我有这个rdf .ttl文件示例:@前缀ns1:http://schema.org/。@前缀xsd:http://www.w3.org/2001/XMLSchema#。
<http://example.org/crime/100010117.0> ns1:beat "308" ;
ns1:crime "AUTO_THEFT" ;
ns1:date "1/1/2010" ;
ns1:lat 3.369307e+01 ;
ns1:location "960 CONSTITUTION RD SE" ;
ns1:long -8.435805e+01 ;
ns1:neighborhood "Norwood Manor" ;
ns1:npu "Z" ;
ns1:number 1.000101e+08 .
<http://example.org/crime/100010121.0> ns1:beat "309" ;
ns1:crime "LARCENY-FROM_VEHICLE" ;
ns1:date "1/1/2010" ;
ns1:lat 3.368274e+01 ;
ns1:location "2685 METROPOLITAN PKWY SW" ;
ns1:long -8.440902e+01 ;
ns1:neighborhood "Perkerson" ;
ns1:npu "X" ;
ns1:number 1.000101e+08 .
<http://example.org/crime/100010127.0> ns1:beat "208" ;
ns1:crime "LARCENY-FROM_VEHICLE" ;
ns1:date "1/1/2010" ;
ns1:lat 3.385211e+01 ;
ns1:location "3600 PIEDMONT RD NE" ;
ns1:long -8.438044e+01 ;
ns1:neighborhood "Buckhead Forest" ;
ns1:npu "B" ;
ns1:number 1.000101e+08 .
<http://example.org/crime/100010147.0> ns1:beat "512" ;
ns1:crime "ROBBERY-PEDESTRIAN" ;
ns1:date "1/1/2010" ;
ns1:lat 3.375104e+01 ;
ns1:location "FORSYTH ST SW / NELSON ST SW" ;
ns1:long -8.439479e+01 ;
ns1:neighborhood "Downtown" ;
ns1:npu "M" ;
ns1:number 1.000101e+08 .
<http://example.org/crime/100010149.0> ns1:beat "311" ;
ns1:crime "BURGLARY-RESIDENCE" ;
ns1:date "1/1/2010" ;
ns1:lat 3.367399e+01 ;
ns1:location "2950 SPRINGDALE RD SW" ;
ns1:long -8.441557e+01 ;
ns1:neighborhood "Hammond Park" ;
ns1:npu "X" ;
ns1:number 1.000101e+08 .
<http://example.org/crime/100010186.0> ns1:beat "501" ;
ns1:crime "BURGLARY-RESIDENCE" ;
ns1:date "1/1/2010" ;
ns1:lat 3.378988e+01 ;
ns1:location "288 16TH ST NW" ;
ns1:long -8.439713e+01 ;
ns1:neighborhood "Home Park" ;
ns1:npu "E" ;
ns1:number 1.000102e+08 .我正在试图计算不同类型的犯罪(NS1:犯罪)
例如,我想要这样的结果
[
{
"crime": "AUTO_THEFT",
"count": 1
},
{
"crime": "LARCENY-FROM_VEHICLE",
"count": 2
},
{
"crime": "ROBBERY-PEDESTRIAN",
"count": 1
},
{
"crime": "ROBBERY-PEDESTRIAN",
"count": 2
}
]所以不同类型的犯罪(不同的)和他们的数量。
我试过这样做:
def countTypes(g):
crimes = []
q = g.query(
"""
PREFIX ns1: <http://schema.org/>
SELECT ?crime (count(distinct ?crime) as ?crimeCount) WHERE {
?s ns1:crime ?crime .
}""")
for row in q:
crimes.append(row)
return crimes但它不能正常工作。知道怎么做吗?谢谢
发布于 2021-11-29 13:01:20
您可以使用GROUP BY子句根据犯罪值对结果进行分组,然后使用COUNT (没有不同)来计算每个组中有多少结果:
PREFIX ns1: <http://schema.org/>
SELECT ?crime (count(*) as ?crimeCount) WHERE {
?s ns1:crime ?crime .
}
GROUP BY ?crimehttps://stackoverflow.com/questions/70153034
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