I'm calling on http://fr.dbpedia.org/sparql
the following SPARQL query:
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
select distinct ?lcs where {
{
?lcs ^(rdf:type/rdfs:subClassOf*) <http://fr.dbpedia.org/resource/Honoré_Daumier> ,
<http://fr.dbpedia.org/resource/Auguste_Rodin>;
a owl:Class .
filter not exists {
?llcs ^(rdf:type/rdfs:subClassOf*) <http://fr.dbpedia.org/resource/Honoré_Daumier> ,
<http://fr.dbpedia.org/resource/Auguste_Rodin>;
a owl:Class ;
rdfs:subClassOf+ ?lcs .
}
}
}
On some call I've http://dbpedia.org/ontology/Person as result, on others call I'm getting http://dbpedia.org/ontology/Person and http://dbpedia.org/ontology/Agent and with others the previous answers plus http://www.w3.org/2002/07/owl#Thing
without nothing to know that a response isn't complete. How can I use the result, if it is a little randomized
The main reason for your query not working as expected is that the data the data is i) split into separate graphs and ii) not all graphs were added to the default graph.
To keep it short, the instance data is contained inside the graph http://fr.dbpedia.org whereas the schema triples will be accessible via http://dbpedia.org graph only. Sometimes, if no graph is given, then the union of some graphs is used as the default graph which will be the dataset at query time. Unfortunately, this doesn't hold for the French DBpedia endpoint, only the instance data graph will be used.
You can check this with
DESCRIBE <http://dbpedia.org/ontology/Person>
which is empty when using either no graph explicitly or the graph http://fr.dbpedia.org, but non-empty for graph http://dbpedia.org.
The way to define the default graph is using the keyword FROM. For your query, it should therefore be
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
select distinct ?lcs
from <http://fr.dbpedia.org>
from <http://dbpedia.org>
where {
?lcs ^(rdf:type/rdfs:subClassOf*) <http://fr.dbpedia.org/resource/Honoré_Daumier> ,
<http://fr.dbpedia.org/resource/Auguste_Rodin>;
a owl:Class .
filter not exists {
?llcs ^(rdf:type/rdfs:subClassOf*) <http://fr.dbpedia.org/resource/Honoré_Daumier> ,
<http://fr.dbpedia.org/resource/Auguste_Rodin>;
a owl:Class ;
rdfs:subClassOf+ ?lcs .
}
}
Note, while this seems to return the correct result, you should also consider the comment from #TallTed regarding possible differences among language chapters (e.g. English vs French Wikipedia as source), release dumps (2016 vs 2018 or even the DBpedia Live) as well as Virtuoso versions used as backend.
Related
I am trying to list all properties created in on a wikibase I installed, using docker-compose, based on this install.
Now, want to list all properties that are available in this wikibase, similar to getting that list available through:
<wikibase.url>wiki/Special:ListProperties
I have also extracted that list through SPARQL with the following SPARQL query:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX wikibase: <http://wikiba.se/ontology#>
PREFIX schema: <http://schema.org/>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
SELECT DISTINCT
?property
?propertyType
?propertyLabel
?propertyAltLabel
WHERE {
?property a wikibase:Property ;
rdfs:label ?propertyLabel ;
wikibase:propertyType ?propertyType .
OPTIONAL {?property skos:altLabel ?propertyAltLabel .}
}
Running that SPARQL query is expensive though and I need to run that query often, so I would very much like to get that list of properties to the core wikibase API.
Is that possible?
If you know the namespace number for properties on the target wiki (it’s usually 122 if the wiki has an Item: namespace, or 120 if, like on Wikidata, items are in the main namespace), you can use the core allpages API: https://www.wikidata.org/w/api.php?action=query&list=allpages&apnamespace=120
To also get the labels at the same time, use it as a generator and combine it with the entityterms API (new in 1.35; looks like it’s not documented yet, but see T257658): https://www.wikidata.org/w/api.php?action=query&generator=allpages&gapnamespace=120&prop=entityterms&wbetterms=label
I have setup GraphDB SE trial version and trying out inference functionality with OWL2-RL ruleset. I have built a simple SKOS knowledge with a single broader relationship. Some how, when I try to query for narrower relationship am not getting any results. Am I going wrong in the usage ?
Insertion:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
INSERT DATA {
ex:mammals rdf:type skos:Concept;
skos:prefLabel "mammals"#en;
ex:animals rdf:type skos:Concept;
skos:prefLabel "animals"#en;
skos:broader ex:mammals .
}
Query:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
select * where {
?s skos:narrower ?o .
}
In the query result I don't see any response. Shouldn't it return below result -
ex:mammals skos:narrower ex:animals
I just came across this question and the OP is correct: Although the official SKOS reference explicitly states that skos:broader and skos:narrower are inverse, the actual RDF implementation does not include this statement. However, properties skos:broaderTransitive and skos:narrowerTransitive are declared as inverse.
And, although not relevant to the original question, the implementation does not state properties skos:topConceptOf and skos:hasTopConcept as inverse either.
I have this query
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX foaf: <http://xmlns.com/foaf/0.1/>
PREFIX dbpedia: <http://dbpedia.org/>
PREFIX dbpedia_property: <http://dbpedia.org/property/>
PREFIX dbpedia_ontology: <http://dbpedia.org/ontology/>
PREFIX yago: <http://dbpedia.org/class/yago/>
PREFIX schema: <http://schema.org/>
SELECT * WHERE
{
{
SELECT ?school
WHERE
{
?school rdf:type yago:EducationalInstitution108276342 .
FILTER ( contains(str(?school), "Australia") )
}
ORDER BY ?school
}
}
Don't mind the extra brackets as this is part of a larger query.
What I want to know is why thi http://dbpedia.org/page/Academic_structure_of_the_Australian_National_University is included in the results since I specify rdf:type yago:EducationalInstitution108276342. This property is not included in the resource page. I'm using this endpoint: http://dbpedia.org/sparql
Looks like a bug in the Pubby Web interface or in the query that is used to get the data that will be shown.
The query
SELECT * WHERE{
<http://dbpedia.org/resource/Academic_Structure_of_the_Australian_National_University> ?p ?o
}
returns the necessary rdf:type statement.
The other strange thing is that even a SPARQL DESCRIBE query does not return the rdd:type triples:
DESCRIBE <http://dbpedia.org/resource/Academic_Structure_of_the_Australian_National_University>
Although DESCIBE is not really defined in the specs, a user would expect those triples for sure. And maybe this kind of query is used to retrieve the data for the Web pages of resources.
I installed Jena Fuseki and want to be able to get classes from my own OWL file.
The following query returns the classes from owl and rdfs but not from ont. How can I retrieve them? I eventually want to add data to TDB using parts from my own OWL ontology when querying with Fuseki. I'm using Fuseki version: 2.3.1.
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
prefix ont: <http://wad.nistorandrei.com/ontology.owl#>
SELECT ?class ?label ?description
WHERE {
?class a owl:Class.
OPTIONAL { ?class rdfs:label ?label}
OPTIONAL { ?class rdfs:comment ?description}
}
The description of your issue makes me suspect that you did not load your ontology in Fuseki.
Declaring prefix ont: http://wad.nistorandrei.com/ontology.owl# will not allow you to query the distant ontology.owl file.
You have to create a dataset (in manage datasets -> create dataset)
and add data to it (existing dataset -> upload data or with a SPARQL UPDATE query)
Then only you can query your data.
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.