Traverse the graph returning all nodes with at least 1 matching property.
For each node returned add up the matching properties to node A and
put that in a list.
Sort the list.


On Thu, Feb 24, 2011 at 1:49 AM, Cedric Hurst <[email protected]> wrote:
> 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.
>> *
>> *
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>> [email protected]
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>>
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