SPARQL package? about wikidataqueryservicer

[bearloga]

First, thank you very much! :)

Second, good question! I specifically wanted to use what I think is the best package for HTTP requests (httr) -- and one I am most comfortable with -- but especially because I wanted my package to have an MIT license, rather than GPL-3 (which I would be forced to use if I had depended on SPARQL).

But yeah, you can totally even just use the SPARQL package instead:

sparql_query <- 'PREFIX wd: <http://www.wikidata.org/entity/>
PREFIX wdt: <http://www.wikidata.org/prop/direct/>
PREFIX wikibase: <http://wikiba.se/ontology#>
SELECT DISTINCT ?genre ?genreLabel WHERE {
  SERVICE wikibase:label { bd:serviceParam wikibase:language "en". }
  ?film wdt:P31 wd:Q11424.
  ?film rdfs:label "The Cabin in the Woods"@en.
  ?film wdt:P136 ?genre.
}'

x <- SPARQL::SPARQL(url = "https://query.wikidata.org/bigdata/namespace/wdq/sparql", query = sparql_query)
x$results

genre	genreLabel
http://www.wikidata.org/entity/Q200092	"horror film"@ en
http://www.wikidata.org/entity/Q471839	"science fiction film"@ en
http://www.wikidata.org/entity/Q224700	"comedy horror"@ en
http://www.wikidata.org/entity/Q859369	"comedy-drama"@ en
http://www.wikidata.org/entity/Q1342372	"monster film"@ en

compared to

y <- WikidataQueryServiceR::query_wikidata(sparql_query)
y

genre	genreLabel
http://www.wikidata.org/entity/Q200092	horror film
http://www.wikidata.org/entity/Q471839	science fiction film
http://www.wikidata.org/entity/Q224700	comedy horror
http://www.wikidata.org/entity/Q859369	comedy-drama
http://www.wikidata.org/entity/Q1342372	monster film

Hope that helps!

[briatte]

That helps a lot, thanks—and it confirms that your package produces more ergonomic code, too!

Thanks again.