,
       SUM(CASE WHEN rn > 1 AND EXISTS(SELECT prev 
                                       INTERSECT 
                                       SELECT Direction) THEN 0 ELSE 1 END) OVER (ORDER BY id) AS GroupNumber
FROM   T 
ORDER BY ID

当具有[m(1,2,[v(1,y),v(1,y)]),m(1,6,[v(1,y),v(5,x)]),m(1,4,[v(3,x),v(1,y)]),m(1,8,[v(3,x),v(5,x)])]元素的子列表中的符号相等时（如此处v）：

或[v(1,y),v(1,y)]此处：

我必须对左边的数字求和并重建列表，如下所示：

[v(3,x),v(5,x)]

我有一个类似的谓词可以在主列表上运行，但我没有把这个正确。

这是另一个有效的谓词：

[m(1,2,[v(2,y)]),m(1,6,[v(1,y),v(5,x)]),m(1,4,[v(3,x),v(1,y)]),m(1,8,[v(8,x)])]

它以这种方式行事：

simplify([], []) :- !.
simplify([X], [X]) :- !.
simplify([m(C, TD, Var), m(C2, TD, Var)| Xs], Ys) :- 
    sumxp(C, C2, K), simplify([m(K, TD, Var)|Xs], Ys), !.
simplify([X, Y|Xs], [X|Ys]) :- !, simplify([Y|Xs], Ys).

sumxp(Power1, Power2, TotalPower) :- TotalPower is Power1 + Power2.

Answer 1

simplify_01规则适用于列表中的每个项目，例如m(1,6,[v(1,y),v(5,x)])

simplify_01(m(A,B,[v(X,Var1),v(Y,Var2)]),m(A,B,[v(X,Var1),v(Y,Var2)])) :-
    Var1 \= Var2.
simplify_01(m(A,B,[v(X,Var1),v(Y,Var1)]),m(A,B,[v(Z,Var1)])) :-
    Z is X + Y.

并且maplist只会将simplify_01应用于列表中的所有项目。

simplify(L,R) :-
    maplist(simplify_01,L,R).

使用SWI-Prolog

运行时

simplify([m(1,2,[v(1,y),v(1,y)]),m(1,6,[v(1,y),v(5,x)]),m(1,4,[v(3,x),v(1,y)]),m(1,8,[v(3,x),v(5,x)])],R).
R = [m(1, 2, [v(2, y)]), m(1, 6, [v(1, y), v(5, x)]), m(1, 4, [v(3, x), v(1, y)]), m(1, 8, [v(8, x)])] ;
false.

SWIProlog条件求和列表并重建

1 个答案: