Finding the relative position of elements in a list using SPARQL

问题

I'm trying to return subjects based on the relative position of their subjects in an ordered list.

A subject can be associated with multiple objects (via a single predicate) and all objects are in an ordered list. Given a reference object in this list I'd like to return the subjects in order of relative distance of their objects from the reference object.

:a : :x     
:b : :v
:b : :z
:c : :v
:c : :y

:ls :list (:v :w :x :y :z)

Taking x as our starting object in the list, the code below returns

:a :x :0
:c :y :1
:b :v :2
:b :z :2
:c :v :2

Instead of returning all positions I would like only the objects relating to the subject's minimum object 'distance' to be returned (which may mean up to two objects per subject - both up and down the list). So I'd like to return

:a :x :0
:c :y :1
:b :v :2
:b :z :2

The code so far... (with a lot of help from Find lists containing ALL values in a set? and Is it possible to get the position of an element in an RDF Collection in SPARQL?)

SELECT ?s ?p (abs(?refPos-?pos) as ?dif) 
WHERE {
      :ls :list/rdf:rest*/rdf:first ?o .
      ?s : ?o .
      {
      SELECT ?o (count(?mid) as ?pos) ?refPos 
      WHERE {
            [] :list/rdf:rest* ?mid . ?mid rdf:rest* ?node .
            ?node rdf:first ?o .
            {
            SELECT ?o (count(?mid2) as ?refPos)
            WHERE {
                  [] :list/rdf:rest* ?mid2 . ?mid2 rdf:rest* ?node2 .
                  ?node2 rdf:first :x .
                  }
            }
            }
            GROUP BY ?o
      }
      }
      GROUP BY ?s ?o
      ORDER BY ?dif

I've been trying to get a minimum ?dif (difference/distance) by grouping by ?s but because I then have to apply this (something like ?dif = ?minDif) to the ?s ?o grouping from earlier I don't know how to go back and forward between these two groupings.

Thanks for any assistance you can provide

回答1:

All you needed to compound a solution is yet another one Joshua Taylor's answer: this or this.

Here below I'm using Jena functions, but I hope the idea is clear.

Query 1

PREFIX list: <http://jena.hpl.hp.com/ARQ/list#>
SELECT ?s ?el ?dif {
    ?s : ?el .
    :ls :list/list:index (?pos ?el) .
    :ls :list/list:index (?ref :x) .
    BIND (ABS(?pos -?ref) AS ?dif) 
    {
    SELECT ?s (MIN (?dif_) AS ?dif) WHERE {
        ?s : ?el_ .
        :ls :list/list:index (?pos_ ?el_) .
        :ls :list/list:index (?ref_ :x) .
        BIND (ABS(?pos_ - ?ref_) AS ?dif_)
        } GROUP by ?s
    }
}

Query 2

PREFIX list: <http://jena.apache.org/ARQ/list#>    
SELECT ?s ?el ?dif {
    ?s : ?el .
    :ls :list/list:index (?pos ?el) .
    :ls :list/list:index (?ref :x) .
    BIND (ABS(?pos -?ref) AS ?dif) 
    FILTER NOT EXISTS {
        ?s : ?el_ .
        :ls :list/list:index (?pos_ ?el_) .
        BIND (ABS(?pos_ - ?ref) AS ?dif_) .
        FILTER(?dif_ < ?dif)                                           
    }
}

Update

Query 1 can be rewritten in this way:

PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

SELECT ?s ?el ?dif {
  ?s : ?el 
  { select (count(*) as ?pos) ?el {[] :list/rdf:rest*/rdf:rest*/rdf:first ?el} group by ?el }
  { select (count(*) as ?ref)     {[] :list/rdf:rest*/rdf:rest*/rdf:first :x} } 
  BIND (ABS(?pos - ?ref) AS ?dif) 
  {
  SELECT ?s (MIN(?dif_) AS ?diff) {
    ?s : ?el_ 
    { select (count(*) as ?pos_) ?el_ {[] :list/rdf:rest*/rdf:rest*/rdf:first ?el_} group by ?el_ }
    { select (count(*) as ?ref_)      {[] :list/rdf:rest*/rdf:rest*/rdf:first :x} } 
    BIND (ABS(?pos_ - ?ref_) AS ?dif_)
    } GROUP by ?s
  }
  FILTER (?dif = ?diff)
}

Notes

As you can see, this is not what SPARQL was designed for. For example, Blazegraph supports Gremlin...
Possibly this is not what RDF was designed for. Or try other modeling approach: do you really need RDF lists?
I haven't tested the above query in Virtuoso.

来源：https://stackoverflow.com/questions/50014338/finding-the-relative-position-of-elements-in-a-list-using-sparql

标签

sparql