如何使用findall/3(仅使用一个谓词)内联目标? [英] How to inline a goal with findall/3, (use just one predicate)?

问题描述

我有一个看起来像这样的知识库

I have a knowledgebase that looks something like this

fact1(1, _, a, _, _).
fact1(2, _, c, _, _).
fact1(3, _, d, _, _).
fact1(4, _, f, _, _).

fact2(_, 1, b, _, _).
fact2(_, 2, c, _, _).
fact2(_, 4, e, _, _).

对于每个fact1& fact2，其中(在此示例中)数字匹配，我希望有一个对应的字母列表作为元组. 我想为此使用findall/3和一个谓词.

For every fact1 & fact2, where (in this example) the numbers match up, I want to have a list of the corresponding letters as tuples. I would like to use findall/3 and only one predicate for this.

在此之前，我曾问过一个问题，该问题是如何解决类似问题的，答案是使用两个谓词.该解决方案如下所示:

I have asked a question here before on how to solve something similar, where the answer was using two predicates. That solution looked like this:

find_item((Val1,Val2)):-
    fact1(A, _, Val1, _, _),
    fact2(_, A, Val2, _, _).`

test(Items) :-
    findall(Item,find_item(Item),Items).

给定事实示例的结果应如下所示:

The result for the given example of facts, should look like this:

[(a, b),  (c, c),  (f, e)]

是否可以仅使用 findall/3 ?

推荐答案

您可以内联过程find_item/1作为

You can inline procedure find_item/1 as the goal of findall/3 (use a conjunction of goals instead of a single goal):

test(Items):-
  findall((Val1, Val2), (fact1(A, _, Val1, _, _), fact2(_, A, Val2, _, _)), Items).

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