如何通过CONSTRUCT在图上创建聚合 [英] how to create aggregation on a graph from CONSTRUCT
本文介绍了如何通过CONSTRUCT在图上创建聚合的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
这是我的查询:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX rs: <http://www.welovethesemanticweb.com/rs#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
construct {
?subject0 rs:similarityValue ?similairty0.
?subject1 rs:similarityValue ?similairty1
}
WHERE {
{
?subject0 ?predicate0 ?object0.
rs:Impromptu_No._1 ?predicate0 ?object0.
?predicate0 rs:hasSimilarityValue ?similairty0Helper.
BIND(?similairty0Helper * (4/9) AS ?similairty0)
FILTER (?subject0 != rs:Impromptu_No)
}
union {
?subject1 ?predicate ?object.
?object ?predicate1 ?object1.
?predicate1 rs:hasSimilarityValue ?similairty1Helper.
rs:Impromptu_No._1 ?predicateHelper ?objectHelper.
?objectHelper ?predicate1 ?object1
BIND(?similairty1Helper * (1/9) AS ?similairty1)
FILTER (?subject1 != rs:Impromptu_No._1)
}
}
结果是:
rs:5th_Symphony
rs:similarityValue
0.011111111111111112e0 .
rs:Polonaise_heroique
rs:similarityValue
0.011111111111111112e0 , 0.17777777777777778e0 , 0.26666666666666666e0 .
rs:Preludes
rs:similarityValue
0.011111111111111112e0 , 0.26666666666666666e0 , 0.17777777777777778e0 .
rs:Requiem_Sequentia
rs:similarityValue
0.011111111111111112e0 .
rs:Le_nozze_di_Figaro
rs:similarityValue
0.011111111111111112e0 .
rs:Symphony_No._29_in_A_major
rs:similarityValue
0.011111111111111112e0 .
rs:Piano_Concerto_No._24
rs:similarityValue
0.011111111111111112e0 .
rs:Impromptu_No._1
rs:similarityValue
0.26666666666666666e0 , 0.17777777777777778e0 .
rs:Sonata_Pathetique
rs:similarityValue
0.011111111111111112e0 .
rs:Dies_Irae
rs:similarityValue
0.011111111111111112e0 .
rs:Piano_Sonata_No._31
rs:similarityValue
0.011111111111111112e0 , 0.26666666666666666e0 .
rs:Violin_Concerto_No._5_in_A_major
rs:similarityValue
0.011111111111111112e0 .
如您所见，对于每个实例，有很多值，我想将它们聚合并为每个实例制作它们的SUM.我会使用SELECT来执行此操作，但是使用CONSTRUCT时，我不知道如何应用聚合.
As you see, for each instance, there are many values, I want to aggregate them and make the SUM of them for each instance. I would do that with SELECT, but with CONSTRUCT, I didn't know how to apply the aggregation.
阅读后，我发现我们不能直接从CONSTRUCT使用聚合，但是我需要一起使用SELECT和CONSTRUCT，看来我必须使用一个名为命名图"的东西但是即使我尝试了，我也不知道该怎么做.
After reading, i found that we can't use the aggregation directly from CONSTRUCT, but i'd need to use SELECT and CONSTRUCT together, it seems i have to use something named "named graph" but i didn't know how to do that even i tried.
我们非常感谢您的帮助.
your help is highly appreciated.
非常感谢，最诚挚的问候，
Many thanks, best regards,
我尝试过的方法之一是:
One of the ways I've tried is:
construct {
?subject0 rs:similarityValue ?similairty0.
?subject1 rs:similarityValue ?similairty1
}
WHERE {
GRAPH ?g {?subject0 rs:similarityValue ?similairty0}.
{
?subject0 ?predicate0 ?object0.
....
但是我得到的结果是空的
but i got empty results
推荐答案
首先，最好确保可以选择要检索的所有信息.看来您的目标是这样的:
First, it's probably better to make sure that you can select all the information that you're trying to retrieve. It looks like you're aiming for something like this:
prefix rs: <http://www.welovethesemanticweb.com/rs#>
select distinct ?s ?weight ?factor where {
#-- ?x is the special value of interest. This
#-- is pulled out into a VALUES block just for
#-- convenience; there's just one place to change
#-- rs:Impromptu_No._1, now.
values ?x { rs:Impromptu_No._1 }
#-- find ?s which are "one step" away from
#-- a common property/value with ?x, and
#-- take 4/9 as ?weight.
{
?s ?p ?o .
?x ?p ?o .
bind(4/9 as ?weight)
}
union
#-- find ?s which are are "two steps" away from
#-- a common property/value with ?x, and take
#-- 1/9 as ?weight
{
?s ?a ?b . ?b ?p ?o .
?x ?c ?d . ?d ?p ?o .
bind(1/9 as ?weight)
}
#-- get the similarity factor of the property
#-- and make sure that ?s is different from ?x.
?p rs:hasSimilarityValue ?factor .
filter(?s != ?x)
}
-----------------------------------------------------------------------------------------------------------------------
| s | weight | factor |
=======================================================================================================================
| rs:5th_Symphony | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Dies_Irae | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Le_nozze_di_Figaro | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Piano_Concerto_No._24 | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Piano_Sonata_No._31 | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Piano_Sonata_No._31 | 0.444444444444444444444444 | "0.6"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Polonaise_heroique | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Polonaise_heroique | 0.444444444444444444444444 | "0.4"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Polonaise_heroique | 0.444444444444444444444444 | "0.6"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Preludes | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Preludes | 0.444444444444444444444444 | "0.4"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Preludes | 0.444444444444444444444444 | "0.6"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Requiem_Sequentia | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Sonata_Pathetique | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Symphony_No._29_in_A_major | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
| rs:Violin_Concerto_No._5_in_A_major | 0.111111111111111111111111 | "0.1"^^<http://www.w3.org/2001/XMLSchema#double> |
-----------------------------------------------------------------------------------------------------------------------
现在，看来是在此之后，您要按?s 的值分组并加权相似度的总和:
Now, it seems like after this, you want to group by by the value of ?s and sum the weighted similarities:
select distinct ?s (sum(?weight * ?factor) as ?similarity) where {
values ?x { rs:Impromptu_No._1 }
{
?s ?p ?o .
?x ?p ?o .
bind(4/9 as ?weight)
}
union
{
?s ?a ?b . ?b ?p ?o .
?x ?c ?d . ?d ?p ?o .
bind(1/9 as ?weight)
}
?p rs:hasSimilarityValue ?factor .
filter(?s != ?x)
}
group by ?s
----------------------------------------------------------------
| s | similarity |
================================================================
| rs:5th_Symphony | 0.044444444444444446e0 |
| rs:Piano_Concerto_No._24 | 0.044444444444444446e0 |
| rs:Requiem_Sequentia | 0.044444444444444446e0 |
| rs:Dies_Irae | 0.044444444444444446e0 |
| rs:Piano_Sonata_No._31 | 0.31111111111111117e0 |
| rs:Symphony_No._29_in_A_major | 0.044444444444444446e0 |
| rs:Le_nozze_di_Figaro | 0.044444444444444446e0 |
| rs:Violin_Concerto_No._5_in_A_major | 0.044444444444444446e0 |
| rs:Sonata_Pathetique | 0.044444444444444446e0 |
| rs:Preludes | 0.48888888888888893e0 |
| rs:Polonaise_heroique | 0.48888888888888893e0 |
----------------------------------------------------------------
最后，由于您已找到所需的值，因此现在可以构造所需的三元组:
Finally, since you've got the values that you're looking for, you can now construct the triples that you want:
construct {
?s rs:similarityValue ?similarity
}
where {{
select distinct ?s (sum(?weight * ?factor) as ?similarity) where {
values ?x { rs:Impromptu_No._1 }
{
?s ?p ?o .
?x ?p ?o .
bind(4/9 as ?weight)
}
union
{
?s ?a ?b . ?b ?p ?o .
?x ?c ?d . ?d ?p ?o .
bind(1/9 as ?weight)
}
?p rs:hasSimilarityValue ?factor .
filter(?s != ?x)
}
group by ?s
}}
@prefix : <http://www.semanticweb.org/rs#> .
@prefix rs: <http://www.welovethesemanticweb.com/rs#> .
@prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix owl: <http://www.w3.org/2002/07/owl#> .
@prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
rs:5th_Symphony rs:similarityValue 0.044444444444444446e0 .
rs:Polonaise_heroique
rs:similarityValue 0.48888888888888893e0 .
rs:Preludes rs:similarityValue 0.48888888888888893e0 .
rs:Requiem_Sequentia rs:similarityValue
0.044444444444444446e0 .
rs:Le_nozze_di_Figaro
rs:similarityValue 0.044444444444444446e0 .
rs:Symphony_No._29_in_A_major
rs:similarityValue 0.044444444444444446e0 .
rs:Piano_Concerto_No._24
rs:similarityValue 0.044444444444444446e0 .
rs:Sonata_Pathetique rs:similarityValue
0.044444444444444446e0 .
rs:Dies_Irae rs:similarityValue 0.044444444444444446e0 .
rs:Piano_Sonata_No._31
rs:similarityValue 0.31111111111111117e0 .
rs:Violin_Concerto_No._5_in_A_major
rs:similarityValue 0.044444444444444446e0 .
这篇关于如何通过CONSTRUCT在图上创建聚合的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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