SPAQRL：选择项目并计算其标签的出现次数

Question

我有<一个href="https://www.orkg.org/orkg/triplestore?query=%0APREFIX%20orkgr%3A%20%3Chttp%3A%2F%2Forkg.org%2F orkg%2Fresource%2F%3E%0APREFIX%20orkgc%3A%20%3Chttp%3A%2F%2Forkg.org%2Forkg%2Fclass%2F%3E%0APREFIX%20 orkgp%3A%20%3Chttp%3A%2F%2Forkg.org%2Forkg%2Fpredicate%2F%3E%0APREFIX%20rdfs%3A%20%3Chttp%3A%2F%2Fww w.w3.org%2F2000%2F01%2Frdf-schema%23%3E%0APREFIX%20xsd%3A%20%3Chttp%3A%2F%2Fwww.w3.org%2F2001%2FXMLSc hema%23%3E%0APREFIX%20rdf%3A%20%3Chttp%3A%2F%2Fwww.w3.org%2F1999%2F02%2F22-rdf-syntax-ns%23%3E%0A%0A SELECT%20%3Fo1Label%20(COUNT(%3Fo1Label)%20AS%20%3Fo1LabelCount)%0AWHERE%20%7B%0A%20%20%3Fo1%20a%20或kgc%3APaper.%0A%20%20%3Fo1%20rdfs%3Alabel%20%3Fo1Label.%0A%20%20%0A%20%20FILTER%20(strlen(%3Fo1Label )%20%3E%201).%0A%20%20%0A%7D%0A%0AGROUP%20BY%20%3Fo1Label%0AORDER%20BY%20DESC(%3Fo1LabelCount)%0A%0A" rel="nofollow noreferrer">此 SPARQL 查询 定向到开放研究知识图 (ORKG)：

PREFIX orkgr: <http://orkg.org/orkg/resource/>
PREFIX orkgc: <http://orkg.org/orkg/class/>
PREFIX orkgp: <http://orkg.org/orkg/predicate/>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

SELECT ?o1Label (COUNT(?o1Label) AS ?o1LabelCount)
WHERE {
  ?o1 a orkgc:Paper.
  ?o1 rdfs:label ?o1Label.
  
  FILTER (strlen(?o1Label) > 1).
}

GROUP BY ?o1Label
ORDER BY DESC(?o1LabelCount)

结果是标签 (?o1Label) 以及该标签出现的次数 (<代码>？o1LabelCount）。

如何扩展此查询以包含实际项目的列 (?o1)？

由于可能有多个候选项（当 o1LabelCount > 1 时），因此每个项目都应占一行（具有相同的标签和相同的标签计数）。

Answer 1

我看到两个选择：

第一个（可能更好）是使用 GROUP_CONCAT< /code> 并将实体收集到一个字段中，以便在应用程序端再次解析。这可能看起来像这样（链接)：

PREFIX orkgr: <http://orkg.org/orkg/resource/>
PREFIX orkgc: <http://orkg.org/orkg/class/>
PREFIX orkgp: <http://orkg.org/orkg/predicate/>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

SELECT ?o1Label (GROUP_CONCAT(?o1, "\t") AS ?o1s) (COUNT(?o1Label) AS ?o1LabelCount)
WHERE {
  ?o1 a orkgc:Paper.
  ?o1 rdfs:label ?o1Label.
  
  FILTER (strlen(?o1Label) > 1).
}

GROUP BY ?o1Label
ORDER BY DESC(?o1LabelCount)

---

另一种方法是使用 嵌套查询并收到您所描述的结果（链接）：

PREFIX orkgr: <http://orkg.org/orkg/resource/>
PREFIX orkgc: <http://orkg.org/orkg/class/>
PREFIX orkgp: <http://orkg.org/orkg/predicate/>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

SELECT ?o1Label ?o1 ?o1LabelCount
WHERE {
  ?o1 rdfs:label ?o1Label .

  {
    SELECT ?o1Label (COUNT(?o1Label) AS ?o1LabelCount)
    WHERE {
      [
        a orkgc:Paper;
        rdfs:label ?o1Label
      ]
      FILTER (strlen(?o1Label) > 1).
    }
  }
}

GROUP BY ?o1Label
ORDER BY DESC(?o1LabelCount)