rdfs

It is possible to use aggregate function when inserting data in SPARQL?

大城市里の小女人 提交于 2019-12-02 05:29:53
I have the following query which aims to prevent the insertion in a named graph when the amount of triples is of a given number, e.g., 5. I tried to reproduce the example showed here where the insertion of data is possible but I need to use the same named graph and, most important an aggregate function, i.e., COUNT. INSERT { GRAPH <http://example/g1> { ?s ?p ?o } } WHERE { GRAPH <http://example/g1> { ?s ?p ?o . FILTER ( COUNT(?p) < 5) } } Jena Fuseki raises the following error: Aggregate expression not legal at this point Yes, this is possible: INSERT { GRAPH <http://example/g2> { ?s ?p ?o } }

RDF Vocabulary and predicates

别等时光非礼了梦想. 提交于 2019-12-01 19:20:18
I am trying to convert some data to RDF format. I am able to create the basic ontologies using the dc and foaf types. The problem is that for more complex ontologies I need some more specific predicates. I am looking for some more specific vocabularies but I am not sure about what a vocabulary means.It is just a URI (maybe some website) with information on terms I want to use or it must be written in RDF format? An example: Can I use the IMDB terminology? e.g http://www.imdb.com/glossary/C unor You shouldn't use external URIs for RDF predicates that are not part of a vocabulary (defined by the

RDF Vocabulary and predicates

大城市里の小女人 提交于 2019-12-01 18:12:54
问题 I am trying to convert some data to RDF format. I am able to create the basic ontologies using the dc and foaf types. The problem is that for more complex ontologies I need some more specific predicates. I am looking for some more specific vocabularies but I am not sure about what a vocabulary means.It is just a URI (maybe some website) with information on terms I want to use or it must be written in RDF format? An example: Can I use the IMDB terminology? e.g http://www.imdb.com/glossary/C

How does schema.org usage fit into Linked Data principles?

别来无恙 提交于 2019-12-01 04:17:49
I am starting to learn schema.org schema. I come from the RDF/OWL community. To my surprise I could not indeed find an RDFS or OWL vocabulary representation of schema.org. More specifically, I realised that it was just a HTML page describing the schema. I also found an RDFa representation of the schema. Not sure that it can be a legal RDFS representation by the way. All of this got me a bit confused. In the meantime I found the following works: http://topbraid.org/schema/ and http://schema.rdfs.org/ . Hence I have the following few questions: 1- Are both work achieving the same thing? if not

How does schema.org usage fit into Linked Data principles?

谁都会走 提交于 2019-12-01 02:02:31
问题 I am starting to learn schema.org schema. I come from the RDF/OWL community. To my surprise I could not indeed find an RDFS or OWL vocabulary representation of schema.org. More specifically, I realised that it was just a HTML page describing the schema. I also found an RDFa representation of the schema. Not sure that it can be a legal RDFS representation by the way. All of this got me a bit confused. In the meantime I found the following works: http://topbraid.org/schema/ and http://schema

RDF Graph Entailment

橙三吉。 提交于 2019-11-30 19:30:11
I just read about the concept of entailment for RDF (Resource Description Framework).Can anyone tell me an example of entailment for two RDF graphs and explain them a bit. Thanks Suppose you have the following : ex:book1 rdf:type ex:Publication . ex:book2 rdf:type ex:Article . So a Sparql query like SELECT ?s { ?s rdf:type ex:Publication } will return only ex:book1 If you add the fact (or a graph in your data set with the fact) that states : ex:Article rdfs:subClassOf ex:Publication If your sparql engine processes entailments, it should deduce that a ex:Article is also a ex:Publication so

Ordering of entities in an ontology

眉间皱痕 提交于 2019-11-30 18:20:37
问题 I have a system that models some domain data in an ontology, the usual sort of triplestore. I've been looking around for a way to express plurality and ordering but haven't found anything via the googles. My main use case is something like one entity in the domain can be a list of tasks, (get groceries, cook meal, eat meal, something like that) but in general I feel like having the ability to 'weight' your edges might be useful across the board. Is there an accepted way of doing this? Just go

Check if an underlying representation is OWL or RDFS

蓝咒 提交于 2019-11-30 09:31:03
问题 Imagine you are querying a data source via an SPARQL endpoint and you want to know if the underlying representation of this data source is OWL or RDF/XML. Is there anyway that you would be able to do that via a SPARQL query? My personal line of thought was to write a query that uses one of the OWL properties and see if that returns any result, however the disadvantage of using such approach is that if you use an OWL property that doesn't appear in the data source even if the underlying

How to query Classes with Object Property in Sparql

安稳与你 提交于 2019-11-29 18:23:15
问题 Does any one know how to query Classes with Object Property in Sparql? Assume we have an OWL file which contains follows Human ----(hasPizza)---> Pizzas Human and Pizzas are classes (or concepts). In SPARQL, this query returns nothing: select ?x ?y where { ?x hasPizza ?y } But if I add two individuals (or entities) under those concepts like Human:Jim ----(hasPizza)---> Pizzas:cheesePizza that query will return ?x=Jim and ?y=cheesePizza How can I get ?x=Human and ?y=Pizza using SPARQL? 回答1:

Check if an underlying representation is OWL or RDFS

旧时模样 提交于 2019-11-29 16:21:28
Imagine you are querying a data source via an SPARQL endpoint and you want to know if the underlying representation of this data source is OWL or RDF/XML. Is there anyway that you would be able to do that via a SPARQL query? My personal line of thought was to write a query that uses one of the OWL properties and see if that returns any result, however the disadvantage of using such approach is that if you use an OWL property that doesn't appear in the data source even if the underlying representation is OWL you would not get a response. The assumption here is that your don't have access to the