Sorted. I don't need the first check - "if
(!callingPartyNode.hasRelationship(RelTypes.CALLS,
Direction.OUTGOING))" and the wrapper class for each relationship
object.
The 2nd check with the boolean variable works fine.
Arijit
On 21 July 2010 17:13, Arijit Mukherjee <ariji...@gmail.com> wrote:
> Ran into another problem this time -
>
> I'm trying to populate a graph with about 3300 links. I've a CSV file
> as input which has approx 3300 records - of the form (A, B, n) -
> specifying a link from A to B, with a weight n. I am reading each line
> of the file, and creating each node (A and B). If the nodes are new, a
> new link is created. If both A and B exist, the link between them is
> updated with the new weight. Now here is the problem. Say I have a
> dataset as follows -
>
> A, B, 3
> C, D, 5
> D, E, 2
> E, A, 7
>
> My algorithm is something like this -
>
> check whether node1 already exists or not, create if it doesn't exist;
> check whether node2 already exists or not, create if it doesn't exist;
> if (either node1 or node2 or both are new) {
> create a new relationship between them;
> }
> if (both node1 and node2 exist) {
> if (! node1.hasRelationship(RelTypes.KNOWS, Direction.OUTGOING) {
> // there is no link between node1 and node2 because node1 has
> no outgoing links
> create a new relationship;
> } else {
> // node1 has outgoing links, but none of them may link with node2
> check whether there is any link from node1 to node2;
> update if found, else create new;
> }
> }
>
> The trouble seems to occur because of the "hasRelationship" statement.
> The data I have has many cases where node1 and node2 have been created
> before separately - which means, in many cases, this "hasRelationship"
> check is done and a new link is created. It's working ok with small
> dataset, but for a dataset with many such cases (like the last line in
> the sample data: E, A, 7) I am getting an OutOfMemory error with "GC
> overhead limit exceeded".
>
> This is the code -
>
> public CallInfo createOrUpdateCallInfo(final Subscriber callingParty,
> final Subscriber calledParty,
> final long count,
> final long duration) {
> CallInfo callInfo = null;
> if (callingParty == null) {
> throw new IllegalArgumentException("Null CallingParty");
> }
> if (calledParty == null) {
> throw new IllegalArgumentException("Null CalledParty");
> }
> final Node callingPartyNode = ((SubscriberImpl)
> callingParty).getUnderlyingNode();
> final Node calledPartyNode = ((SubscriberImpl)
> calledParty).getUnderlyingNode();
> if (!callingPartyNode.hasRelationship(RelTypes.CALLS,
> Direction.OUTGOING)) {
> //no outgoing relationships - bound to be a new one
> final Relationship rel =
> callingPartyNode.createRelationshipTo(calledPartyNode,
> RelTypes.CALLS);
> callInfo = new CallInfoImpl(rel);
> callInfo.setCount(count);
> callInfo.setDuration(duration);
> } else {
> // could be an update or a create
> // get all the outgoing relationships and check if there
> is one which
> // ends with the calledParty node
> final Iterable<Relationship> currentRels =
> callingPartyNode.getRelationships(RelTypes.CALLS, Direction.OUTGOING);
> for (Relationship rel : currentRels) {
> final boolean found = rel.getEndNode().equals(calledPartyNode);
> if (found) {
> // update the existing relationship
> callInfo = new CallInfoImpl(rel);
> callInfo.updateCount(count);
> callInfo.updateDuration(duration);
> break;
> } else {
> final Relationship newRel =
> callingPartyNode.createRelationshipTo(calledPartyNode,
> RelTypes.CALLS);
> callInfo = new CallInfoImpl(newRel);
> callInfo.setCount(count);
> callInfo.setDuration(duration);
> }
> }
> }
> return callInfo;
> }
>
> I've tried to allocate 3GB to this application, but that didn't help.
> It crawls for 10-15 mins, then throws the OutOfMemory error...
>
> Can anyone give me some pointers please?
>
> Regards
> Arijit
>
> --
> "And when the night is cloudy,
> There is still a light that shines on me,
> Shine on until tomorrow, let it be."
>
--
"And when the night is cloudy,
There is still a light that shines on me,
Shine on until tomorrow, let it be."
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