在序言中返回一个列表 [英] Returning a list in prolog

问题描述

我想问一个关于返回列表的问题...事实:

i wanna ask a question about returning a list... Facts:

团队(团队名称、董事、国籍、总体目标)

TEAM(TEAMNAME,DIRECTOR,NATIOANALITY,OVERALLGOAL)

team (milan,allegri,italy, 8.5).
team (inter,benitez,italy,7.6).
team (barcelona,guardiola,spain,7.8).
team (realmadrid,mourinho,spain,7.2).

我想创建一个谓词:find(T,N,G) : T 是团队名称，N 是团队的国籍，这个团队的总目标必须大于 G.输出必须是这样的:

and i want to create a predicate: find(T,N,G) : T is name of team, N is nationality of team and this team's overallgoal must be greater than G. and outputs must be like these:

find([], spain,9).返回真

find(X, spain,6).返回 X=[巴塞罗那，皇家马德里]

我尝试这样做:

find(T,N,G):-find1(T,N,G),is_set(T).

find1([]).

find1([T|Ts],N,G):-team(T,_,N,Gs),Gc>G,find1(Ts).

它给出了结果，但不像上面的输出......如果我的目标是 find([],spain,9).然后给假...如果我的目标是 find(X,spain,6).然后给出第一个 X=barcelona 并等待;"之后给 X=realmadrid ......但我想要一个像上面那样的列表......

it gives results but not like output above... if my goal is find([],spain,9). then give false... if my goal is find(X,spain,6). then give first X=barcelona and wait for ";" after that give X=realmadrid... but i want to a list like above...

非常感谢...

推荐答案

要从子句数据库中提取满足谓词的项目列表，应使用 findall 谓词.例如，您的代码可以重写如下:

To extract a list of items satisfying a predicate from a database of clauses, one should use findall predicate. For example, your code could be rewritten as follows:

find(T, N, G) :- findall(X, (team(X, _, N, G0), G0 > G), T).

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