I'm trying to make a SPARQL query using Prolog and DBpedia. My objective is to tag in text all Persons, so for retrieving famous people I made this query that remove all results like Music groups(Band) and Organization, since I want to tag only real people and not abstract
select ?person where{
{
?person a dbpedia-owl:Person; rdfs:label "Name Surname" #it.
}
UNION
{
?person a dbpedia-owl:Person; foaf:name "Name"#it; foaf:surname "Surname"#it.
}
UNION
{
?person a dbpedia-owl:Person; foaf:name "Name Surname"#it.
}
FILTER NOT EXISTS {
{ ?subject <http://airpedia.org/ontology/type_with_conf#10> dbpedia-owl:Band .
?subject rdfs:label ?artistName .
FILTER ( str(?artistName) = "Name Surname" )
}
UNION
{
?subject <http://airpedia.org/ontology/type_with_conf#10> dbpedia-owl:Organisation .
?subject rdfs:label ?artistName .
FILTER ( str(?artistName) = "Name Surname" )
}
}
}
I use It. version of Dbpedia if you run this query use this version although the results will not be good for me.
So for example if I search "Metallica" as a person i don't want to get results cause is it a Band or(for me, but in this case is Metallica are an Organisation too) an Organisation
and it works good this are the results Metallica Query Results and those are for "Michael Jackson" Michael Jackson Query results
My problem is when i put someone that is not a Singer or a Music band for example if i try something like "Jim Carrey" i get " error transction timed out Jim Carrey.
I think I got this problem because those properties are Undefined for Jim Carrey, but i tried an to put an OPTIONAL marker in each subquery in the first filter, but i get too the same error
I put the code in a pastebin file so you can find all three query
I know that i should not use Static String in a query or there are a lot of better mode but i need that since i compose the query with prolog and than send to sparql online so i must do in this way.
TO #Joshua I tried to remove the FILTER(String) in the NOT EXIST (Filter) But I will not work anymore thanks however for helping me
Excuse me for too much editing but i resolved some part of the starting problem but didn't find a solution
First problem :Filtering results based on specific properties with specific values. (Works)
Second : The first works only for Things with that specific property (as show above) like(Metallica,Michael Jackson, The Beatles, ...) but not for thos without the properties in the filter.
(i can't use more than two link because I'm a newbe so i will put a link in the comments with a pastebin links with the 3 Query and the results of they)
Related
Looking for SPARQL query to do the following:
For example, I have the word Apple. Apple may refer to the organization Apple_Inc or the Species of Plants class as per the ontology. Owl: Thing has a subclass called Species, so I want to return those most relevant/maximum-hit URIs where the keyword Apple does not belong to the Species subclass. So when you return all the URIs, http://dbpedia.org/page/Apple should not be one of them, neither must ANY relevant link that comes under Species subclass.
By maximum-hit/most relevant I mean the top returned results that match the query! Like when you access the PrefixSearch (i.e. Autocomplete) API, it has the parameter called MaxHits.
For example http://lookup.dbpedia.org/api/search/PrefixSearch?QueryClass=&MaxHits=2&QueryString=berl is a link where you want to return the top 2 URIs that match the QueryString=berl.
Like I'm actually really struggling to even explain the work I've done so far because I'm not able to understand the structure and how to formulate a proper query..
with respect to negation in SPARQL, I found a relevant portion of the documentation in the link here.. But I do not know how and where to proceed from there, and cannot understand why keywords like ?person are used.. I can understand the person is used to selected well.. PEOPLE names, but I would like to know how and where to find these keywords like ?person, ?name to represent a specific entity..
SELECT ?uri ?label
WHERE {
?uri rdfs:label ?label .
filter(?label="car"#en)
}
I would really appreciate if someone could link me the part of the documentation I can clearly read and understand that ?uri is used to select a URI in the form www.dbpedia.org'/page/SomeEntity and what these ?person, ?name, ?label represent.
I'm actually so lost.. I will go up and start eating one elephant at a time. For now, I'll be very grateful if I get an answer to this.
If there is anyway you know where I can avoid learning and using SPARQL, that would work too! I know Python well enough, so leveraging an API to pull this information is also fine by me. This question was posted by me.
Answer posted by #Stanislav-Kravin --
SELECT DISTINCT ?s
WHERE
{ ?s a owl:Thing .
?s rdfs:label ?label .
FILTER ( LANGMATCHES ( LANG ( ?label ), 'en' ) )
?label bif:contains '"apple"' .
FILTER NOT EXISTS { ?s rdf:type/rdfs:subClassOf* dbo:Species }
}
I am new to SPARQL and trying to fetch a movie adapted from specific book from dbpedia. This is what I have so far:
PREFIX onto: <http://dbpedia.org/ontology/>
SELECT *
WHERE
{
<http://dbpedia.org/page/2001:_A_Space_Odyssey> a ?type.
?type onto:basedOn ?book .
?book a onto:Book
}
I can't get any results. How can I do that?
When using any web resource, and in your case the property :basedOn, you need to make sure that you have declared the right prefix. If you are querying from the DBpedia SPARQL endpoint, then you can directly use dbo:basedOneven without declaring it, as it is among predefined. Alternatively, if you want to use your own, or if you are using another SPARQL client, make sure that whatever short name you choose for this property, you declare the prefix for http://dbpedia.org/ontology/.
Then, first, to get more result you may not restrict the type of the subject of this triple pattern, as there could be movies that actually not type as such. So, a query like this
select distinct *
{
?movie dbo:basedOn ?book .
?book a dbo:Book .
}
will give you lots of good results but not all. For example, the resource from your example will be missing. You can easily check test the available properties between these two resource with a query like this:
select ?p
{
{<http://dbpedia.org/resource/2001:_A_Space_Odyssey_(film)> ?p <http://dbpedia.org/resource/2001:_A_Space_Odyssey> }
UNION
{ <http://dbpedia.org/resource/2001:_A_Space_Odyssey> ?p <http://dbpedia.org/resource/2001:_A_Space_Odyssey_(film)>}
}
You'll get only one result:
http://www.w3.org/2000/01/rdf-schema#seeAlso
(note that the URI is with 'resource', not with 'page')
Then you may search for any path between the two resource, using the method described here, or find a combination of other patterns that would increase the number of results.
I'm trying to extract alumni lists for universities using SPARQL.
I've identified the ontologies I need:
http://mappings.dbpedia.org/server/ontology/classes/University
http://mappings.dbpedia.org/server/ontology/classes/Person
I tried this query, which you can examine here:
SELECT * WHERE {
?University dbpedia2:alumni ?Person .
}
Which seemed to make sense, except this returns counts instead of people, as the ontology says the property contains.
I found this query somewhere which seemed to do a better job finding universities, but was very slow.
SELECT * WHERE {
{ <http://dbpedia.org/ontology/University> ?property ?hasValue }
UNION
{ ?isValueOf ?property <http://dbpedia.org/ontology/University> }
}
I also tried going the other way, start with all people and look for their almae matres, in this form:
SELECT * WHERE {
?person dbpedia2:almaMater ?University
}
But this is much slower, possibly because searching through the people space is too laborious. This does actually work, but it returns a different set of results in application---namely, all people with a listed alma mater, rather than all people listed by universities as alumni. I'd prefer a syntax that gets me the alumni.
How can I phrase this to return all alumni listed for universities?
The performance of DBpedia's SPARQL endpoint can be a bit unreliable at times. After all, it's apublic service, and isn't intended for huge queries. Nonetheless, I think you can get what you're looking for here without too much trouble. First, you can check how many results there are with a query like this at the public SPARQL endpoint:
select (count(*) as ?nResults) where {
?person dbpedia-owl:almaMater ?almaMater
}
SPARQL results (64928)
Now, if you just want the big list, you'd get it like this. The order by helps organize the results for easy consumption, but isn't technically necessary:
select ?almaMater ?person where {
?person dbpedia-owl:almaMater ?almaMater
}
order by ?almaMater ?person
SPARQL results
If you need to place some additional restrictions on ?almaMater, e.g., to ensure that it's a university, then you can add them to the query. For instance:
select ?almaMater ?person where {
?person dbpedia-owl:almaMater ?almaMater .
?almaMater a dbpedia-owl:University .
}
order by ?almaMater ?person
SPARQL results
In your last query, you are almost there. However, you are currently asking for any resource that can take the place of the ?University variable. As you only want universities to take that place, you can use another triple to further restrict that variable:
SELECT * WHERE {
?University a dbpedia-owl:University.
?person dbpedia2:almaMater ?University.
}
This means that ?University can only be an individual of class dbpedia-owl:University (where dbpedia-owl is mapped to http://dbpedia.org/ontology/).
Your first query:
SELECT * WHERE {
?University dbpedia2:alumni ?Person .
}
isn't just returning counts; it's returning both counts and individual alumni. Apparently dbpedia's data here is poor quality and there are a number of triples misusing the dbpedia2:alumni relation.
You can filter out the counts by adding a second condition requiring that an entity satisfying Person be a member of the appropriate class:
SELECT * WHERE {
?university dbpedia2:alumni ?person .
?person rdf:type <http://dbpedia.org/ontology/Person>
}
What you see running this is that there are very few individuals tagged as alumni; the data is surprisingly scant, unfortunately.
What is sparql query that finds count of some entity? For examles, on Linked movie database, if I want find count of actors or films, how can I get it?
I tried this
SELECT (count ( ?Film)){?entity rdf:type ?Film}
but got wrong number.
There's a whole lot missing from this question (e.g., where you ran the query, what you expected as a result, etc.) but I think we can pinpoint the problem even without those details. First, let's rewrite your query using proper syntax (the formatting is optional; the important thing is count(?Film) as ?count):
select (count(?Film) as ?count) {
?entity rdf:type ?Film
}
?Film here is a variable, so you're asking "find me things and their types, and then count how many types were found." If you were trying to count the number of things of some particular film type, though, you probably wanted a query like:
select (count(?entity) as ?numberOfFilms) {
?entity rdf:type :Film .
}
Where :Film is some particular IRI, not a variable. Also note that you can abbreviate rdf:type with a, so you can make this even shorter and fit it nicely on one line again, if you want:
select (count(?entity) as ?numberOfFilms) { ?entity a :Film }
Is it possible to get the list of films in function of their genre?
I tried this:
SELECT DISTINCT ?film_title ?film_abstract ?film_genre
WHERE {
?film_title rdf:type <http://dbpedia.org/ontology/Film> .
?film_title rdfs:comment ?film_abstract .
?film_genre <http://dbpedia.org/ontology/genre> ?film_genre .
FILTER(lang(?film_abstract) = "en" ).
}
LIMIT 20
But probably I've doing something wrong !
Thanks,
Danilo
Looks like a simple typo on your part. The third triple pattern should be the following:
?film_title <http://dbpedia.org/ontology/genre> ?film_genre
Also the FILTER you are using may make the query very slow, try using the following instead:
FILTER(LANGMATCHES(LANG(?film_abstract), "en"))
Though having played with your query there doesn't appear to be any data that actually matches your query in DBPedia. Essentially the genre property you are using appears only to be applied to music and not to films so you should remove the third triple pattern entirely if you actually want to get any results