Protege使用swrl:equal比较数据类型属性的值

问题描述

我正在建立一个3类的本体:

I am building an ontology of 3 classes :

1. 消息
2. 火腿
3. 垃圾邮件

2个dataproperties，domain:消息和范围xsd:string:

2 dataproperties , domain : Messages and range xsd:string :

1. 有兴趣
2. hasCategory

2个SWRL规则:Message(?x)，hasInterest(?x，?a)，hasCategory(?x，?b)，swrl:equal(?a，?b)-> Ham(?x)

2 SWRL Rules:Message(?x),hasInterest(?x,?a),hasCategory(?x,?b), swrl:equal(?a,?b) ->Ham(?x)

Message(?x)，hasInterest(?x，?a)，hasCategory(?x，?b)，swrl:notEqual(?a?b)->垃圾邮件(?x)

Message(?x),hasInterest(?x,?a),hasCategory(?x,?b), swrl:notEqual(?a?b) ->Spam(?x)

我想将Message类的实例分类为Spam或Ham类；如果hasCategory值(邮件类别)等于hasInterest值(用户兴趣)，则邮件为垃圾邮件

I want to classify instances of class Message to class Spam or Ham ; if the hasCategory value ( message category) is equal to the hasInterest value ( user interests) then the message is ham else spam

如果我有1个消息类别和1个兴趣，这可以正常工作前任:m1具有兴趣运动m1具有类别体育

This worked correctly If I have 1 message category and 1 interestex:m1 hasInterests sportsm1 hasCategory sports

那么，如果我有一个清单或类别的清单，例如:每则讯息都有1个以上的兴趣爱好{体育，电影}每封邮件都有1个以上的类别{电影，政治}

So what If I have a list of iterests or categories ex:Each message has more than 1 interests {sports, movies}Each message has more than 1 category {movies , politics}

我想说如果两个列表都相交，则消息是火腿，因此swrl:equal无效，我如何定义它以比较所有个体

I want to say if both lists intersect then the message is ham so the swrl:equal did not work how can i define it to compare all the individuals

我所做的是根据个人意思重复hasInterests和hasCategory，我的意思是手动定义列表并起作用，是否有另一种自动方式使用字符串列表以及如何在swrl中进行比较?

What I did is repeating the hasInterests and hasCategory depending on the individual values I mean defining manually the list and it worked , is there another automatic way using a list of strings and how to compare them in swrl ?

推荐答案

SWRL字符串内置程序( http://www.daml.org/rules/proposal/builtins.html )仅支持简单的字符串函数.在模型中，您可以为具有多个兴趣和许多类别的单个m1消息建模，例如:

SWRL Built-Ins for Strings (http://www.daml.org/rules/proposal/builtins.html) only support simple string functions.In your model you can model a message individual m1 with many interests and many categories like this:

m1 hasInterests "sports", m1 hasInterests "movies"
m1 hasCategory "sports", m1 m1 hasCategory "movies"

并按照您的规则

Message(?x),hasInterest(?x,?a),hasCategory(?x,?b), swrl:equal(?a,?b) ->Ham(?x)

每条具有至少一个等于类别的兴趣的消息将变为Ham.

every message with at least one interest equal to a category becomes Ham.

使用SQWRL查询可能有用的提示来找到感兴趣的数量

Perhaps useful hint to find number of interests but with the SQWRL Query:

Message(?x) ^ hasInterest(?x,?a) → sqwrl:select(?x) ^ sqwrl:count(?a)

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