rdf - Count resources having a property of a certain type in SPARQL -


i have data in triple store , compute following:

how many resources 'x' have object property 'op' @ least 2 different resources 'r' of similar type 'r' values?

here example of such data in turtle syntax:

prefix ex: <http://example.com>  ex:doc1    ex:document ;   ex:mentions p1, p2, p3 .  ex:doc2    ex:document ;   ex:mentions p4, p5 .  ex:p1   ex:person ;   ex:hasrole ex:r1 .  ex:p2   ex:person ;   ex:hasrole ex:r1 .  ex:p3   ex:person ;   ex:hasrole ex:r2 .  ex:p4   ex:person ;   ex:hasrole ex:r1 .  ex:p5   ex:person ;   ex:hasrole ex:r2 .  ex:r1   ex:role1 .   ex:r2   ex:role2 .

the objective count resources such ex:doc1 has 2 ex:mentions having similar roles (r1 of type ex:role1). here result 1, leaving aside ex:doc2`.

the strategy be:

  1. identify resources having desired property, i.e. documents (doc) having object properties (mentions) pointing on resources (person), these resources having properties (hasrole) of similar values (the role)).

  2. count them.

i have difficulties step 1. example, query returns docs having p1 role1, if there 1 p (p1) having property.

select distinct ?doc {     ?doc ex:document .     ?doc ex:mentions ?p1 .     ?doc ex:mentions ?p2 .     ?p1 ex:hasrole ?r1 .     ?p2 ex:hasrole ?r1 .     ?r1 ex:role1 . }

many help.

your data wasn't quite usable (there no prefixes on p1, p2, etc., resources), after fixing that, able use following query. pretty close; trick need filter(?p1 != ?p2) ensure you're getting different values of ex:mentions property. can check have role common type ?p1 ex:hasrole/a ?roletype . ?p2 ex:hasrole/a ?roletype, or more concisely, ?roletype ^(a/ex:hasrole) ?p1, ?p2. then, in counting, want count distinct values of ?document, need (count(distinct ?document) ?ndocuments):

prefix ex: <http://example.com>  select   #-- count ?document, count *distinct* values   #-- of ?document.   (count(distinct ?document) ?ndocuments)  {   #-- documents have 2 distinct    #-- values ex:mentions property   ?document ex:document ; ex:mentions ?p1, ?p2   filter(?p1 != ?p2)    #-- check have common role type   ?roletype ^(a/ex:hasrole) ?p1, ?p2 } 
-------------- | ndocuments | ============== | 2          | --------------

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