在SPARQL中删除不需要的超类答案 [英] Removing unwanted superclass answers in SPARQL

问题描述

我有一个包含分类学层次结构的OWL文件，我想编写一个查询，其中答案包括每个人及其直系分类学父级.这是一个示例(完整的查询更加混乱).

I have an OWL file that includes a taxonomic hierarchy that I want to write a query where the answer includes each individual and its immediate taxonomic parent. Here's an example (the full query is rather messier).

@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
@prefix rdf: <http:://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix : <urn:ex:> .

:fido rdf:type :Dog .
:Dog rdfs:subClassOf :Mammal .
:Mammal rdfs:subClassOf :Vertebrate .
:Vertebrate rdfs:subClassOf :Animal .
:fido :hasToy :bone

:kitty rdf:type :Cat .
:Cat rdfs:subClassOf :Mammal .
:kitty :hasToy :catnipMouse .

此查询可以满足我的要求.

And this query does what I want.

prefix rdf: <http:://www.w3.org/1999/02/22-rdf-syntax-ns#> .
prefix : <urn:ex:> .

SELECT ?individual ?type 
WHERE {
   ?individual :hasToy :bone .
   ?individual rdf:type ?type .
}

问题在于，我宁愿使用OWL文件的合理版本，当然也包括其他语句:

The problem is that I'd rather use a reasoned-over version of the OWL file, which unsurprisingly includes additional statements:

:fido rdf:type :Mammal .
:fido rdf:type :Vertebrate .
:fido rdf:type :Animal .
:kitty rdf:type :Mammal .
:kitty rdf:type :Vertebrate .
:kitty rdf:type :Animal .

现在查询产生有关Fido是哺乳动物等的其他答案.我可以放弃使用文件的合理版本，或者，由于SPARQL查询是从Java调用的，因此我可以做很多其他查询以查找出现的最小包容性类型.我的问题是，是否只有返回Dog方法才有合理的纯SPARQL解决方案.

And now the query results in additional answers about Fido being a Mammal, etc. I could just give up on using the reasoned version of the file, or, since the SPARQL queries are called from java, I could do a bunch of additional queries to find the least inclusive type that appears. My question is whether there is a reasonable pure SPARQL solution to only returning the Dog solution.

推荐答案

一个通用的解决方案是确保您只要求使用 direct 类型.如果满足以下条件，则类C是实例X的直接类型:

A generic solution is that you make sure you ask for the direct type only. A class C is the direct type of an instance X if:

1. X的类型为C
2. 没有C'这样的:
   • X的类型为C'
   • C'是C
   的子类
   • C'不等于C

1. X is of type C
2. there is no C' such that:
   • X is of type C'
   • C' is a subclass of C
   • C' is not equal to C

(顺便说一句，最后一个条件是必要的，因为在RDF/OWL中，子类关系是自反的:每个类都是其自身的子类)

(that last condition is necessary, by the way, because in RDF/OWL, the subclass-relation is reflexive: every class is a subclass of itself)

在SPARQL中，它变成这样:

In SPARQL, this becomes something like this:

PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX : <urn:ex:> .

SELECT ?individual ?type 
WHERE {
   ?individual :hasToy :bone .
   ?individual a ?type .
   FILTER NOT EXISTS { ?individual a ?other .
                       ?other rdfs:subClassOf ?type .
                       FILTER(?other != ?type)
   }
}

根据用于执行这些查询的API/三重存储/库，还可能存在其他特定于工具的解决方案.例如，Sesame API(公开:我在Sesame开发团队中)可以选择出于单个查询的目的而禁用推理:

Depending on which API/triplestore/library you use to execute these queries, there may also be other, tool-specific solutions. For example, the Sesame API (disclosure: I am on the Sesame dev team) has the option to disable reasoning for the purpose of a single query:

TupleQuery query = conn.prepareTupleQuery(SPARQL, "SELECT ...");
query.setIncludeInferred(false); 

TupleQueryResult result = query.evaluate();

Sesame还提供了一个可选的附加推断器(称为直接类型推断器")，它引入了可以查询的附加虚拟"属性，例如sesame:directType，sesame:directSubClassOf等.其他工具无疑也具有相似的选项.

Sesame also offers an optional additional inferencer (called the 'direct type inferencer') which introduces additional 'virtual' properties you can query, such as sesame:directType, sesame:directSubClassOf, etc. Other tools will undoubtedly have similar options.

这篇关于在SPARQL中删除不需要的超类答案的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！