Alison,
If by a maximum value, you mean the highest degree earned, you could write
two functions - one to get the highest degree earned, another the associated
college. If, however, a person earned two masters at two different colleges,
the result of the second function will be uncertain - sometimes it may
return one college, sometimes the other. Same if he earned a BS degree and a
BA degree and they are both at the same level, making it unclear which one
is the highest.
In this case, your question boils down to how to write a function that will
return the highest degree earned. In SPARQL, this can be accomplished by the
following query:
SELECT ?maxDegree
WHERE {?person :degree ?maxDegree.
?maxDegree hasRank ?maxDegreeRank.
NOT EXISTS {?person :degree ?degree.
?degree hasRank ?degreeRank.
FILTER (?degreeRank < ?maxDegreeRank)}
}
You will need to have all this information in the model and Kennedys example
does not cover it. You could try with Kennedy's to get the oldest of a
person's parents.
Regards,
Irene Polikoff
From: [email protected]
[mailto:[email protected]] On Behalf Of Alison Callahan
Sent: Thursday, May 16, 2013 9:17 AM
To: Scott Henninger; [email protected]
Subject: Re: [topbraid-users] defining a SPIN function that returns multiple
variables
Hi again Scott,
Please see my follow-up question below.
Thanks.
On Thu, May 9, 2013 at 2:26 PM, Scott Henninger <[email protected]>
wrote:
Allison; Since the AnotherExample:ScoreAttendedCollege returns one value,
not a set of values (multiple rows) or a list of values (multiple columns),
then the best choice is to use a SPIN function. Change the type of
:ScoreAttendedCollege to spin:Functions and use in a BIND or projection:
SELECT ?score
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
...and the projection version would be as follows:
SELECT (AnotherExample:ScoreCollege("John") AS ?score)
WHERE {}
These translate into the exact same SPARQL algebra, so it's a matter of
preference which one you use.
Because kennedys:JohnKennedy has kennedys:almaMater more than one college,
is it not correct that making AnotherExample:ScoreCollege a spin:function
and not magic property would mean that only one of the colleges he attended
would be evaluated and its score returned? How could one, for example, have
a function evaluate all colleges he attended and return the maximum value
and the associated college? Apologies for the very specific example, but
this is really the behaviour I'm going after.
For your magic property, I think you just got it backwards the subject of
the magic property are the variables bound in the WHERE clause. The object
are the selected variables. So I think the following will work:
SELECT *
WHERE {
("John") AnotherExample:ScoreCollege (?score) .
}
Yes, I had it backwards. This worked.
On Wed, May 15, 2013 at 3:37 PM, Alison Callahan <[email protected]>
wrote:
Hi again Scott,
Please see my follow-up question below.
Thanks.
On Thu, May 9, 2013 at 2:26 PM, Scott Henninger <[email protected]>
wrote:
Allison; Since the AnotherExample:ScoreAttendedCollege returns one value,
not a set of values (multiple rows) or a list of values (multiple columns),
then the best choice is to use a SPIN function. Change the type of
:ScoreAttendedCollege to spin:Functions and use in a BIND or projection:
SELECT ?score
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
...and the projection version would be as follows:
SELECT (AnotherExample:ScoreCollege("John") AS ?score)
WHERE {}
These translate into the exact same SPARQL algebra, so it's a matter of
preference which one you use.
Because John Kennedy attended multiple colleges, is it not correct that
making AnotherExample:ScoreCollege a spin:function and not magic property
would mean that only one of the colleges he attended would be evaluated and
its score returned? How could I, for example, have a function evaluate all
colleges he attended and return the maximum value and the associated
college? I apologize for the very specific example, but this is really the
behaviour I'm going after.
For your magic property, I think you just got it backwards the subject of
the magic property are the variables bound in the WHERE clause. The object
are the selected variables. So I think the following will work:
SELECT *
WHERE {
("John") AnotherExample:ScoreCollege (?score) .
}
Yes, I had it backwards. This worked.
-- Scott
On 5/9/2013 10:42 AM, Alison Callahan wrote:
Hi Scott,
Thanks for your detailed reply. I successfully followed your
MagicExample:FindNamesOfCollegeGrad example.
I have a follow-up question, based on another thread
<http://mail-archive.com/[email protected]/msg02089.html> : If
I use a magic property in the spin:body of another magic property, and there
is more than one result when the magic property is executed, how can I
retrieve all of the results?
For example, when John Kennedy is passed in to the FindNamesOfCollegeGrad
magic property in a SPARQL query, i.e.
SELECT *
WHERE
{ ?person a kennedys:Person .
?person kennedys:almaMater ?college .
(?person ?college) MagicExample:FindNamesOfCollegeGrad ("John" ?middle
?last)
}
there are two results for ?college: kennedys:Harvard and kennedys:Princeton.
Now let's say I have another magic property called
'AnotherExample:ScoreAttendedCollege' where a first name is passed in as
?arg1 and the spin:body of the property is:
SELECT ?collegeScore
WHERE
{
(?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?middle
?last) .
BIND(IF((?college = kennedys:Princeton), 1, 0) AS ?collegeScore) .
}
In other words, if the person with the specified first name attended
Princeton, return a score of 1, else return a score of 0.
I created this magic property, but when I use it in a SPARQL query as
follows:
SELECT *
WHERE {
(?score) AnotherExample:ScoreCollege("John") .
}
there are no results. However, if I use it in a SPARQL query this way:
SELECT ?score
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
the result of the query is 0. My hunch is that this is because only the
first bound result (kennedys:Harvard) was evaluated by the
'AnotherExample:ScoreCollege' magic property, i.e. the kennedys:Princeton
result is never evaluated because the iteration seems to break. Is this
correct?
How can I use AnotherExample:ScoreCollege to return the score for each value
bound to the ?college variable returned by the
MagicExample:FindNameOfCollegeGrad magic property? If I can't use BIND
within a magic property to assign a value based on the result of an embedded
magic property, what alternatives exist?
Thanks again!
Alison
On Wed, May 8, 2013 at 4:11 PM, Scott Henninger <[email protected]>
wrote:
Allison, probably the best way to learn magic properties is by example. In
the Composer navigator, go to TopBraid/Examples/kennedysSPINMagic.ttl and
open that.
In the Classes view find spin:MagicProperties. A Magic property is a
subclass of spin:MagicProperties. You can see the body of the definition
for age (after running inferences), grandfather, and grandMother. These are
used in SPARQL as a property that instead calls the body of the query (hence
a "magic" property).
An example, let's say you want to find a grandfather:
SELECT *
WHERE
{ ?person a kennedys:Person .
?person kspin:grandFather ?gf
}
Note you can use this magic property either way. So to find all
grandchildren of Joe Kennedy:
SELECT *
WHERE
{ ?gchild kspin:grandFather kennedys:JosephKennedy
}
I know, I haven't answered the question yet. For multiple parameters of
values, use a list structure (see, for example top:files in Help > TopBraid
Composer > Reference > SPARQL Property Functions). As an additional
example, I've added an example in the attached file to define a magic
property named MagicExample:FindNamesOfCollegeGrad (kinda a silly example,
buyt it gets the idea across)
You can use it this way for example:
SELECT *
WHERE
{ ?person a kennedys:Person .
?person kennedys:almaMater ?college .
(?person ?college) MagicExample:FindParentsOfCollegeGrad (?first ?middle
?last)
}
You can bind/not bind any of the input parameters or output values. For
example to find all John's try this.
SELECT *
WHERE
{ ?person a kennedys:Person .
?person kennedys:almaMater ?college .
(?person ?college) MagicExample:FindParentsOfCollegeGrad ("John" ?middle
?last)
}
Given that as an interesting example to explore, let us know if that answers
the question or if you have follows.
-- Scott
On 5/8/2013 11:51 AM, Alison Callahan wrote:
Hi Holger,
I have looked at Magic Properties, and thought they were relevant to my
question, but I couldn't find any examples of how to create a magic property
that returns multiple values. On the page you linked to, it says "Magic
properties can also take multiple arguments and result values using a
(rather obscure) list syntax - these cases are technically supported but
complex to represent in the SPIN RDF syntax".
Could you provide more detail or a link that describes this list syntax for
creating a magic property that returns multiple values?
Thanks!
Alison
On Mon, Apr 29, 2013 at 7:00 PM, Holger Knublauch <[email protected]>
wrote:
Hi Allison,
yes take a look at Magic Properties
http://spinrdf.org/spin.html#spin-magic
These may not only return multiple "rows" but also multiple variables per
row.
Please take a look at the online material and get back to us if you have
specific follow-up questions.
HTH
Holger
On 4/30/2013 1:01, Alison Callahan wrote:
Hello all,
I would like to be able to define the body of a SPIN function that returns
two variables, e.g.
SELECT ?x ?y
WHERE {
?example test:x ?x .
?example text:y ?y .
}
My question is: if such a function is possible, how are the results bound
when the function is called? In my experience with SPIN I have written
functions that return one variable, and thus the result is bound to a single
variable when the function is called. For example, if I have a SPIN function
called "functionA" where the spin:body is:
SELECT ?a
WHERE {
?example test:a ?a .
}
I would call this function and bind the result as
BIND(:functionA(?aVariable) AS ?returnedA) .
Is it possible to write a similar function that returns two (or more)
variables?
Any help is appreciated!
Alison
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