Convert xsd:date to xsd:gYear in SPARQL - sparql

Question

Hello I'm trying to retrieve from DBpedia persons that are born in Lyon after 1900 ("1900"^^xsd:gYear)
Here is my piece of code :
prefix dbo: <http://dbpedia.org/ontology/>
prefix foaf: <http://xmlns.com/foaf/0.1/>
prefix dbr: <http://dbpedia.org/resource/>
prefix xsd: <http://www.w3.org/2001/XMLSchema#>
select $n $birthDate
{
$p a dbo:Person.
$p dbo:birthPlace dbr:Lyon.
$p foaf:name $n.
$p dbo:birthDate $birthDate.
filter($birthDate > "1980"^^xsd:gYear).
}
And I'm getting the following results :
query results

Answer 1

From my understanding, the problem is in the data: many dates do not have valid xsd:date literals, thus, the comparison fails. It also fails for filter($birthDate > "1980-01-01"^^xsd:date). According to XML Schema, xsd:date must provide the literal in form "YYYY-MM-DD". And that's unfortunately not the case for many dates in DBpedia. For valid dates, the comparison works perfectly.
Workaround:
prefix dbo: <http://dbpedia.org/ontology/>
prefix foaf: <http://xmlns.com/foaf/0.1/>
prefix dbr: <http://dbpedia.org/resource/>
prefix xsd: <http://www.w3.org/2001/XMLSchema#>
select $n $birthDate ?year
{
$p a dbo:Person.
$p dbo:birthPlace dbr:Lyon.
$p foaf:name $n.
$p dbo:birthDate $birthDate.
filter($birthDate > "1980-01-01"^^xsd:date).
bind(replace(str($birthDate),"(\\d+)-\\d*-\\d*", "$1") as ?year)
filter(xsd:integer(?year) > 1980)
}
Note, it might fail for other kinds of ill-formed dates. I didn't check all corner cases.
Somebody should report this to the DBpedia community. It should be fixed.