Please help me with the SPARQL query. I have an ontology with the class 'Verse' as a subclass of owl:Thing.
The Individuals look like this:
I need to do a SPARQL query, which will retrieve all the individuals according to the Verse (class name).
I have tried this, but it seems like this query is wrong:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
SELECT ?individual ?class
WHERE {
?individual rdf:type owl:NamedIndividual .
?class rdf:type owl:Class .
}
ps: for download my ontology data, here
You can directly use the class name from your ontology.
PREFIX ontobible: <ur ontology IRI>
SELECT ?individual
WHERE {
?individual a ontobible:Verse .
}
Related
I am trying to get a full subgraph, with all relations intact from the following ontology: http://purl.obolibrary.org/obo/NCBITaxon_9443
The NCBI ontology is very large and we only want the subset of all the primates for example. I currently use the following query:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT DISTINCT ?s2 ?s3 ?s4 ?s5 FROM <http://purl.obolibrary.org/obo/merged/NCBITAXON> WHERE {
SERVICE <http://sparql.hegroup.org/sparql>
{
?s rdf:type ?o .
?s2 rdfs:subClassOf ?s.
?s3 rdfs:subClassOf ?s2.
?s4 rdfs:subClassOf ?s3.
?s5 rdfs:subClassOf ?s4.
FILTER ( ?s = <http://purl.obolibrary.org/obo/NCBITaxon_9443> )
}
}
I don't think this is extensive enough and it also does not return a "nice" format. I would preferably get a subgraph in OWL or RDF(S) format back.
Can anyone help me with this? Thanks!
I have a question about polymorphism in RDF schema.
I created a class "MobilePhone":
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#prefix schema: <http://schema.org/> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
#prefix my: <http://localhost/> .
my:MobilePhone rdf:type rdfs:Class .
my:MobilePhone rdfs:subClassOf schema:Product .
my:MobilePhone rdfs:label "MobilePhone" .
Then I executed two queries:
PREFIX : <http://schema.org/>
PREFIX my: <http://localhost/>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT ?instance
WHERE {
?instance a my:MobilePhone .
}
and
PREFIX : <http://schema.org/>
PREFIX my: <http://localhost/>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT ?instance
WHERE {
?instance a :Product .
}
I expected the second query to retrieve all Products. Even "MobilePhone" products. But that didn't happen. I had to use:
PREFIX : <http://schema.org/>
PREFIX my: <http://localhost/>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT ?instance
WHERE {
?class rdfs:subClassOf* :Product .
?instance rdf:type ?class .
}
Does anyone know if it is possible to update my schema so the first query for searching all Products would work?
It is possible to update your data (not your schema [or ontology]) to deliver the desired result, by "forward chaining" -- i.e., by explicitly including statements that each entity which is a my:MobilePhone is also a :Product.
It's also possible to use "backward chaining" to deliver the desired result; the specifics of this will vary with your triple/quad store and other software. (My employer's solution, Virtuoso, uses a DEFINE input:inference pragma to activate an inference rule set for each query.)
Both of these may be considered as "inferencing" or "reasoning", though common usage generally applies these terms only to the dynamic action -- which happens with every query in "backward chaining", and once, pre-query, typically during data load, in "forward chaining".
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 .
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 .
}
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 .
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.
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:
X is of type C
there is no C' such that:
X is of type C'
C' is a subclass of C
C' is not equal to C
(that last condition is necessary, by the way, because in RDF/OWL, the subclass-relation is reflexive: every class is a subclass of itself)
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)
}
}
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 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.
The first ontology has the following:
Issue Ontology members(classes):
<http://www.issueonto.com/ontologies/issues#issues>
<http://www.issueonto.com/ontologies/issues#products>
Predicate/Properties:
<http://www.issueonto.com/ontologies/issues#hasIssues>
Triple store for this ontology (raw data), I show it here in Turtle format:
#prefix : <http://www.issueonto.com/ontologies/issues#> .
#prefix owl: <http://www.w3.org/2002/07/owl#> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
#prefix xml: <http://www.w3.org/XML/1998/namespace> .
#prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#base <http://www.issueonto.com/ontologies/issues> .
:Fido rdf:type :products ,
owl:NamedIndividual ;
:productName "FidoProdCEO_12"^^xsd:string ;
:hasIssues :issue_1239 .
### http://www.issueonto.com/ontologies/issues#issue_1239
:issue_1239 rdf:type :issues ,
owl:NamedIndividual ;
:issueName "FeatureIssue"^^xsd:string .
The Second ontology has the following:
Project Ontology members (classes):
<http://www.projectexample.com/ontology/project#GroupProject>
<http://www.projectexample.com/ontology/project#Project>
<http://www.projectexample.com/ontology/project#ProjectVersion>
Predicate/Properties:
<http://www.projectexample.com/ontology/project#belongsTo>
<http://www.projectexample.com/ontology/project#dependsOn>
Triple store for the ontology (raw data), I show it here in Turtle format:
#prefix : <http://www.projectexample.com/ontology/project#> .
#prefix owl: <http://www.w3.org/2002/07/owl#> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
#prefix xml: <http://www.w3.org/XML/1998/namespace> .
#prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#base <http://www.projectexample.com/ontology/project> .
### http://www.projectexample.com/ontology/project#Apple
:Apple rdf:type :ProjectVersion ,
owl:NamedIndividual ;
:hasProjectName "AppleTowandOne"^^xsd:string ;
:belongsTo :RedBlueCompany .
### http://www.projectexample.com/ontology/project#Fido
:Fido rdf:type :ProjectVersion ,
owl:NamedIndividual ;
:hasProjectName "FidoProdCEO"^^xsd:string ;
:dependsOn :Apple .
### http://www.projectexample.com/ontology/project#RedBlueCompany
:RedBlueCompany rdf:type :GroupProject ,
owl:NamedIndividual ;
:groupName "RedGroupCompant lmt"^^xsd:string .
Question
1- I would like to say, project:projectversion from ontology project same as issues:product from ontology issues, is that possible and how?
2- if question (1) is yes, how could I infer the similar individuals from the shared concepts, i.e, if we say projectversion is same as product it does not mean all the individuals are similar, in the example, i would like to automatically infer the individual issues:Fido type of issue:products is same as the individual prject:Fido type of project:projectversion. From that inferred fact, i would infer automatically that project:Fido issue:hasissue issues:issues_1239. Finally, I would like to run SPARQL query as follow:
SELECT ?product ?issue FROM <namegraph>
WHERE{
?product issues:hasIssues ?issue.
}
The results that I should get as follow:
?product ?issue
--------------------------------------------------------------------------
<http://www.projectexample.com/ontology/project#Fido> <http://www.issueonto.com/ontologies/issues#issue_1239>
<http://www.issueonto.com/ontologies/issues#Fido> <http://www.issueonto.com/ontologies/issues#issue_1239>
I would like to say, project:projectversion from ontology project same
as issues:product from ontology issues, is that possible and how?
All you need is the triple
project:projectversion owl:equivalentClass issues:product
I don't know how you're combining these ontologies; whether you're just loading the data from both into a triple store, or creating a third ontology that imports both and loading that (along with its imports) into a triple store, but somewhere you need that axiom. For a "merging" ontology like this, I'd usually create a third ontology that imports both (but leaving them unchanged) and add the axiom to that third ontology.
2- if question (1) is yes, how could I infer the similar individuals
from the shared concepts, i.e, if we say projectversion is same as
product it does not mean all the individuals are similar, in the
example, i would like to automatically infer the individual
issues:Fido type of issue:products is same as the individual
prject:Fido type of project:projectversion. From that inferred fact, i
would infer automatically that project:Fido issue:hasissue
issues:issues_1239.
You still have told us what criteria you would use to decide that issues:Fido and project:Fido are the same individual. The only apparent similarity that they have is the strings "FidoProdCEO_12" and "FidoProdCEO". Is that what the decision is supposed to be based on? If so, then you could do something like the following. I've created a minimal amount of data for convenience:
#prefix o1: <urn:ex:ont1#> .
#prefix o2: <urn:ex:ont2#> .
#prefix owl: <http://www.w3.org/2002/07/owl#> .
o1:A a o1:Product ;
o1:productName "ProductA_1234" ;
o1:hasIssue o1:issue42 .
o2:B a o2:ProjectVersion ;
o2:projectName "ProductA" .
o1:Product owl:equivalentClass o2:ProjectVersion .
prefix o1: <urn:ex:ont1#>
prefix o2: <urn:ex:ont2#>
prefix owl: <http://www.w3.org/2002/07/owl#>
select ?product ?issue where {
#-- A *product* is something that's an instance of
#-- o1:Product or another class that's equivalent
#-- to it.
?product a/(owl:equivalentClass|^owl:equivalentClass)* o1:Product
#-- The issues of a product are any of its
#-- o1:hasIssue values, or the o1:hasIssue
#-- value of any product that has a name
#-- beginning with its o2:projectName.
{ ?product o1:hasIssue ?issue }
union
{ ?product o2:projectName ?projectName .
?_product o1:productName ?productName ;
o1:hasIssue ?issue .
filter strstarts(?productName,?projectName)
}
}
------------------------
| product | issue |
========================
| o2:B | o1:issue42 |
| o1:A | o1:issue42 |
------------------------
Of course, the fact that you still end up having to examine projectName and productName values means that the equivalent class axiom isn't actually buying you all that much (at least in terms of this query). That is, it would be sufficient to just ask for "products (and projects with matching names) and their issues." That is, you get the same results from this query, which is just the second part of the first query:
prefix o1: <urn:ex:ont1#>
prefix o2: <urn:ex:ont2#>
prefix owl: <http://www.w3.org/2002/07/owl#>
select ?product ?issue where {
{ ?product o1:hasIssue ?issue }
union
{ ?product o2:projectName ?projectName .
?_product o1:productName ?productName ;
o1:hasIssue ?issue .
filter strstarts(?productName,?projectName)
}
}
The solution here is an owl:equivalentClass relation. To make this work you have to perform the following tasks:
Create and RDF document into which you place the owl:equivalentClass relation -- alternatively you can use SPARQL 1.1 INSERT to place the relation into a Virtuoso hosted Named Graph
Create an Inference Rule that functions as a Context-Lense when viewing the data in Virtuoso -- be it via SPARQL Query Results of an Entity Description Page etc.
Command (issued via SQL CommandLine of Conductor UI) for associating a Named Graph with an Inference Rule:
RDFS_RULE_SET ('{rule-name}', '{named-graph-uri-or-rdf-document-url}');
Links:
Sample file (note SQL part is commented out re., Rules and Named Graph association)
Live Example -- showcasing owl:equivalentClass reasoning via Schema.org, FOAF, and GoodRelations ontologies.
I am trying to extract the hierarchy of Wikipedia category or Yago classification for DBpedia resources using the SPARQL endpoint. For instance, I would like to find out all the possible categories and classes in hierarchical form of entity, say, http://dbpedia.org/resource/Nokia, like Thing → Organization → Company → … → Nokia.
A simple SPARQL select can retrieve the information that you're interested in, though it won't be arranged hierarchically. You're interested in getting all the types of a resource, as well as the rdfs:subClassOf relations between them. Here's a very simple query for Nokia that can be run on the DBpedia SPARQL endpoint
SELECT * WHERE {
dbpedia:Nokia a ?c1 ; a ?c2 .
?c1 rdfs:subClassOf ?c2 .
}
SPARQL results
If you treat each pair of classes in that result set as a directed edge and perform a topological sort , then you'll see the hierarchy of the classes to which the Nokia resource belongs. In fact, since it is probably convenient to treat this as a graph, you can get it in the form of an RDF graph by using a SPARQL construct query.
CONSTRUCT WHERE {
dbpedia:Nokia a ?c1 ; a ?c2 .
?c1 rdfs:subClassOf ?c2 .
}
SPARQL results
The construct query produces this graph (in N3 format):
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#prefix dbpedia-owl: <http://dbpedia.org/ontology/> .
#prefix owl: <http://www.w3.org/2002/07/owl#> .
#prefix yago: <http://dbpedia.org/class/yago/> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
#prefix dbpedia: <http://dbpedia.org/resource/> .
dbpedia-owl:Agent rdfs:subClassOf owl:Thing .
dbpedia-owl:Company rdfs:subClassOf dbpedia-owl:Organisation .
dbpedia-owl:Organisation rdfs:subClassOf dbpedia-owl:Agent .
yago:CompaniesBasedInEspoo rdfs:subClassOf yago:Company108058098 .
dbpedia:Nokia rdf:type yago:CompaniesListedOnTheHelsinkiStockExchange ,
owl:Thing ,
yago:CompaniesBasedInEspoo ,
dbpedia-owl:Agent ,
yago:DisplayTechnologyCompanies ,
yago:ElectronicsCompaniesOfFinland ,
dbpedia-owl:Company ,
dbpedia-owl:Organisation ,
yago:Company108058098 ,
yago:CompaniesEstablishedIn1865 .
yago:CompaniesEstablishedIn1865 rdfs:subClassOf yago:Company108058098 .
yago:CompaniesListedOnTheHelsinkiStockExchange rdfs:subClassOf yago:Company108058098 .
yago:DisplayTechnologyCompanies rdfs:subClassOf yago:Company108058098 .
yago:ElectronicsCompaniesOfFinland rdfs:subClassOf yago:Company108058098 .
Remarks
The queries above retrieve the rdf:type hierarchy for Nokia. In the question, you also mention Wikipedia categories. DBpedia resources are associated with the Wikipedia categories to which their corresponding articles belong by the dcterms:subject property. Those Wikipedia categories are then structured hierarchically by skos:broader. These really are not types for the individuals though. For instance, the data contain:
dbpedia:Nokia dcterms:subject category:Finnish_brands
category:Finnish_brands skos:broader category:Brands_by_country
While it probably makes sense to say that Nokia is a Finnish_brand, it makes much less sense to say that Nokia is a Brand_by_country.