- From: jorge perez <jorge.perez.rojas@gmail.com>
- Date: Wed, 9 Feb 2011 22:19:35 -0300
- To: Andy Seaborne <andy.seaborne@epimorphics.com>
- Cc: public-rdf-dawg-comments@w3.org

Thank you Andy, your email answers my question. I am not very comfortable with the need of using a subquery + DISTINCT to obtain the data that I wanted to obtain, but since considering multiple paths is a design decision in SPARQL 1.1 I am OK with that. I would advice the group to include that example (subquery + DISTINCT) in the specification, and also include the following example SELECT SUM(?A) WHERE { :me (:friend)+ ?F ?F :age ?A } which does not give the same answer and is also a possible way that a first-time user of property paths would try to obtain the data of friend ages (instead of using a sub query). Both examples would be a perfect way of making users aware of the way paths are being evaluated in SPARQL. Best regards, - jorge On Sun, Feb 6, 2011 at 5:55 PM, Andy Seaborne <andy.seaborne@epimorphics.com> wrote: > Jorge, > > You asked: > >> Aggregation is actually another reason of why multiple paths to the >> same endpoint *do not* have to be considered. >> >> Consider a network of friends, and assume that you want to obtain the >> SUM of the age of all your network (friends of your friends). Then a >> very natural way to do this is with the query (simplified syntax) >> >> SUM (?A) >> :me (:friend)+/:age ?A >> >> The query is navigating to all the friends of my friends, then to the >> age value of every one, and then taking the SUM. Isn't this natural? >> But, consider the following data >> >> :me :friend :f1 >> :me :friend :f2 >> :f1 :friend :f2 >> :f1 :age 20 >> :f2 :age 20 >> >> I would expect 40 as the result of the above query, but the expression >> >> :me (:fiend)+/:age ?A >> >> returns >> >> ?A >> 20 (for the path :me->:f1) >> 20 (for the path :me->:f2) >> 20 (for the path :me->:f1->:f2) >> >> and thus, the answer of the SUM would be 60. How do you explain the >> result of this query to a user? Notice that using DISTINCT does not >> solve the problem, since with DISTINCT you would obtain 20 as the SUM >> which is also wrong. >> >> Is there a way to correctly answer the above query with the current >> design of property paths? > > One way to query to get the sum of ages is: > > SELECT SUM(?A) > WHERE > { ?F :age ?A > { SELECT DISTINCT ?F > WHERE > { :me (:friend)+ ?F } }} > > This puts the uniqueness on the friends, then combines it with the ages and > calculates the SUM. There is a split in the property path because your query > requires distinctness on one part but not another. > > > Consider the following simplified purchase order, which includes two units > of :item1, by different paths (part of :compound and directly as a entry on > the purchase order). > > Data: > > @prefix : <http://example/> . > > :order :contains :thing1 . > :order :contains :compound1 . > > :thing1 :unitOf :item1 . > :thing2 :unitOf :item2 . > :thing3 :unitOf :item1 . > > :item2 :price 2 . > :item1 :price 2 . > > :compound1 :contains :thing2 . > :compound1 :contains :thing3 . > > Query: > > PREFIX : <http://example/> > > SELECT (SUM(?itemPrice) AS ?price) > { > :order :contains+/:unitOf/:price ?itemPrice . > } > > This returns 6 for ?price. Making the path match with DISTINCT would results > in 2. Here, all the prices are the same but we wish to retain duplicates as > they relate to different parts of the :order. > > We would be grateful if you would acknowledge that your comment has been > answered by sending a reply to this mailing list. > > Andy > On behalf of the SPARQL working group. >

Received on Thursday, 10 February 2011 01:20:08 UTC