识别Prolog中的句子结构 [英] Recognize sentence structure in Prolog

问题描述

我遇到一种情况，正在阅读三种不同类型的句子.它们将是以下形式之一:

I have a situation where I'm reading in three different types of sentences. They will be of one of the following forms:

_ is a _
A _ is a _
Is _ a _?

我需要能够识别输入了哪种类型的句子，然后添加或查询我的知识库.

I need to be able to recognize which type of sentence was entered and then add to or query my knowledge base.

例如，用户可以输入:

Fido is a dog.

然后我将该事实添加到我的知识库中.然后，用户可以输入:

I would then add that fact to my knowledge base. The user could then enter:

Is Fido a dog?

该程序将回答是.到目前为止，我认识事实的唯一想法是将句子拆分为空格并将其存储在列表中.然后检查以查看关键字在列表中的位置.这不是最佳解决方案，因为它假定"_"字符始终是单个单词.

And the program would answer yes. So far my only idea of recognizing the facts is splitting the sentences on spaces and storing them in a list. Then checking to see where the keywords appear in the list. This is not the best solution since it assumes the "_" characters will always be a single word.

有人有更好的解决方案吗?

Does anyone have a better solution?

推荐答案

将输入拆分为令牌列表的最简单方法， 是使用atomic_list_concat/3.这是一些示例:

The easiest way to split an input into a list of tokens, is to use atomic_list_concat/3. Here is are some example:

?- atomic_list_concat(X,' ','Fido is a dog .').
X = ['Fido', is, a, dog, '.'].

?- atomic_list_concat(X,' ','Is Fido a dog ?').
X = ['Is', 'Fido', a, dog, ?].

然后您可以使用DCG解析句子，返回解析 句子的树.最简单的是:

You can then use a DCG to parse sentences, return a parse tree of the sentence. The simplest is:

s(fact(isa(X,Y))) --> [X, is, a, Y, '.'].
s(query(isa(X,Y))) --> ['Is', X, a, Y, '?'].

要实际解析，请使用谓词短语/2.你可以 然后通过解析树解释器将所有内容粘合在一起， 称为谓词句柄/1:

To actually parse use the predicate phrase/2. You can then glue everything together by a parse tree interpreter, call this predicate handle/1:

do(X) :-
   atomic_list_concat(Y,' ',X),
   phrase(s(Z),Y),
   handle(Z).

:- dynamic isa/2.

handle(fact(X)) :- assertz(X), 
    write('Knowledege base extended'), nl.
handle(query(X)) :- 
    (X -> write('Query succeeded'), nl; 
          write('Query failed'), nl).

这是一个示例运行:

 ?- do('Is Fido a dog ?').
 Query failed

 ?- do('Fido is a dog .').
 Knowledege base extended

 ?- do('Is Fido a dog ?').
 Query succeeded

再见

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