我有一个问题问你。
我想编写一个查询，检索与给定字符串“Londn”相似(给定相似度函数，例如Lev)的值，以与DBPedia 的谓词“RDFS:label”进行比较。例如，在输出中，我想获取“伦敦”的值。
我读过一种可用的方法可能是使用 iSPARQL(“Imprecise SPARQL”)，尽管它在文献中的使用不是很广泛。
我可以使用 iSPARQL 还是有一些 SPARQL 方法来执行相同的操作？
最佳答案
简短版本——您可以在纯 SPARQL 中执行其中的一些操作
您可以使用这样的查询来查找名称类似于“伦敦”的城市，并按(一种度量)相似性对它们进行排序。答案的其余部分解释了这是如何工作的:
select ?city ?percent where {
?city a dbpedia-owl:City ;
rdfs:label ?label .
filter langMatches( lang(?label), 'en' )
bind( replace( concat( 'x', str(?label) ), "^x[^Londn]*([L]?)[^ondn]*([o]?)[^ndn]*([n]?)[^dn]*([d]?)[^n]*([n]?).*$", '$1$2$3$4$5' ) as ?match )
bind( xsd:float(strlen(?match))/strlen(str(?label)) as ?percent )
}
order by desc(?percent)
limit 100
SPARQL results
city percent
----------------------------------------------
http://dbpedia.org/resource/London 0.833333
http://dbpedia.org/resource/Bonn 0.75
http://dbpedia.org/resource/Loudi 0.6
http://dbpedia.org/resource/Ladnu 0.6
http://dbpedia.org/resource/Lonar 0.6
http://dbpedia.org/resource/Longnan 0.571429
http://dbpedia.org/resource/Longyan 0.571429
http://dbpedia.org/resource/Luoding 0.571429
http://dbpedia.org/resource/Lodhran 0.571429
http://dbpedia.org/resource/Lom%C3%A9 0.5
http://dbpedia.org/resource/Andong 0.5
计算字符串相似度度量
SPARQL 中没有内置任何内容来计算字符串匹配距离，但您可以使用 SPARQL 中的正则表达式替换机制来完成其中的一些工作。假设您想匹配某些字符串中的序列“cat”。然后，您可以使用这样的查询来确定序列“cat”中存在多少给定字符串:
select ?string ?match where {
values ?string { "cart" "concatenate" "hat" "pot" "hop" }
bind( replace( ?string, "^[^cat]*([c]?)[^at]*([a]?)[^t]*([t]?).*$", "$1$2$3" ) as ?match )
}
-------------------------
| string | match |
=========================
| "cart" | "cat" |
| "concatenate" | "cat" |
| "hat" | "at" |
| "pot" | "t" |
| "hop" | "" |
-------------------------
通过检查字符串和匹配的长度，您应该能够计算出一些不同的相似性度量。作为使用您提到的“伦敦”输入的更复杂示例。百分比列是与输入匹配的字符串的百分比。
select ?input
?string
(strlen(?match)/strlen(?string) as ?percent)
where {
values ?string { "London" "Londn" "London Fog" "Lando" "Land Ho!"
"concatenate" "catnap" "hat" "cat" "chat" "chart" "port" "part" }
values (?input ?pattern ?replacement) {
("cat" "^[^cat]*([c]?)[^at]*([a]?)[^t]*([t]?).*$" "$1$2$3")
("Londn" "^[^Londn]*([L]?)[^ondn]*([o]?)[^ndn]*([n]?)[^dn]*([d]?)[^n]*([n]?).*$" "$1$2$3$4$5")
}
bind( replace( ?string, ?pattern, ?replacement) as ?match )
}
order by ?pattern desc(?percent)
--------------------------------------------------------
| input | string | percent |
========================================================
| "Londn" | "Londn" | 1.0 |
| "Londn" | "London" | 0.833333333333333333333333 |
| "Londn" | "Lando" | 0.6 |
| "Londn" | "London Fog" | 0.5 |
| "Londn" | "Land Ho!" | 0.375 |
| "Londn" | "concatenate" | 0.272727272727272727272727 |
| "Londn" | "port" | 0.25 |
| "Londn" | "catnap" | 0.166666666666666666666666 |
| "Londn" | "cat" | 0.0 |
| "Londn" | "chart" | 0.0 |
| "Londn" | "chat" | 0.0 |
| "Londn" | "hat" | 0.0 |
| "Londn" | "part" | 0.0 |
| "cat" | "cat" | 1.0 |
| "cat" | "chat" | 0.75 |
| "cat" | "hat" | 0.666666666666666666666666 |
| "cat" | "chart" | 0.6 |
| "cat" | "part" | 0.5 |
| "cat" | "catnap" | 0.5 |
| "cat" | "concatenate" | 0.272727272727272727272727 |
| "cat" | "port" | 0.25 |
| "cat" | "Lando" | 0.2 |
| "cat" | "Land Ho!" | 0.125 |
| "cat" | "Londn" | 0.0 |
| "cat" | "London" | 0.0 |
| "cat" | "London Fog" | 0.0 |
--------------------------------------------------------
更新
上面的代码在 Apache Jena 中有效，但在 Virtuoso 中失败，因为模式可以匹配空字符串。例如，如果您在 DBpedia 的端点(由 Virtuoso 提供支持)上尝试以下查询，您将收到以下错误:
select (replace( "foo", ".*", "x" ) as ?bar) where {}
这让我感到惊讶，但是 replace 的规范说它基于 XPath fn:replace 。 fn:replace 的文档说:
不过，我们可以通过在模式和字符串的开头添加一个字符来解决这个问题:
select ?input
?string
(strlen(?match)/strlen(?string) as ?percent)
where {
values ?string { "London" "Londn" "London Fog" "Lando" "Land Ho!"
"concatenate" "catnap" "hat" "cat" "chat" "chart" "port" "part" }
values (?input ?pattern ?replacement) {
("cat" "^x[^cat]*([c]?)[^at]*([a]?)[^t]*([t]?).*$" "$1$2$3")
("Londn" "^x[^Londn]*([L]?)[^ondn]*([o]?)[^ndn]*([n]?)[^dn]*([d]?)[^n]*([n]?).*$" "$1$2$3$4$5")
}
bind( replace( concat('x',?string), ?pattern, ?replacement) as ?match )
}
order by ?pattern desc(?percent)
--------------------------------------------------------
| input | string | percent |
========================================================
| "Londn" | "Londn" | 1.0 |
| "Londn" | "London" | 0.833333333333333333333333 |
| "Londn" | "Lando" | 0.6 |
| "Londn" | "London Fog" | 0.5 |
| "Londn" | "Land Ho!" | 0.375 |
| "Londn" | "concatenate" | 0.272727272727272727272727 |
| "Londn" | "port" | 0.25 |
| "Londn" | "catnap" | 0.166666666666666666666666 |
| "Londn" | "cat" | 0.0 |
| "Londn" | "chart" | 0.0 |
| "Londn" | "chat" | 0.0 |
| "Londn" | "hat" | 0.0 |
| "Londn" | "part" | 0.0 |
| "cat" | "cat" | 1.0 |
| "cat" | "chat" | 0.75 |
| "cat" | "hat" | 0.666666666666666666666666 |
| "cat" | "chart" | 0.6 |
| "cat" | "part" | 0.5 |
| "cat" | "catnap" | 0.5 |
| "cat" | "concatenate" | 0.272727272727272727272727 |
| "cat" | "port" | 0.25 |
| "cat" | "Lando" | 0.2 |
| "cat" | "Land Ho!" | 0.125 |
| "cat" | "Londn" | 0.0 |
| "cat" | "London" | 0.0 |
| "cat" | "London Fog" | 0.0 |
--------------------------------------------------------
关于sparql - 使用 iSPARQL 来比较使用相似性度量的值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/24557020/