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Re: What is it that's wrong with rdf:List

From: Paul Gearon <gearon@ieee.org>
Date: Wed, 23 Jun 2010 06:08:10 -0700
Message-ID: <AANLkTin2KAWm68auPm0yPbiX3EPA3ZOOv2dP1s1tpXx4@mail.gmail.com>
To: Pierre-Antoine Champin <swlists-040405@champin.net>
Cc: Andy Seaborne <andy.seaborne@talis.com>, Semantic Web <semantic-web@w3.org>
On Wed, Jun 23, 2010 at 3:59 AM, Pierre-Antoine Champin
<swlists-040405@champin.net> wrote:
> On 20/06/2010 13:23, Andy Seaborne wrote:
>>
>>
>> On 19/06/2010 10:05 PM, Pat Hayes wrote:
>>>
>>> On Jun 19, 2010, at 11:54 AM, Andy Seaborne wrote:
>>>
>>>>
>>>>
>>>> On 17/06/2010 11:35 PM, Paul Gearon wrote:
>>>>> The main problem with an RDF list is that there is no mechanism in
>>>>> SPARQL to query or update them. SPARQL only allows you to form queries
>>>>> that explicitly describe connections, while an arbitrary list can have
>>>>> any number of elements down its length. That means that you can refer
>>>>> to, say, the 2nd, 3rd, or 4th elements in the list, but there is no
>>>>> way to refer to ALL the elements in the list, since you don't know how
>>>>> long the list is.
>>>>>
>>>>> SPARQL 1.1 will fix this problem with "property paths".
>>>>
>>>> with
>>>>
>>>> { <list> rdf:rest*/rdf:first ?x }
>>>>
>>>> this is only a partial solution:
>>>>
>>>> 1/ Order in a list is not preserved
>>>
>>> ? Preserved by what? Lists in RDF have an order (that is, the elements
>>> of the list are put into an order by the list.) What is it that doesn't
>>> 'preserve' this?
>>
>> The SPARQL property path expression that Paul mentioned does not provide
>> any guarantee on order of results returned, nor do further operations of
>> the SPARQL algebra for graph patterns preserve order.  SPARQL looses the
>> ordering.  It can't properly return lists either, only members of a list.
>
> By the way, is there a clean way in SPARQL to retrieve *all* items of an
> rdf:Seq, which would preserve the order ?
>
> The only way I can think of involves FILTER, regex on URIs, and would
> put item #11 before item #2...

I have to check, but I believe that substring will be a function in
SPARQL 1.1 (perhaps with a domain). Some triplestores already have
this. Once you have it you can say something like:

SELECT ?seq ?element WHERE {
 ?seq ?prop ?element FILTER regex(?prop,
"^http://www.w3.org/1999/02/22-rdf-syntax-ns#_")
}
ORDER BY xsd:int(substring(?prop, 45))


(The rdf:_ string is 44 characters long)

Regards,
Paul Gearon
Received on Wednesday, 23 June 2010 13:08:43 GMT

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