I need to add, in an empty class, individuals obtained from an sqwrl query.
This is possible with an swrl rule but I can't do that with sqwrl.
Is there any chances to put in a class the results obtained from an SQWRL query?
For example in this image I have this result:
How can i put in a class the column v?
Can someone help me?
Related
I have an exercise in Semantic Web. I must extract some individuals from the DBpedia. These individuals must be inserted into an ontology that I must create. My question is. Can I retrieve individuals from the DBedia?
Let me clarify !
When I send this sparql query
PREFIX dbo: <http://dbpedia.org/ontology>
SELECT distinct * WHERE
{
?album a dbo:Album .
} LIMIT 10
I get 10 URIs. Should I get whole instances ? I mean, label, object properties, data properties etc. in order to insert them to my ontology?
I want them as a complete instance. I don't want to add more variables e.g
?album dbo:artist ?artist .
Can I use a java api e.g. Jena ?
EDIT:
Let me give you an example. Suppose that you get an Album with URI
http://dbpedia.org/resource/...Baby_One_More_Time_(album)
This album has also some properties with their values e.g.
dbo:artist dbr:Britney_Spears
dbo:releaseDate 1999-01-12 (xsd:date)
...
How could I get all of them in order to create an indivual album for my ontology with properties artist and releaseDate and values Britney_Spears and 1999-01-12 respectively ?
Well, a good point always to start is your requirements! What do you exactly need? There is scientific plethora research on Ontology Module Extraction (see for example here).
My rule of thumb is that: the amount you extract must align with the required constraints of soundness and completeness of results, which in turn, aligns with your requirements. To make it clear, consider the following: A DBpedia Artist is a subClassOf Person. Now consider that you extract all the instances of Artist from DBPedia, without the piece of information that Artist is a subClassOf Person. Now if you query your dataset asking for Person, you will get nothing. Is this a sound result? yes, but is it complete? No! However, if you don't care about the fact that each Artist is a Person, then it's okay. A mentioning worthy thing is that it depends on the DBpedia endpoint itself and what kind of reasoning it performs as well.
Concluding: Specify what you really need. While you can suffice for a couple of classes with their instances, you can as well extract the whole DBpedia.
Regarding getting the data, there are multiple ways; again depending on your requirements. For simple purposes, you can use Jena TDB for triples storage and access them via Jena. You can even store your data simply in an RDF file. You can, for example, use a construct query on DBpedia endpoint and specify the results format as RDF and then insert them to your RDF engine. Another option, for example, this answer, states how to use an INSERT query to perform the insert task into a local graph.
You can retrieve instances from DBpedia with whatever metadata you want, but it depends on your ontology that you would like to create. Please take a look at this document, it will help you to understand some notions.
Should you get whole instances? I think you are asking if you should take all the proporties and objects depending on the subject. Not necessarily..It depends on your ontology as stated in first step and you decide what to take.
Should you use Jena? You can but you don't have to! If you pose a CONSTRUCT query to the endpoint you can get the data but as far as I understood you don't want to add variables. So you can pose a query as follows by asking all the metadata regarding to the instance.
CONSTRUCT { ?album ?p ?o } WHERE {
?album a dbo:Album .
?album ?p ?o
}
If you would like to get a limited number of instances then you can add limit again at the end of this query.
I am working on Panama dataset using Neo4J graph database 1.1.5 web version. I identified Ion Sturza, former Prime Minister of Moldova on the database and want to make a map of his related network. I used following code to query using Cypher (creating a variable 'IonSturza'):
MATCH (IonSturza {name: "Ion Sturza"}) RETURN IonSturza
I identified that the entity 'CONSTANTIN LUTSENKO' linked differently to entities like 'Quade..' and 'Kinbo...' with a name in small letters as in this picture. I hence want to map a relationship 'SAME_COMPANY_AS' between the capslock and the uncapped version. I tried the following code based on this answer by #StefanArmbruster:
MATCH (a:Officer {name :"Constantin Lutsenko"}),(b:Officer{name :
"CONSTANTIN LUTSENKO"})
where (a:Officer{name :"Constantin Lutsenko"})-[:SHAREHOLDER_OF]->
(b:Entity{id:'284429'})
CREATE (a)-[:SAME_COMPANY_AS]->(b)
Instead of indexing, I used the 'where' statement to specify the uncapped version which is linked only to the entity bearing id '284429'.
My code however shows the cartesian product error message:
This query builds a cartesian product between disconnected patterns.If a part of a query contains multiple disconnected patterns, this will build a cartesian product between all those parts. This may produce a large amount of data and slow down query processing. While occasionally intended, it may often be possible to reformulate the query that avoids the use of this cross product, perhaps by adding a relationship between the different parts or by using OPTIONAL MATCH (identifier is: (b))<<
Also when I execute, there are no changes, no rows!! What am I missing here? Can someone please help me with inserting this relationship between the nodes. Thanks in advance!
The cartesian product warning will appear whenever you're matching on two or more disconnected patterns. In this case, however, it's fine, because you're looking up both of them by what is likely a unique name, s your result should be one node each.
If each separate part of that pattern returned multiple nodes, then you would have (rows of a) x (rows of b), a cartesian product between the two result sets.
So in this particular case, don't mind the warning.
As for why you're not seeing changes, note that you're reusing variables for different parts of the graph: you're using variable b for both the uppercase version of the officer, and for the :Entity in your WHERE. There is no node that matches to both.
Instead, use different variables for each, and include the :Entity in your match. Also, once you match to nodes and bind them to variables, you can reuse the variable names later in your query without having to repeat its labels or properties.
Try this:
MATCH (a:Officer {name :"Constantin Lutsenko"})-[:SHAREHOLDER_OF]->
(:Entity{id:'284429'}),(b:Officer{name : "CONSTANTIN LUTSENKO"})
CREATE (a)-[:SAME_COMPANY_AS]->(b)
Though I'm not quite sure of what you're trying to do...is an :Officer a company? That relationship type doesn't quite seem right.
I tried the answer by #InverseFalcon and thanks to it, by modifying the property identifier from 'id' to 'name' and using the property for both 'a' and 'b', 4 relationships were created by the following code:
MATCH (a:Officer {name :"Constantin Lutsenko"})-[:SHAREHOLDER_OF]->
(:Entity{name:'KINBOROUGH PORTFOLIO LTD.'}),(b:Officer{name : "CONSTANTIN
LUTSENKO"})-[:SHAREHOLDER_OF]->(:Entity{name:'Chandler Group Holdings Ltd'})
CREATE (a)-[:SAME_NAME_AS]->(b)
Thank you so much #InverseFalcon!
i have just started using Neo4j, and after creating the whole graph i'm trying to get all the nodes related to another one by a relatioship.
Match (n)-[Friendship_with]->({Name:"Gabriel"}) return n
That should give me the nodes that are friend of Gabriel, what i'm doing wrong?
I have used too this:
Match n-[r:Friendship_with]->n1 where n1.Name="Gabriel" return n
That give me some nodes, but some of then aren't directly related to Gabriel (for example, Maria is friend of Gabriel, she appears when i write that, but Alex who is friend of Maria and not from Gabriel, appear too)
This is weird.
Your query is correct.
I would suggest to check your data. Are you sure there isn't any direct connection between Alex and Gabriel ?
You could visualize your graph and see what is happening exactly in the neo4j browser. Just type a query with a bit more info like:
MATCH (n)-[f:Friendship_with]->(p {Name:"Gabriel"}) return n,f,p
and use the graph view to inspect your data.
EDIT:
As Pointed out by Michael, your first query is missing a colon in front of the specified relationship label "Friendship_with". So neo4j thinks it is a (rather long) variable name for your relationships, just as 'n' or 'n1'. It will thus retrieve anything that is connected to Gabriel without filtering by relationship label.
It doesn't explain though why you:
get the same results with the first and second query
get a 2nd degree relation as a result
so check your data anyway :)
You forgot the colon before :Friendship_with
Don't forget to provide labels, e.g. (n1:Person {Name:"Gabriel"})
Also some of your friendships might go in the other direction, so leave off the direction-arrow: Match (n:Person)-[Friendship_with]-(:Person {Name:"Gabriel"}) return n
I'm trying to query Book nodes for recommendation by Cypher.
I want to recommend A:Book and C:Book for A:User.
i'm sorry I need some graph to explain this question, but I could't up graph image because my lepletion lacks for upload function.
I wrote query below.
match (u1:User{uid:'1003'})-->(o1:Order)-->(b1:Book)<--(o2:Order)
<--(u2:User)-->(o3:Order)-->(b2:Book)
return b2
This query return all Books(A,B,C,D) dispite cypher's Uniqueness.
I expect to only return A:Book and C:Book.
Is this behavior Neo4j' specification?
How do I get expected return? Thanks, everyone.
environment:
Neo4j ver.v2.0.0-RC1
Using Neo4j Server with REST API
Without the sample graph its hard to say why you get something back when you expected something else. You can share a sample graph by including a create statement that would generate said graph, or by creating it in Neo4j console and putting the link in your question. Here is an example of the latter: console.neo4j.org/r/fnnz6b
In the meantime, you probably want to declare the type of the relationships in your pattern. If a :User has more than one type of outgoing relationships you will be excluding those other paths based on the labels of the nodes on the other end, which is much less efficient than to only traverse the right relationships to begin with.
To my mind its not clear whether (u:User)-->(o:Order)-->(b:Book) means that a user has one or more orders, and each order consists of one or more books; or if it means only that a user ordered a book. If you can share a sample, hopefully that will be clear too.
Edit:
Great, so looking at the graph: You get B and D back because others who bought B also bought D, and others who bought D also bought B, which is your criterion for recommendation. You can add a filter in the WHERE clause to exclude those books that the user has already bought, something like
WHERE NOT (u1)-[:BUY]->()-[:CONTAINS]->(b2)
This will give you A, C, C back, since there are two matching paths to C. It's probably not important to get two result items for C, so you can either limit the return to give only distinct values
RETURN DISTINCT(b2)
or group the return values by counting the matching paths for each result as a 'recommendation score'
RETURN b2, COUNT(b2) as score
Also, if each order only [CONTAINS] one book, you could try modelling without order, just (:User)-[:BOUGHT]->(:Book).
I have a domain class which has around 20 properties. A findBy on the domain class results in a select query which has all the columns selected from the database, which can be a performance hit when the required column could be only 1.
So I thought of using, withCriteria.
def sampleDomainInst = SampleDomain.withCriteria{
projections {
property('fieldOne')
}
eq('id', idVal)
}
The value returned is a list. But what I need is an instance of SampleDomain How do I do that?
Thank You.
Regards,
Jay Chandran
The goal of projections is IMHO not to have domain instances returned. In theory you might add 'id' to the projections closure and then you can do a DomainClass.get(id). But that is the same as working completely without the projection.
If your domain class has so many properties that a you're using projections to get only parts of them, you should consider splitting up the domain class in multiple joined classes. A good design practice is that each class should only represent one single abstraction.
Try withCriteria(uniqueResult: true) {...} or, longer, SampleDomain.createCriteria().get {...}.
OTOH, how can you select only 1 column, if you're selecting whole SampleDomain object (unless most of its properties are lazily-fetched)? That sounds unclear.
// And I believe you'll get more performance hits then selecting 20 fields for 1 record.
def whatYouWant = sampleDomainInst[0]
unless I'm missing something.