python 查询中的 RDF 文件计数不同

Question

我有这个 rdf .ttl 文件示例： @prefix ns1: http://schema.org/ 。 @prefix 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 .

我正在统计不同类型的犯罪 (ns1:crime)

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
    }
]

因此，不同类型的犯罪（不同）及其计数。

我试过这个：

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。关于如何做的任何想法？谢谢

Answer 1

您可以使用 GROUP BY 子句按犯罪值对结果进行分组，然后 COUNT（不区分）计算每组中有多少个结果：

PREFIX ns1: <http://schema.org/>
SELECT ?crime (count(*) as ?crimeCount) WHERE {
    ?s ns1:crime ?crime .
}
GROUP BY ?crime

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