Question

I am trying to filter a turtle file using pyrdf with sparql. But I noticed that sparql queries lose information over the context of the elements. I would like then to re-print the result of the query as a turtle file, is it possible to do this without manually scanning all the subfields of the element? We have data about locations formatted like this:

:pt0001
     vcard:category "Poste e Telegrafi"
    ; vcard:fn "Ufficio Bologna 1"
    ; vcard:extended-address "Via Cairoli 9, Bologna BO, Italy"
    ; vcard:latitude "44.504192"
    ; vcard:longitude "11.338661"
    ; vcard:tel "051 243425"
    ; vcard:fax "051 244459"
    ; cs:opening "Mon, Tue, Wed, Thu, Fri: 0800-1330. Sat: 0800-1230."
    ; cs:closing "01-01, 01-06, P, LA, 04-25, 05-01, 06-02, 08-15, 11-01, 12-08, 12-25, 12-26: .".

For example we want only locations that have a name (fn). Thanks for any tip you find...

Was it helpful?

Solution

To get back all locations which have a name, you could do something simple like:

SELECT DISTINCT ?location
WHERE { 
   ?location vcard:fn [].
}

This will give you back back the identifier (:pt0001 in your example), but of course the query can be adapted to return all the property values as well.

(Edit I added additional examples based on your clarification, I think this is what you're after)

Like so:

SELECT ?location ?prop ?value
WHERE { 
   ?location vcard:fn [];
             ?prop ?value .
} ORDER BY ?location

The result of this query will be a table of results of the form:

?location    ?prop           ?value
:pt0001      vcard:category  "Poste e Telegrafi"
:pt0001      vcard:name      "Ufficio Bologna 1"
:pt0001      vcard:tel       "051 243425"
(etc...)   

Or, as yet another alternative, you can explicitly formulate your query to get specific property values for each location:

SELECT ?location ?name ?cat ?tel
WHERE { 
   ?location vcard:fn ?name ;
             vcard:category ?cat ;
             vcard:tel ?tel .
} ORDER BY ?location

which will give you back a result table of the form:

?location ?name               ?cat                ?tel
:pt0001   "Ufficio Bologna 1" "Poste e Telegrafi" "051 243425"

Take your pick.

The trick with SPARQL queries is to think in terms of triples. Your data contains subject-predicate-object triples, SPARQL queries formulate patterns over those triples.

As a further remark: I see you are thinking about re-printing the result of the query as turtle. In that case, a CONSTRUCT query might be what you're after. While the result of a SELECT query (as shown above) is a table structure, the result of a CONSTRUCT query is a collection of RDF triples:

 CONSTRUCT { ?subject ?predicate ?object }
 WHERE { 
   ?subject ?predicate ?object ;
            vcard:fn [] 
 }
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