Not sure if its the best one, but one possible strategy would be to define a comparator with a constructor that takes an argument of your comparison node (A), and then implement the compareTo() function to retrieve the number of common properties for one node against the number of common properties for other node and return the difference. Then, you would simply iterate through all the nodes using graph.getAllNodes(), load them into an arraylist, sort it using your comparator, and navigate it in reverse. This doesn't leverage any features of the graph, really, but it does give you a solution. I'm curious to know if there's a "graphier" way of doing this.
On Thu, Feb 24, 2011 at 12:36 AM, Agam Dua <[email protected]> wrote: > Hey > > I'm a graph database and Neo4j newbie and I'm in a bit of a fix: > > *Problem Description* > Let's say I have 'n' nodes in the graph, representing the same type of > object. They have certain undirected links between them. > Now each of these 'n' nodes has the same 10 properties, the *values* of > which may differ. > > *Problem Statement* > Take starting node A. I need to find a way to traverse all the nodes of the > graph and print out which nodes have the most properties in common with A. > For example, if A, C, D, E, F, G have 'x' properties in common I want to > print the nodes. > Then, I want to print the nodes which have 'x-1' properties with the same > value. Then 'x-2', and so on. > > *Question* > Now my question is, is this possible? If so, what would be the best way to > go about it? > > Thanks in advance! > Agam. > * > * > _______________________________________________ > Neo4j mailing list > [email protected] > https://lists.neo4j.org/mailman/listinfo/user > _______________________________________________ Neo4j mailing list [email protected] https://lists.neo4j.org/mailman/listinfo/user
