Question

是否可以创建一个SPARQL查询，该查询根据两个谓词生成排名结果？我搜索了类似的查询，我可以找到最近的解决方案thread。这个解决方案给了我超时，因为我有大约600三倍，所以一个有效的解决方案将非常感激。考虑下面的RDF：

const items = {
    "questionlist": [{
            "question": "Test question 1",
            "option1": "Answer1A",
            "option2": "Answer1B",
            "option3": "Answer1C"
        },
        {
            "question": "Test question 2",
            "option1": "Answer2A",
            "option2": "Answer2B",
            "option3": "Answer2C"
        },
        {
            "question": "Test question 3",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        },
        {
            "question": "Test question 4",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        },
        {
            "question": "Test question 5",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        },
        {
            "question": "Test question 6",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        },
        {
            "question": "Test question 7",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        },
        {
            "question": "Test question 8",
            "option1": "Answer3A",
            "option2": "Answer3B",
            "option3": "Answer3C"
        }
    ]
}

function random_item(questions, n) {
    var arr = [];
    for (let i = 0; i < n; i++) {
        arr.push(questions[Math.floor(Math.random() * questions.length)])
    }
    return arr

}

console.log(random_item(items.questionlist, 2))

我想得到以下结果：

@prefix : <http://example.org#> .

:A :groupid 119;
   :date 2018-04-01.
:B :groupid 119 ;
   :date 2018-03-01.
:C :groupid 119 ;
   :date 2018-05-01.
:D :groupid 120  ;
   :date 2018-02-01.
:E :groupid 120 ;  
   :date 2018-03-01.

Answer 1

正如AKSW的回答，这个脚本完美无缺：

PREFIX : <http://example.org#> 
SELECT ?id ?date (COUNT(*) as ?ranking)
{
   ?x :groupid ?id ; :date ?date .
   ?y :groupid ?id ; :date ?date2 .
    filter(?date2 <= ?date) 
}
GROUP BY ?id ?date
ORDER BY ?id ?date ?ranking

两个谓词的SPARQL排名

1 个答案: