Re: [Neo4j] Node.hasRelationship question
2010/10/28 Walaa Eldin Moustafa wa.moust...@gmail.com Hi Mattias, Even if I pass nulls, I will still have to build an index on some key (property). I do not want to do that because it adds the property index overhead for something that I am not actually using. I think it's definately worth that tiny extra overhead since lookups are very fast (no looping at all, but instead direct index lookups). By the way, does the underlying code for index.get(kay, value, start, end) iterate over all the relationships of start and checks if one is connected to end, or it does perform a direct look up to get the edges between (start, end)? No looping, just a direct lookup. Maybe RelationshipIndex should have a method: add( Relationship relationship ); which indexes that relationship (with its start/end node as usual) but w/o any additional property. Would that be a good idea? Thanks. On Wed, Oct 27, 2010 at 4:50 PM, Mattias Persson matt...@neotechnology.com wrote: 2010/10/27, Walaa Eldin Moustafa wa.moust...@gmail.com: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. you can actually leave out key/value (pass in null) if you pass in start/end node, or you can index the relationship type as a key/value pair for each relationship and then do: get( type, knows, person1, person2 ); would that help? ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
I already used the regular index to lookup for existing relationships as workaround. it's definitely faster then looping specially when database is growing. it will be really useful to have an index for relationships. -- From: Mattias Persson matt...@neotechnology.com Sent: Thursday, October 28, 2010 10:48 AM To: Neo4j user discussions user@lists.neo4j.org Subject: Re: [Neo4j] Node.hasRelationship question 2010/10/28 Walaa Eldin Moustafa wa.moust...@gmail.com Hi Mattias, Even if I pass nulls, I will still have to build an index on some key (property). I do not want to do that because it adds the property index overhead for something that I am not actually using. I think it's definately worth that tiny extra overhead since lookups are very fast (no looping at all, but instead direct index lookups). By the way, does the underlying code for index.get(kay, value, start, end) iterate over all the relationships of start and checks if one is connected to end, or it does perform a direct look up to get the edges between (start, end)? No looping, just a direct lookup. Maybe RelationshipIndex should have a method: add( Relationship relationship ); which indexes that relationship (with its start/end node as usual) but w/o any additional property. Would that be a good idea? Thanks. On Wed, Oct 27, 2010 at 4:50 PM, Mattias Persson matt...@neotechnology.com wrote: 2010/10/27, Walaa Eldin Moustafa wa.moust...@gmail.com: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. you can actually leave out key/value (pass in null) if you pass in start/end node, or you can index the relationship type as a key/value pair for each relationship and then do: get( type, knows, person1, person2 ); would that help? ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
By the way, does the underlying code for index.get(kay, value, start, end) iterate over all the relationships of start and checks if one is connected to end, or it does perform a direct look up to get the edges between (start, end)? No looping, just a direct lookup. Maybe RelationshipIndex should have a method: add( Relationship relationship ); which indexes that relationship (with its start/end node as usual) but w/o any additional property. Would that be a good idea? Yes, definitely that is a very good idea. ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
Hi, You could try Gremlin -- http://gremlin.tinkerpop.com ... g:count($n1/ou...@label='t']/i...@id=$n2_id) 0 It will iterate over all out edges of n1, but, you will find more flexibility in determining if particular paths exist. That is, if you plan more complicated expressions. If you are doing this from within Java, see: http://github.com/tinkerpop/gremlin/wiki/Using-Gremlin-through-Java Good luck, Marko. http://markorodriguez.com On Oct 27, 2010, at 2:09 PM, Walaa Eldin Moustafa wrote: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
2010/10/27, Walaa Eldin Moustafa wa.moust...@gmail.com: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. you can actually leave out key/value (pass in null) if you pass in start/end node, or you can index the relationship type as a key/value pair for each relationship and then do: get( type, knows, person1, person2 ); would that help? ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
Thanks Marko. I was actually referring to something more efficient than iterating over all the nodes. Gremlin is something that I definitely want to try, but not for that project. On Wed, Oct 27, 2010 at 4:16 PM, Marko Rodriguez okramma...@gmail.com wrote: Hi, You could try Gremlin -- http://gremlin.tinkerpop.com ... g:count($n1/ou...@label='t']/i...@id=$n2_id) 0 It will iterate over all out edges of n1, but, you will find more flexibility in determining if particular paths exist. That is, if you plan more complicated expressions. If you are doing this from within Java, see: http://github.com/tinkerpop/gremlin/wiki/Using-Gremlin-through-Java Good luck, Marko. http://markorodriguez.com On Oct 27, 2010, at 2:09 PM, Walaa Eldin Moustafa wrote: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] Node.hasRelationship question
Hi Mattias, Even if I pass nulls, I will still have to build an index on some key (property). I do not want to do that because it adds the property index overhead for something that I am not actually using. By the way, does the underlying code for index.get(kay, value, start, end) iterate over all the relationships of start and checks if one is connected to end, or it does perform a direct look up to get the edges between (start, end)? Thanks. On Wed, Oct 27, 2010 at 4:50 PM, Mattias Persson matt...@neotechnology.com wrote: 2010/10/27, Walaa Eldin Moustafa wa.moust...@gmail.com: I am looking for a method that is if given a node n1, can answer the query if n1 has a relationship of type t with node n2, something along the lines of: boolean b = n1.hasRelatioship(n2,t); I know we can answer this question by iterating over n1's getRelationships() and testing them. However, I was looking for something more efficient that does not have to iterate over the node's relationships, especially that I noticed that the RelationshipIndex interface has a get() method that does something like that; however we have to supply it with a key and a value for an attribute for the edge between the two nodes. So I was looking for something like this method, but without having to pass a key/value pair. you can actually leave out key/value (pass in null) if you pass in start/end node, or you can index the relationship type as a key/value pair for each relationship and then do: get( type, knows, person1, person2 ); would that help? ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user -- Mattias Persson, [matt...@neotechnology.com] Hacker, Neo Technology www.neotechnology.com ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
