Question

当我喜欢解决以下问题时，我正在解决一些序言练习：考虑一下你有关于object的事实基础：

object(obj1). 
object(obj2). 
object(obj3). 
object(obj4). 
object(obj5). 
material(obj1,wood). 
material(obj2,wood). 
material(obj3, glass). 
material(obj4, glass). 
material(obj5, iron). 
type(obj1, able). 
type(obj2, chair). 
type(obj3, mesa). 
type(obj4, jar). 
type(obj5, rattle). 
weight(obj1, 10.5). 
weight(obj2, 1.5). 
weight(obj3, 1.6). 
weight(obj4, 0.5). 
weight(obj5, 1.8).

现在的想法是创建谓词object_description（List），其中List是每个对象与它的特征的连接，如：

([obj1-wood-table-10.5, obj2-wood-chair-1.5, …, obj5-iron-rattle-1.8] )

我尝试使用bagof并找到了但却找不到合适的答案。

提前谢谢

Answer 1

 ?- findall(O-M-T-W,(object(O),material(O,M),type(O,T),weight(O,W)),Res).
Res = [obj1-wood-able-10.5, obj2-wood-chair-1.5, obj3-glass-mesa-1.6, obj4-glass-jar-0.5, obj5-iron-rattle-1.8].

Answer 2

我改变了输入格式。现在搜索更容易一些。我希望这样没事。

obj(obj1, material, wood).
obj(obj2, material, wood).
obj(obj3, material, glass).
obj(obj4, material, glass).
obj(obj5, material, iron).
obj(obj1, type, table).
obj(obj2, type, chair).
obj(obj3, type, mesa).
obj(obj4, type, jar).
obj(obj5, type, rattle).
obj(obj1, weight, 10.5).
obj(obj2, weight, 1.5).
obj(obj3, weight, 1.6).
obj(obj4, weight, 0.5).
obj(obj5, weight, 1.8).

鉴于此输入格式，您现在可以将其映射到列表（列表），例如像这样：

object_description(List) :-
    findall(Id-TmpList, bagof(Type-Value, obj(Id, Type, Value), TmpList), List).

这不会产生问题中的确切输出格式，但会提供类似的东西（并且可能更容易进一步处理）。

用法：

?- object_description(List).
List = [obj1-[material-wood, type-table, weight-10.5],
        obj2-[material-wood, type-chair, weight-1.5],
        obj3-[material-glass, type-mesa, weight-1.6],
        obj4-[material-glass, type-jar, weight-0.5],
        obj5-[material-iron, type-rattle, ... - ...]].

Answer 3

classic style of prolog :

    member(_, []):-!,fail.
    member(X, [X| _]).
    member(X, [_|T]):- member(X,T).

    object_description(R):-
           get_all_objects([], R).

    get_all_objects(T, [H|R]):-
          get_object(H),
          not(member(H,T)),
          get_all_objects([H|T], R).
    get_all_objects([], []).

    get_object(Obj):-
          object(X),
          material(X,M),
          type(X, T),
          weight(X, W),
          concat(X, "-", R1),
          concat(R1, M, R2),
          concat(R2, "-", R3),
          concat(R3, T, R4),
          concat(R4, "-", R5),
          concat(R5, W, Obj).

% // concat(str1, str2, str3) if your compilator have'nt you must make it or use other %//analog, idea is  str1+str2=str3

如何在Prolog中获取对象列表

3 个答案: