The following code queries DBpedia for places within a bounded geographic area and returns the name, lat, and long of the place. I'd also like the query to return the category of the place--e.g., park, restaurant, museum, etc.
The following code works fine.
sparql = SPARQLWrapper("http://dbpedia.org/sparql")
sparql.setQuery("""
PREFIX geo: <http://www.w3.org/2003/01/geo/wgs84_pos#>
PREFIX dbo: <http://dbpedia.org/ontology/>
PREFIX category: <http://dbpedia.org/resource/Category:>
SELECT * WHERE {
?s a dbo:Place .
?s geo:lat ?lat .
?s geo:long ?long .
I tried to add the following code to get categories for places, but this doesn't work:
?s category:cat ?cat .
What should I add/change? Thanks.
You can get the category of a place (assuming you mean the type) by finding the type (rdfs:type) or the subject (dcterms:subject) of a resource. In DBPedia the first relates to the DBPedia and Yago ontologies and the second is a SKOS hierarchy in DBPedia. Here is an example query:
PREFIX geo: <http://www.w3.org/2003/01/geo/wgs84_pos#>
PREFIX dbo: <http://dbpedia.org/ontology/>
PREFIX dcterms: <http://purl.org/dc/terms/>
SELECT * WHERE {
?s a dbo:Place .
?s geo:lat ?lat .
?s geo:long ?long .
?s a ?type .
?s dcterms:subject ?sub
}
Note that you will get multiple types and subjects for each place.
Related
Hello I am trying to find the total number of municipalities a region has along with the name of each region and the total number of municipalities a regional unit has along with the name of the regional unit. A region consists of regional units and a regional unit consists of municipalities. Below is my query that unfortunately returns wrong results. What I am basically trying to do is group by region and get the name and the total municipalities of each region and group by regional unit and take the name and the total municipalities of each unit. Any suggestions to the right direction would be appreciated. Cheers!:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema
PREFIX strdf: <http://strdf.di.uoa.gr/ontology
PREFIX gag: <http://geo.linkedopendata.gr/gag/ontology/>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
SELECT ?region ?municipality_region ?unit ?municipality_unit
WHERE
{
{ SELECT ?region (COUNT(?municipality) AS ?municipality_region)
{?m rdf:type gag:Δήμος .
?m gag:έχει_επίσημο_όνομα ?municipality .
?m gag:ανήκει_σε ?reg_un .
?reg_un gag:ανήκει_σε ?reg .
?reg gag:έχει_επίσημο_όνομα ?region .
}GROUP BY ?region}
{ SELECT ?unit (COUNT(?municipality_un) AS ?municipality_unit)
{ ?m rdf:type gag:Δήμος .
?m gag:έχει_επίσημο_όνομα ?municipality_un .
?m gag:ανήκει_σε ?reg_un .
?reg_un gag:έχει_επίσημο_όνομα ?unit .
} GROUP BY ?unit}
};
Below I am giving a mapping of properties in english:
Δήμος = municipality
έχει_επίσημο_όνομα = has name
ανήκει_σε = belongs to
And here is the ontology I am working with:
link
I want to extract a list o players and a list of clubs where it has played, separated by commas.
SELECT DISTINCT ?playerLabel
(GROUP_CONCAT(?teamLabel ; separator=',') as ?teams)
WHERE {
?player wdt:P106 wd:Q937857 .
?player wdt:P2574 ?team
SERVICE wikibase:label { bd:serviceParam wikibase:language "[AUTO_LANGUAGE],en". }
}
GROUP BY ?playerLabel
I have two problems:
I don't get a list of teams for each player, only the name, and variable ?teams empty.
If I don't use GROUP CONCAT and GROUP BY I obtain the team id, but I prefer the label of team.
For example 2 players...:
playerLabel teams
Cristiano Ronaldo Sporting Portugal, Manchester U, Real Madrid, Juventus, Manchester U
Leo Messi Barcelona, PSG
At least I need the Concat and group by, even with code...
thanks
You use P2574, which is "National-Football-Teams.com player ID". While National-Football-Teams.com lists all teams a player played for, this data is not accessible through the Wikidata Query Service. But Wikidata itself has a dedicated property for sports team member: P54.
So write ?player wdt:P54 ?team instead of ?player wdt:P2574 ?team.
Additionaly, you need to add ?team rdfs:label ?teamLabel . filter (lang(?teamLabel)='en') to be able to use ?teamLabel in GROUP_CONCAT().
Thus, the full working query looks like this (restricted to US players to avoid query time outs):
SELECT DISTINCT ?playerLabel (GROUP_CONCAT(?teamLabel ; separator=',') as ?teams)
WHERE {
?player wdt:P106 wd:Q937857 .
?player wdt:P27 wd:Q30 .
?player wdt:P54 ?team .
?team rdfs:label ?teamLabel . filter (lang(?teamLabel)='en')
SERVICE wikibase:label { bd:serviceParam wikibase:language "[AUTO_LANGUAGE],en". }
}
GROUP BY ?playerLabel
I have an "EquivalentTo" definition in Protege of a class EquivClass as (hasObjProp some ClassA) and (has_data_prop exactly 1 rdfs:Literal)
Is there a form of SPARQL query for GraphDB 9.4 to get the "direct" answer to a select query of an equivalent class without having to collect and traverse all the constituent blank nodes explicitly? Basically, I'm looking for a short cut. I'm not looking to get instances of the equivalent class, just the class definition itself in one go. I've tried to search for answers, but I'm not really clear on what possibly related answers are saying.
I'd like to get something akin to
(hasObjProp some ClassA) and (has_data_prop exactly 1 rdfs:Literal)
as an answer to the SELECT query on EquivClass. If the answer is "not possible", that's enough. I can write the blank node traversal with the necessary properties myself.
Thanks!!
Files are -
Ontology imported into GraphDB: tester.owl - https://pastebin.com/92K7dKRZ
SELECT of all triples from GraphDB *excluding* inferred triples: tester-graphdb-sparql-select-all-excl-inferred.tsv - https://pastebin.com/fYdG37v5
SELECT of all triples from GraphDB *including* inferred triples: tester-graphdb-sparql-select-all-incl-inferred.tsv - https://pastebin.com/vvqPH1FZ
Added sample query in response to #UninformedUser. I use "select *" for example, but really I'm interested in the "end results", ie, ?fp, ?fo, ?rop, ?roo. Essentially, I'm looking for something simpler and more succinct than what I have below.The example I posted only has a single intersection ("and" clause). In my real world set, there are multiple equiv classes with different numbers of "and" clauses.
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX : <http://www.semanticweb.org/ontologies/2020/9/tester#>
select * where {
:EquivClass owl:equivalentClass ?bneq .
?bneq ?p ?bnhead .
?bnhead rdf:first ?first .
?first ?fp ?fo .
?bn3 rdf:rest ?rest .
?rest ?rp ?ro .
?ro ?rop ?roo .
filter(?bn3 != owl:Class && ?ro!=rdf:nil)
}
You can unroll the list using a property path:
prefix : <http://www.semanticweb.org/ontologies/2020/9/tester#>
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix xsd: <http://www.w3.org/2001/XMLSchema#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
select * {
:EquivClass owl:equivalentClass/owl:intersectionOf/rdf:rest*/rdf:first ?restr.
?restr a owl:Restriction .
optional {?restr owl:onProperty ?prop}
optional {?restr owl:cardinality ?cardinality}
optional {?restr owl:someValuesFrom ?class}
}
This returns:
| | restr | prop | cardinality | class |
| 1 | _:node3 | :hasObjProp | | :ClassA |
| 2 | _:node5 | :has_data_prop | "1" ^^xsd:nonNegativeInteger | |
I am trying to select one label and comment of classes but group by doesn't work as expected. The following query is an example of the select.
SELECT ?class ?label ?comment WHERE
{
{SELECT DISTINCT ?class WHERE { {?uri rdf:type ?class}UNION {?class rdf:type owl:Class} UNION {?class rdf:type rdfs:Class} }OFFSET 0 LIMIT 100}
.optional{?class rdfs:label ?label}
.optional{?class rdfs:comment ?comment}
}GROUP BY ?class
The goal is to to have every class uri with one label and comment
But am getting results as :
http://dbpedia.org/ontology/Activity "attività"#it
http://dbpedia.org/ontology/Activity "活動"#ja
Any idea ?
Your query is actually illegal - it uses non-aggregate, non-group key in the outer SELECT.
You need to use "SAMPLE" to one pick (random) item from a group if you think there might be multiple labels or comments.
SELECT ?class (sample(?labelX) as ?label) (sample(?commentX) as ?comment) WHERE
{
SELECT DISTINCT ?class {
{?uri rdf:type ?class} UNION
{?class rdf:type owl:Class} UNION
{?class rdf:type rdfs:Class}
} LIMIT 100
optional{?class rdfs:label ?labelX}
optional{?class rdfs:comment ?commentX}
} GROUP BY ?class
At the moment, I am getting rows with Unicode decode issues, while using SPARQL on Dbpedia (using Virtuoso servers). This is an example of what I am getting Knut %C3%85ngstr%C3%B6m.
The right name is Knut Ångström. Cool, now how do I fix this? My crafted query is:
select distinct (strafter(str(?influencerString),str(dbpedia:)) as ?influencerString) (strafter(str(?influenceeString),str(dbpedia:)) as ?influenceeString) where {
{ ?influencer a dbpedia-owl:Person . ?influencee a dbpedia-owl:Person .
?influencer dbpedia-owl:influenced ?influencee .
bind( replace( str(?influencer), "_", " " ) as ?influencerString )
bind( replace( str(?influencee), "_", " " ) as ?influenceeString )
}
UNION
{ ?influencee a dbpedia-owl:Person . ?influencer a dbpedia-owl:Person .
?influencee dbpedia-owl:influencedBy ?influencer .
bind( replace( str(?influencee), "_", " " ) as ?influenceeString )
bind( replace( str(?influencer), "_", " " ) as ?influencerString )
}
}
The DBpedia wiki explains that the identifiers for resources in the English DBpedia dataset use URIs, not IRIs, which means that you'll end up with encoding issues like this.
3. Denoting or Naming “things”
Each thing in the DBpedia data set is denoted by a de-referenceable
IRI- or URI-based reference of the form
http://dbpedia.org/resource/Name, where Name is derived from the URL
of the source Wikipedia article, which has the form
http://en.wikipedia.org/wiki/Name. Thus, each DBpedia entity is tied
directly to a Wikipedia article. Every DBpedia entity name resolves to
a description-oriented Web document (or Web resource).
Until DBpedia release 3.6, we only used article names from the English
Wikipedia, but since DBpedia release 3.7, we also provide localized
datasets that contain IRIs like http://xx.dbpedia.org/resource/Name,
where xx is a Wikipedia language code and Name is taken from the
source URL, http://xx.wikipedia.org/wiki/Name.
Starting with DBpedia release 3.8, we use IRIs for most DBpedia entity
names. IRIs are more readable and generally preferable to URIs, but
for backwards compatibility, we still use URIs for DBpedia resources
extracted from the English Wikipedia and IRIs for all other languages.
Triples in Turtle files use IRIs for all languages, even for English.
There are several details on the encoding of URIs that should always
be taken into account.
In this particular case, it looks like you don't really need to break up the identifier so much as get a label for the entity.
## If things were guaranteed to have just one English label,
## we could simply take ?xLabel as the value that we want with
## `select ?xLabel { … }`, but since there might be more than
## one, we can group by `?x` and then take a sample from the
## set of labels for each `?x`.
select (sample(?xLabel) as ?label) {
?x dbpedia-owl:influenced dbpedia:August_Kundt ;
rdfs:label ?xLabel .
filter(langMatches(lang(?xLabel),"en"))
}
group by ?x
SPARQL results
Simplifying your query a bit, we can have this:
select
(sample(?rLabel) as ?influencerName)
(sample(?eLabel) as ?influenceeName)
where {
?influencer dbpedia-owl:influenced|^dbpedia-owl:influencedBy ?influencee .
dbpedia-owl:Person ^a ?influencer, ?influencee .
?influencer rdfs:label ?rLabel .
filter( langMatches(lang(?rLabel),"en") )
?influencee rdfs:label ?eLabel .
filter( langMatches(lang(?eLabel),"en") )
}
group by ?influencer ?influencee
SPARQL results
If you don't want language tags on those results, then add a call to str():
select
(str(sample(?rLabel)) as ?influencerName)
(str(sample(?eLabel)) as ?influenceeName)
where {
?influencer dbpedia-owl:influenced|^dbpedia-owl:influencedBy ?influencee .
dbpedia-owl:Person ^a ?influencer, ?influencee .
?influencer rdfs:label ?rLabel .
filter( langMatches(lang(?rLabel),"en") )
?influencee rdfs:label ?eLabel .
filter( langMatches(lang(?eLabel),"en") )
}
group by ?influencer ?influencee
SPARQL results