defining a SPIN function that returns multiple variables
Alison Callahan
SELECT ?x ?y
?example test:x ?x .
}
Holger Knublauch
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
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Alison Callahan
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Scott Henninger
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
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Alison Callahan
WHERE
{ ?person a kennedys:Person .
?person kennedys:almaMater ?college .
}
WHERE {
(?score) AnotherExample:ScoreCollege("John") .
}
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
Scott Henninger
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
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.
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) .
}
-- Scott
Alison Callahan
Thanks.
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...and the projection version would be as follows:
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
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.
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) .
}
Hi again Scott,Please see my follow-up question below.
Thanks.
On Thu, May 9, 2013 at 2:26 PM, Scott Henninger <shenn...@topquadrant.com> 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...and the projection version would be as follows:
WHERE {
BIND(AnotherExample:ScoreCollege("John") AS ?score).
}
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.
Irene Polikoff
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
Alison Callahan
WHERE {
WHERE {
Alison Callahan
WHERE {
WHERE {
Scott Henninger
So let's say you create a property named AnotherExample:collegeScore, which is associated with instances of kennedys:College.� Then the SPIN function query you are looking for, I believe, is the following:
SELECT (max(?score) AS ?maxScore)
WHERE {
��� ?arg1 kennedys:almaMater ?college .
��� ?college AnotherExample:collegeScore ?score .
}
...which gives the max college score for any resource passed to ?arg1
Give that a try and let us know if it works for you.
-- Scott
Hello again all,
I wanted to ask a question again (below), as this thread has gotten quite long and I didn't get a reply to my latest email. Thanks,
Alison----------------------
I have created a magic property AnotherExample:ScoreCollege that assigns a score to a Kennedy based on the college they attended - 1 if they attended Princeton, and 0 otherwise.
The spin:body of this property is as follows:
SELECT ?collegeScore ?collegeWHERE{ ��� (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?middle ?last) .���BIND(IF((?college = kennedys:Princeton), 1, 0) AS ?collegeScore) .}
The magic property MagicExample:FindNamesOfCollegeGrad was provided by Scott and defined as:
SELECT ?fname ?mname ?lname
WHERE {
� � �?arg1 kennedys:almaMater ?arg2 .� � �?arg1 kennedys:firstName ?fname .� � �?arg1 kennedys:middleName ?mname .� � �?arg1 kennedys:LastName ?lname .
}
When I use AnotherExample:ScoreCollege in a SPARQL query, as in:
SELECT ?college ?score
WHERE {
� ���("John")�AnotherExample:ScoreCollege�(?score ?college) .
}
The result is two scores - 0 for Harvard, and 1 one Princeton.
My question is: how can I use this magic property to get all colleges attended and then return the maximum score and the college that corresponds to it. Based on Irene's above example, I created a magic property with this spin:body:
SELECT ?collegeScore ?collegeWHERE {
� � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?collegeScore) .� � NOT EXISTS {� � � � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � � � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?lowerCollegeScore .� � � � FILTER(?lowerCollegeScore < ?collegeScore).� �}
}
When I use this property as above, I get two results: 0 for Harvard and 1 for Princeton.
When I make a corresponding spin:Function called AnotherExample:ScoreCollegeFunction that returns only ?collegeScore e.g.
SELECT ?collegeScoreWHERE {� � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?collegeScore) .� � NOT EXISTS {� � � � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � � � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?lowerCollegeScore .� � � � FILTER(?lowerCollegeScore < ?collegeScore).� �}}
and call it in a SPARQL query as in:
SELECT * score {
� � BIND(AnotherExample:ScoreCollegeFunction("John") AS ?score) .
}
it returns one value: 0. I think that this is because the spin:Function is only executed on the first result of the magic property MagicExample:FindNamesOfCollegeGrad and then stops.
Is it possible to write a function that uses a magic property to get a set of values, and then evaluates the entire set in some way? (In this case, to return the maximum score for an attended college?) Do I need to incorporate a foreach to iterate over the results?
On Thu, May 16, 2013 at 11:09 AM, Alison Callahan <alison....@gmail.com> wrote:
Hi Irene,
Thanks for your reply.
Earlier in this email thread I specified a magic property AnotherExample:ScoreCollege that assigns a score to a Kennedy based on the college they attended - 1 if they attended Princeton, and 0 otherwise.
The spin:body of this property is as follows:
SELECT ?collegeScore ?collegeWHERE{ ��� (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?middle ?last) .���BIND(IF((?college = kennedys:Princeton), 1, 0) AS ?collegeScore) .}
The magic property MagicExample:FindNamesOfCollegeGrad was provided by Scott and defined as:
SELECT ?fname ?mname ?lname
WHERE {
� � �?arg1 kennedys:almaMater ?arg2 .� � �?arg1 kennedys:firstName ?fname .� � �?arg1 kennedys:middleName ?mname .� � �?arg1 kennedys:LastName ?lname .
}
When I use AnotherExample:ScoreCollege in a SPARQL query, as in:
SELECT ?college ?score
WHERE {
� ���("John")�AnotherExample:ScoreCollege�(?score ?college) .
}
The result is two scores - 0 for Harvard, and 1 one Princeton.
My question is: how can I use this magic property to get all colleges attended and then return the maximum score and the college that corresponds to it. Based on your above example, I created a magic property with this spin:body:
SELECT ?collegeScore ?collegeWHERE {
� � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?collegeScore) .� � NOT EXISTS {� � � � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � � � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?lowerCollegeScore .� � � � FILTER(?lowerCollegeScore < ?collegeScore).� �}
}
When I use this property as above, I still get two results: 0 for Harvard and 1 for Princeton.
When I make a corresponding spin:Function called AnotherExample:ScoreCollegeFunction that returns only ?collegeScore e.g.
SELECT ?collegeScoreWHERE {� � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?collegeScore) .� � NOT EXISTS {� � � � (?person ?college) MagicExample:FindNamesOfCollegeGrad (?arg1 ?mname ?lname) .� � � � BIND(IF((?college = kennedys:Princeton), 1, 0) as ?lowerCollegeScore .� � � � FILTER(?lowerCollegeScore < ?collegeScore).� �}}
and call it in a SPARQL query as in:
SELECT * score {
� � BIND(AnotherExample:ScoreCollegeFunction("John") AS ?score) .
}
it returns one value: 0. I think that this is because the spin:Function is only executed on the first result of the magic property MagicExample:FindNamesOfCollegeGrad and then stops.
Is it possible to write a function that uses a magic property to get a set of values, and then evaluates the entire set in some way? (In this case, to return the maximum score for an attended college?) Do I need to incorporate a foreach to iterate over the results?
Thanks,
Alison
On Thu, May 16, 2013 at 10:32 AM, Irene Polikoff <ir...@topquadrant.com> wrote:
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: topbrai...@googlegroups.com [mailto:topbrai...@googlegroups.com] On Behalf Of Alison Callahan
Sent: Thursday, May 16, 2013 9:17 AM
To: Scott Henninger; topbrai...@googlegroups.com
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�<shenn...@topquadrant.com>�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 <alison....@gmail.com> wrote:
Hi again Scott,
�
Please see my follow-up question below.
Thanks.
On Thu, May 9, 2013 at 2:26 PM, Scott Henninger <shenn...@topquadrant.com> 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: 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 <shenn...@topquadrant.com> 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 <hol...@topquadrant.com> 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 .
�
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Group "TopBraid Suite Users", the topics of which include Enterprise Vocabulary Network (EVN), TopBraid Composer, TopBraid Live,
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