SPARQL Query for common values

Question

I need an help with a SPARQL query. In my ontology I have the entity item, wich design an article, it has a property called "keyWord" used to store the main arguments treated in a article. In the Triple Store there is a triple (article, keyWord, argument) for each argument of the article. I want to know with a query what are the articles that have the maximum number of triples with the same value of the property keyWord, how can i do? I've tried something like this, but it's wrong because I want to know only the articles that have the maximum number of common arguments:

SELECT ?item (COUNT(?argument) as ?maxArgument)   
WHERE{   
     ?item keyWord ?argument                                                                
}
ORDER BY DESC(?maxArgument)

You have to group by the argument then. Your current query does something different, it returns the number of arguments per each item. That's for sure not what you want. — UninformedUser

vassil_momtchev vassil_momtchev · Accepted Answer · 2018-02-18T14:33:48

According the SPARQL specification in order to calculate aggregate values for a solution, the solution is first divided into one or more groups, and the aggregate value is calculated for each group.

This query returns the number of keywords for each article:

SELECT ?item (COUNT(?argument) as ?maxArgument)   
WHERE{   
     ?item :keyword ?argument                                                                
}
GROUP BY ?item
ORDER BY DESC(?maxArgument)

This query returns the most frequent keywords and their article count:

SELECT (COUNT(?item) as ?articles) ?argument
WHERE{   
     ?item :keyword ?argument                                                                
}
GROUP BY ?argument
ORDER BY DESC(?articles)

The same applies for all other aggregate functions like MAX or MIN.

SPARQL Query for common values

1 Answers