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Re: Comments about the semantics of property paths

From: Andy Seaborne <andy.seaborne@epimorphics.com>
Date: Sun, 06 Feb 2011 20:55:52 +0000
Message-ID: <4D4F0AD8.7030507@epimorphics.com>
To: jorge perez <jorge.perez.rojas@gmail.com>
CC: public-rdf-dawg-comments@w3.org
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 Sunday, 6 February 2011 20:56:29 GMT

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