SV: [RDF] Sorting selections

[Stefan Andersson]

Yo!

Actually, I have been thinking about this issue for some time. I haven't
read anything, though, so supposedly somebody somewhere already have solved
the problem. Oh, well. That won't stop me from thinking loudly for myself:

I was wondering whether this actually is a problem of defining exactly
_what_ you want, not how to pick a certain number more or less randomly.

So, when you say 'I just want _one_ occurence', what do you _really_ say,
i.e. _why_ do you just want one occurence?

And, implicitly, what should we do with the rest of them?

Ok. Maybe banal, maybe naive, but I think resolving the question is so
complex, maybe what you would want to do, is redefine the _question_?

And I have a hunch that the 'dialogue' approach is one way of solving
this...

Just my ha'penny.
/Stefan

> -----Ursprungligt meddelande-----
> Fran: rdf-admin@uxn.nu [mailto:rdf-admin@uxn.nu]For Jonas Liljegren
> Skickat: den 19 februari 2001 14:02
> Till: Wraf development
> Amne: [RDF] Sorting selections
> 
> 
> I started this message two weeks ago.  I have prosponed this topic,
> but may as well send out this now.
> 
> 
> I was thinking about how to sort selections.
> 
> Let's say we have four interfaces and asks for resources with a
> specific property.
> 
> If there are a lot of resources that match the criterion, it would
> generate a very large selection.  But we may only want to list the
> first five.
> 
> In order to save time and memory, we only put the criterion in the
> selection and doesn't compute the result.  Then li($n) is called to
> get the $n entry, the selection entry is expanded.
> 
> If the selection comes from only one interface, we could let the
> interface directly return the $n resource.  But because several
> interfaces is onvolved, none of the interface know which enty is
> number $n.  We have to get one by one until we get to $n.
> 
> If the ordering isn't important, we could return the objects from one
> interface after another.  If we request numer 248, and a size()
> request to the first interface gives 235, we continue with the next
> interface.
> 
> But a size() should not have to count all resources if the point is to
> know in which interface number 248 resides in.
> 
> 
> Dynamic properties could be computed based on many things.  If we want
> all objects of a specific type, we cant let the backend only retrieve
> those resources. anotther interface could say that another class is a
> subClassOf that type.  It could be very complicated.  many selections
> would have to be done.
> 
> 
> If the order is significant, we have to build an internal list of
> resources in the container.  If the conteiner is very larg, we don't
> want to have the whole list in memory.
> 
> We could do some sorts of partial sorting.  For example, if we only
> want the top three items, we wouldn't have to sort the bottom 9997
> items.
> 
> 
> -- 
> / Jonas Liljegren
> 
> The Wraf project http://www.uxn.nu/wraf/
> Sponsored by http://www.rit.se/
> 
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