On 12/13/11 5:22 PM, Liam Kirsher wrote:
Mohamed,Thanks again! I ran that query, but unfortunately it timed out. Meanwhile, I just ran the previous query you sent for the different country/language combinations. Apparently, that will be satisfactory. However, there was one thing that perhaps you or someone else on this list might be able to shed some light on. The numbers of Articles about France seem unusually large in comparison to other countries for all languages except German. Since the request was identical for all combinations, I'm really wondering what could explain it. The pattern is especially suspicious for Articles in Spanish.Anyone have an explanation for why there would be so many articles about France, or is their something wrong with the SPARQL request used to retrieve the data?*Language / Country* *German* *French* *Italian* *English* *Dutch* *Spanish* *Total Per Country**Germany* 13,816 7,909 10,378 13,809 12,982 1,312 *60,206 * *France* 8,504 37,232 36,903 37,324 36,888 36,802 *193,653 * *Italy* 5,736 8,268 8,721 8,800 8,316 8,259 *48,100 * *United Kingdom* 1,851 1,836 2,020 24,086 11,319 709 *41,821 **Netherlands / Belgium* 1,458 3,179 1,142 4,031 3,993 469 *14,272 **Spain* 1,173 3,318 5,586 6,295 3,330 6,170 *25,872 **Total Per Language* *32,538 * *61,742 * *64,750 * *94,345 * *76,828 * *53,721 * *383,924 *This is the SPARQL query I used for the various country/language parameters:sparql.setQuery(""" select count(?place) where { ?place ?p ?region.?place a <http://dbpedia.org/ontology/Place>. filter(?p = <http://dbpedia.org/ontology/country> && ?region = <http://dbpedia.org/resource/%s>) .?place rdfs:label ?placeName. filter( langMatches( lang(?placeName), "%s")) } """ % (country, language))
Please repeat using the SPARQL endpoint at: http://lod.openlinksw.com/sparql. Also include <http://dbpedia.org> in the FROM clause of your SPARQL. This instance is based on a cluster edition of Virtuoso with more iron behind it. The main DBpedia endpoint is configured for mass use too, but it doesn't have the iron behind it, so we have restriction in place to ensure everyone has a good shot at usage etc..
Kingsley
Best, ~Liam On 12/12/2011 02:36 AM, Mohamed Morsey wrote:Hi Liam, On 12/12/2011 04:26 AM, Liam Kirsher wrote:It's the connection part that has me stumped. In SQL it might be done with a subquery. But in SPARQL? Don't know.The subqueries are not a standardized part of SPARQL.I ended up writing a script and querying the various combination of country and language using the query Mohamed supplied, but I'm sure there is a better way.When I run that query, I only get 29 results, so something is not right there. Thanks, though, it is helping improve my understanding of SPARQL.Please try the following query: select ?place ?placeName ?country where { ?place ?p ?country. ?place a <http://dbpedia.org/ontology/Place>. ?country ?p2 ?continent. filter(?p2 = <http://purl.org/dc/terms/subject> && ?continent = <http://dbpedia.org/resource/Category:European_countries>) ?place rdfs:label ?placeName. filter(langMatches( lang(?placeName), "fr" )) } limit 1000On 12/11/2011 02:45 AM, Saeedeh Shekarpour wrote:Hi you should somehow connect the variable ?place to the variable ?country there are some properties in DBpedia you can use like: select count( distinct ?place) where { ?place a <http://dbpedia.org/ontology/Place>. ?place rdfs:label ?placeName. ?country ?p ?o.filter(?p = <http://purl.org/dc/terms/subject> && ?o = <http://dbpedia.org/resource/Category:European_countries>)?place <http://dbpedia.org/ontology/locationCountry> ?country. filter(langMatches( lang(?placeName), "fr" )). } You can find the list of possible properties by this query: select distinct ?x where { ?place a <http://dbpedia.org/ontology/Place>. ?country ?p ?o.filter(?p = <http://purl.org/dc/terms/subject> && ?o = <http://dbpedia.org/resource/Category:European_countries>)?place ?x ?country. } Select one you aim at your query. Best RegardsOn Sun, Dec 11, 2011 at 6:13 AM, Liam Kirsher <[email protected] <mailto:[email protected]>> wrote:I can get the list of the countries, but how to combine with the rest of the query? Okay, I tried this to get the count of all the articles aboutplaces in Europe that are in French. However, it times out. Is this the correct query?select ?country count(?place) where { ?place a <http://dbpedia.org/ontology/Place>. ?place rdfs:label ?placeName. filter(langMatches( lang(?placeName), "fr" )). ?country ?p ?o. filter(?p = <http://purl.org/dc/terms/subject> && ?o = <http://dbpedia.org/resource/Category:European_countries>) } On 12/10/2011 03:44 PM, Mohamed Morsey wrote:Hi Liam, On 12/09/2011 09:21 PM, Liam Kirsher wrote:Mohamed, Thanks so much! That was very helpful, and I now have most of what I needed.Perfect, then good luck with your next step ;)The remaining thing is somewhat similar -- just to get entries in a particular language anywhere in Europe. Simply substituting Europe as the region does not work. I imagine there might be some way to get the list of European countries and filter on membership in the list.You can use the following query to get a list of all European countries: SELECT ?country WHERE { ?country ?p ?o. filter(?p = <http://purl.org/dc/terms/subject> && ?o = <http://dbpedia.org/resource/Category:European_countries>) }Thanks again! Liam-- Kind Regards Mohamed Morsey Department of Computer Science University of Leipzig-- Liam Kirsher PGP:http://liam.numenet.com/pgp/ ------------------------------------------------------------------------------ Systems Optimization Self Assessment Improve efficiency and utilization of IT resources. Drive out cost and improve service delivery. Take 5 minutes to use this Systems Optimization Self Assessment. http://www.accelacomm.com/jaw/sdnl/114/51450054/ _______________________________________________ Dbpedia-discussion mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/dbpedia-discussion
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