change gremlin query titangraph
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Mustafa Tinwala
Mar 31, 2017, 6:52:35 AM3/31/17
to Gremlin-users
first of all i have created a binary tree using titan graph having
code like this of 10000 users
{
TitanGraph graph = TitanFactory.open(conf);
System.out.println("titan graph open");
System.out.println("time= "+ (System.currentTimeMillis()-q));
graph.tx().rollback();
TitanManagement mgmt = graph.openManagement();
EdgeLabel childs = mgmt.makeEdgeLabel("childs").multiplicity(Multiplicity.MULTI).make();
PropertyKey nodeId = mgmt.makePropertyKey("NodeId").dataType(String.class).make();
mgmt.buildIndex("byNodeId", Vertex.class).addKey(nodeId).unique().buildCompositeIndex();
mgmt.commit();
System.out.println(graph);
GraphTraversalSource g = graph.traversal();
System.out.println(g);
System.out.println("vertex creation started");
int n = 10000;
for (int i = 0; i < n; i++)
{
System.out.println(i);
graph.addVertex("NodeId", i);
graph.tx().commit();
}
int k = (n / 2);
for (int j = 0; j < k; j++)
{
Vertex v = g.V().has("NodeId", j).next();
System.out.println(v);
v.addEdge("childs", g.V().has("NodeId", (j * 2 + 1)).next());
v.addEdge("childs", g.V().has("NodeId", (j * 2 + 2)).next());
graph.tx().commit();
}
System.out.println("Graph is created.....");
System.out.println("graph is " + graph);
Vertex fromNode = g.V().has("NodeId", 0).next();
System.out.println("Node for calculating its child nodes : " + fromNode.property("NodeId"));
long st = System.currentTimeMillis();
System.out.println(g.V(fromNode).repeat(out("childs").dedup()).emit().count().next());
//System.out.println(g.V(fromNode).repeat(out("childs")).emit().until(out("childs").count().is(0)).count().next());
System.out.println("time : " + (System.currentTimeMillis() - st) + " milliseconds");
graph.close();
System.exit(0);
}
Now there is a code to retrieve only children node in dynamodb so here is the code with query i have to just change in query highligted with bold letter to decrease for througput
TitanGraph graph = TitanFactory.open(conf);
System.out.println("titan graph open");
System.out.println("time= "+ (System.currentTimeMillis()-q));
graph.tx().rollback();
System.out.println(graph);
GraphTraversalSource g = graph.traversal();
System.out.println(g);
Vertex fromNode = g.V().has("NodeId", 0).next();
System.out.println("Node for calculating its child nodes : " + fromNode.property("NodeId"));
System.out.println(g.V(fromNode).repeat(out("childs").dedup()).emit().count().next());
need some help
thank you in advcance
code like this of 10000 users
{
TitanGraph graph = TitanFactory.open(conf);
System.out.println("titan graph open");
System.out.println("time= "+ (System.currentTimeMillis()-q));
graph.tx().rollback();
TitanManagement mgmt = graph.openManagement();
EdgeLabel childs = mgmt.makeEdgeLabel("childs").multiplicity(Multiplicity.MULTI).make();
PropertyKey nodeId = mgmt.makePropertyKey("NodeId").dataType(String.class).make();
mgmt.buildIndex("byNodeId", Vertex.class).addKey(nodeId).unique().buildCompositeIndex();
mgmt.commit();
System.out.println(graph);
GraphTraversalSource g = graph.traversal();
System.out.println(g);
System.out.println("vertex creation started");
int n = 10000;
for (int i = 0; i < n; i++)
{
System.out.println(i);
graph.addVertex("NodeId", i);
graph.tx().commit();
}
int k = (n / 2);
for (int j = 0; j < k; j++)
{
Vertex v = g.V().has("NodeId", j).next();
System.out.println(v);
v.addEdge("childs", g.V().has("NodeId", (j * 2 + 1)).next());
v.addEdge("childs", g.V().has("NodeId", (j * 2 + 2)).next());
graph.tx().commit();
}
System.out.println("Graph is created.....");
System.out.println("graph is " + graph);
Vertex fromNode = g.V().has("NodeId", 0).next();
System.out.println("Node for calculating its child nodes : " + fromNode.property("NodeId"));
long st = System.currentTimeMillis();
System.out.println(g.V(fromNode).repeat(out("childs").dedup()).emit().count().next());
//System.out.println(g.V(fromNode).repeat(out("childs")).emit().until(out("childs").count().is(0)).count().next());
System.out.println("time : " + (System.currentTimeMillis() - st) + " milliseconds");
graph.close();
System.exit(0);
}
Now there is a code to retrieve only children node in dynamodb so here is the code with query i have to just change in query highligted with bold letter to decrease for througput
TitanGraph graph = TitanFactory.open(conf);
System.out.println("titan graph open");
System.out.println("time= "+ (System.currentTimeMillis()-q));
graph.tx().rollback();
System.out.println(graph);
GraphTraversalSource g = graph.traversal();
System.out.println(g);
Vertex fromNode = g.V().has("NodeId", 0).next();
System.out.println("Node for calculating its child nodes : " + fromNode.property("NodeId"));
System.out.println(g.V(fromNode).repeat(out("childs").dedup()).emit().count().next());
need some help
thank you in advcance
HadoopMarc
Mar 31, 2017, 3:08:09 PM3/31/17
to Gremlin-users
Hi Mustafa,
I see two possible problems in your code:
1. g.V() takes id's as argument, so I would expect:
Int fromNodeId = g.V().has("NodeId", 0).id().next();
2. Infinite loops with repeat() are allowed, but then you must be sure that your traversal does not hit two vertices with mutual out links. Adding a times(10) step won't hurt, just for safety.
HTH, Marc
Op vrijdag 31 maart 2017 12:52:35 UTC+2 schreef Mustafa Tinwala:
I see two possible problems in your code:
1. g.V() takes id's as argument, so I would expect:
Int fromNodeId = g.V().has("NodeId", 0).id().next();
2. Infinite loops with repeat() are allowed, but then you must be sure that your traversal does not hit two vertices with mutual out links. Adding a times(10) step won't hurt, just for safety.
HTH, Marc
Op vrijdag 31 maart 2017 12:52:35 UTC+2 schreef Mustafa Tinwala:
Mustafa Tinwala
Apr 1, 2017, 3:39:02 AM4/1/17
to Gremlin-users
Retrieval code:--
TitanGraph graph = TitanFactory.open(conf);
System.out.println("titan graph open");
System.out.println("time= "+ (System.currentTimeMillis()-q));
graph.tx().rollback();
System.out.println(graph);
GraphTraversalSource g = graph.traversal();
System.out.println(g);
Vertex fromNode = g.V().has("NodeId", 0).next();
System.out.println("Node for calculating its child nodes : " + fromNode.property("NodeId"));
System.out.println(g.V(fromNode).repeat(out("childs").dedup()).emit().count().next());
need some help
thank you in advcance
int fromNodeId = (int) g.V().has("NodeId", 0).id().next();
if i run this code in place of
Vertex fromNode = g.V().has("NodeId", 0).next();
i'm getting the following error
Exception in thread "main" java.lang.ClassCastException: java.lang.Long cannot be cast to java.lang.Integer
at com.graph_retrieve.main(graph_retrieve.java:178)
Please help me how to sought it...!
Exception in thread "main" java.lang.ClassCastException: java.lang.Long cannot be cast to java.lang.Integer
at com.graph_retrieve.main(graph_retrieve.java:178)
Please help me how to sought it...!
Mustafa Tinwala
Apr 1, 2017, 4:36:50 AM4/1/17
to Gremlin-users
also i have another question
if there is n number of nodes means my binary graph is increasing so how to use times() ?
HadoopMarc
Apr 1, 2017, 7:17:49 AM4/1/17
to Gremlin-users
Hi Mustafa,
My error (I onlychecked in TinkerGraph which returns int id's). From the error message you that fromNodeId needs type Long for Titan.
Cheers, Marc
Op zaterdag 1 april 2017 09:39:02 UTC+2 schreef Mustafa Tinwala:
My error (I onlychecked in TinkerGraph which returns int id's). From the error message you that fromNodeId needs type Long for Titan.
Cheers, Marc
Op zaterdag 1 april 2017 09:39:02 UTC+2 schreef Mustafa Tinwala:
HadoopMarc
Apr 1, 2017, 7:25:27 AM4/1/17
to Gremlin-users
HI Mustafa,
times() is an optional part of the repeat() step:
http://tinkerpop.apache.org/docs/current/reference/#repeat-step
So, you can protect your query against unwanted infinite loops by writing:
System.out.println(g.V(fromNodeId).repeat(out("childs").dedup()).times(10).emit().count().next());
Cheers, Marc
Op zaterdag 1 april 2017 10:36:50 UTC+2 schreef Mustafa Tinwala:
times() is an optional part of the repeat() step:
http://tinkerpop.apache.org/docs/current/reference/#repeat-step
So, you can protect your query against unwanted infinite loops by writing:
System.out.println(g.V(fromNodeId).repeat(out("childs").dedup()).times(10).emit().count().next());
Cheers, Marc
Op zaterdag 1 april 2017 10:36:50 UTC+2 schreef Mustafa Tinwala:
Mustafa Tinwala
Apr 1, 2017, 8:11:47 AM4/1/17
to Gremlin-users
if there is n number of nodes means my binary graph is increasing so how to use times() ?.
Surely i will check it
btw thanks a lot for this..! :)
Mustafa Tinwala
Apr 1, 2017, 8:40:12 AM4/1/17
to Gremlin-users
Actually i was telling i have used this query in gremlin console
like this
clockWithResult{ g.V(fromNode).repeat(__.out("childs").dedup()).times(13).emit().count().next()}
==>31.84179384
==>9999
and it gives child vertex count =9999 in 31 sec.
Now my question again is how to decrease time
my system configuration for your information is
pentium g2020
ram = 5.5 gb
o.s = ubuntu 16.04
HadoopMarc
Apr 1, 2017, 10:38:24 AM4/1/17
to Gremlin-users
Hi Mustafa,
I still miss the fromNodeId instead of the fromNode. The results look like Titan interpreted your query as:
clockWithResult{ g.V().repeat(__.out("childs").dedup()).times(13).emit().count().next()}
Still, I do not expect to wait 30 secs for just 10.000 edges traversed once. What are the results for:
clockWithResult{ g.V(fromNodeId).repeat(__.out("childs").dedup()).times(13).emit().count().next()}
clockWithResult{ g.V(fromNodeId).out("childs").count().next()}
clockWithResult{ g.V().out("childs").count().next()}
Cheers, Marc
Op zaterdag 1 april 2017 14:40:12 UTC+2 schreef Mustafa Tinwala:
I still miss the fromNodeId instead of the fromNode. The results look like Titan interpreted your query as:
clockWithResult{ g.V().repeat(__.out("childs").dedup()).times(13).emit().count().next()}
Still, I do not expect to wait 30 secs for just 10.000 edges traversed once. What are the results for:
clockWithResult{ g.V(fromNodeId).repeat(__.out("childs").dedup()).times(13).emit().count().next()}
clockWithResult{ g.V(fromNodeId).out("childs").count().next()}
clockWithResult{ g.V().out("childs").count().next()}
Cheers, Marc
Op zaterdag 1 april 2017 14:40:12 UTC+2 schreef Mustafa Tinwala:
Mustafa Tinwala
Apr 3, 2017, 12:57:14 AM4/3/17
to Gremlin-users
As your post says fromNodeId
but how to declare in gremlin console i dont know
hey but yeah im still trying with this query
clockWithResult{ g.V(fromNode).repeat(out("childs")).times(12).emit().count().next()}
==>18.389357519999997
==>8190
Mustafa Tinwala
Apr 3, 2017, 1:03:24 AM4/3/17
to Gremlin-users
and yeah as you say i have tried like this also in java
long fromNodeId = (long) g.V().has("NodeId", 0).id().next();
but it shows me an error as follows
Exception in thread "main" java.lang.NullPointerException
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep.standardAlgorithm(RepeatStep.java:143)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ExpandableStepIterator.next(ExpandableStepIterator.java:54)
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep$RepeatEndStep.standardAlgorithm(RepeatStep.java:230)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep.standardAlgorithm(RepeatStep.java:143)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ExpandableStepIterator.hasNext(ExpandableStepIterator.java:44)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ReducingBarrierStep.processNextStart(ReducingBarrierStep.java:86)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.next(AbstractStep.java:126)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.next(AbstractStep.java:37)
at org.apache.tinkerpop.gremlin.process.traversal.util.DefaultTraversal.next(DefaultTraversal.java:127)
at com.graph_retrieve.main(graph_retrieve.java:182)
long fromNodeId = (long) g.V().has("NodeId", 0).id().next();
but it shows me an error as follows
Exception in thread "main" java.lang.NullPointerException
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep.standardAlgorithm(RepeatStep.java:143)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ExpandableStepIterator.next(ExpandableStepIterator.java:54)
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep$RepeatEndStep.standardAlgorithm(RepeatStep.java:230)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.branch.RepeatStep.standardAlgorithm(RepeatStep.java:143)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ComputerAwareStep.processNextStart(ComputerAwareStep.java:47)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.hasNext(AbstractStep.java:140)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ExpandableStepIterator.hasNext(ExpandableStepIterator.java:44)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.ReducingBarrierStep.processNextStart(ReducingBarrierStep.java:86)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.next(AbstractStep.java:126)
at org.apache.tinkerpop.gremlin.process.traversal.step.util.AbstractStep.next(AbstractStep.java:37)
at org.apache.tinkerpop.gremlin.process.traversal.util.DefaultTraversal.next(DefaultTraversal.java:127)
at com.graph_retrieve.main(graph_retrieve.java:182)
Mustafa Tinwala
Apr 3, 2017, 1:55:56 AM4/3/17
to Gremlin-users
hey btw your query tells
clockWithResult{ g.V().repeat(__.out("childs")).times(12).emit().count().next()}
tells
WARN com.thinkaurelius.titan.graphdb.transaction.StandardTitanTx - Query requires iterating over all vertices [()]. For better performance, use indexes
HadoopMarc
Apr 3, 2017, 3:59:50 AM4/3/17
to Gremlin-users
Hi Mustafa,
OK, you declared property "NodeId" as string, but you provide the addVertex method with an int 0, this is not consistent. I did not notice that before. If you correct that either way, g.V().has("NodeId", 0).id().next() or g.V().has("NodeId", "0").id().next() should return an id.
Titan is very generous with the index warning, it does not sense ones intent. For a g.V() query you just expect Titan not to use any index and the warning can be ignored.
Cheers, Marc
Op maandag 3 april 2017 07:55:56 UTC+2 schreef Mustafa Tinwala:
Op maandag 3 april 2017 07:55:56 UTC+2 schreef Mustafa Tinwala:
Jason Plurad
Apr 3, 2017, 9:19:18 AM4/3/17
to Gremlin-users
> clockWithResult{ g.V(fromNode).repeat(__.out("childs").dedup()).times(13).emit().count().next()}
> ==>31.84179384
> ==>9999
> ==>31.84179384
> ==>9999
> and it gives child vertex count =9999 in 31 sec.
You are incorrect on the units here. That is 31 milliseconds. For example:
gremlin> clockWithResult{ Thread.sleep(10) }
==>11.67529047
==>null
The method clockWithResults runs the operation 100 times if you don't explicitly specify a number of loops.
gremlin> clockWithResult(5, { Thread.sleep(10) })
==>11.5913306
==>null
Mustafa Tinwala
Apr 3, 2017, 1:23:54 PM4/3/17
to Gremlin-users
What you say might b right but Actually Jason it takes 31 second as I noticed
But I have a question for you
Is it fine that my system configuration takes this much time as I have to deploy it on aws ?
BTW thaks
Robert Dale
Apr 3, 2017, 1:45:14 PM4/3/17
to gremli...@googlegroups.com
Correct, you would have experienced 3 seconds of user time because 31ms average per query * 100 queries = 3.1s.
Robert Dale
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Robert Dale
Apr 3, 2017, 1:45:57 PM4/3/17
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Or even * 1000 queries = 31s
Robert Dale
On Mon, Apr 3, 2017 at 1:45 PM, Robert Dale <rob...@gmail.com> wrote:
Correct, you would have experienced 3 seconds of user time because 31ms average per query * 100 queries = 3.1s.
Robert Dale
Mustafa Tinwala
Apr 3, 2017, 11:28:40 PM4/3/17
to Gremlin-users
Thanks guys
But I have a problem.
What if my graph is increasing and how to set times(n)
How to use accurate times() so that the query run fast and efficiently
Thank a lot guys...! :)
Mustafa Tinwala
Apr 4, 2017, 12:33:49 AM4/4/17
to Gremlin-users
Yeah jason as you said it gives me like this
clockWithResult{ Thread.sleep(10) }
==>10.12367083
==>null
gremlin> clockWithResult(5, { Thread.sleep(10) })
==>10.1140054
==>null
so was i wrong ? in case of 31 seconds of time my query takes ... ! :)
Message has been deleted
Mustafa Tinwala
Apr 4, 2017, 1:39:26 AM4/4/17
to Gremlin-users
On Monday, 3 April 2017 23:15:57 UTC+5:30, Robert Dale wrote:
Or even * 1000 queries = 31s
Robert Dale
On Mon, Apr 3, 2017 at 1:45 PM, Robert Dale <rob...@gmail.com> wrote:
Correct, you would have experienced 3 seconds of user time because 31ms average per query * 100 queries = 3.1s.
Robert Dale
On Mon, Apr 3, 2017 at 1:23 PM, Mustafa Tinwala <mustafa.ti...@gmail.com> wrote:
What you say might b right but Actually Jason it takes 31 second as I noticed
But I have a question for you
Is it fine that my system configuration takes this much time as I have to deploy it on aws ?
BTW thaks
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Oh :) but i cant understand what is the actual time d query takes
but when i run this queruy it shows me like this
t = System.currentTimeMillis();
==>1491284142148
gremlin> g.V(fromNode).repeat(out("childs")).times(12).emit().count().next();
==>8190
gremlin> t = System.currentTimeMillis();
==>1491284171835
well thank you for letting me know this
:)
Message has been deleted
Message has been deleted
Stephen Mallette
Apr 4, 2017, 9:06:37 AM4/4/17
to Gremlin-users
Mustafa - this thread is becoming difficult to follow as a result of these many small, follow-up emails. Please give people on the list a more time to answer you and do a better job summarizing your updates into a single post rather than lots of play-by-play updates of your progress. Thank you.
On Tue, Apr 4, 2017 at 9:00 AM, Mustafa Tinwala <mustafa.ti...@gmail.com> wrote:
On Tuesday, 4 April 2017 11:38:26 UTC+5:30, Mustafa Tinwala wrote:
On Tuesday, 4 April 2017 11:09:26 UTC+5:30, Mustafa Tinwala wrote:
On Monday, 3 April 2017 23:15:57 UTC+5:30, Robert Dale wrote:Or even * 1000 queries = 31sRobert Dale
plz read the last 5 conversation of mine thank you :)
And hey friends i have found another link out of this may be this would be help
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Mustafa Tinwala
Apr 4, 2017, 9:09:09 AM4/4/17
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Yeah sure....! :)
Robert Dale
Apr 4, 2017, 10:40:58 AM4/4/17
to Gremlin-users
Mustafa,
I'm not sure if you ever posted your "clockWithResults" results.
What does clock give you?
clockWithResult{g.V(fromNode).repeat(out("childs")).times(12).emit().count().next()}
What does the profile look like?
g.V(fromNode).repeat(out("childs")).times(12).emit().count().profile()
Your connection to AWS and AWS configuration would probably be the most limiting factor here. You could try all of this locally in the console using either Titan or TinkerGraph to try to isolate query (local) processing from network (remote).
Outside of that, there is nothing fundamentally slow about the graph structure, query, or volume.
I've created a slightly modified script to run with TinkerGraph in the gremlin console - https://gist.github.com/robertdale/8b4e604165852a5344476af940b75e30
Note that you'll have to cut-n-paste the last few statements to see the results.
You'll have to isolate and experiment with different components of your system to determine where the bottleneck is. Let use know what you discover.
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Mustafa Tinwala
Apr 5, 2017, 1:15:12 AM4/5/17
to Gremlin-users
Robert Dale
As you say
g.V(fromNode).repeat(out("
childs")).times(12).emit().count().profile()
it gives me a picture like this
g.V(fromNode).repeat(out("childs")).times(12).emit().count().profile()
==>8190
it gives me a picture like this
g.V(fromNode).repeat(out("childs")).times(12).emit().count().profile()
==>8190
clockWithResult{g.V(fromNode).repeat(out("childs")).times(12).emit().count().next()}
==>39.25605984
==>8190
Now i want to ask is it milliseconds or seconds
:)
btw Nice discovery @ robert dale
Mustafa Tinwala
Apr 5, 2017, 10:28:43 PM4/5/17
to Gremlin-users
Is there someone who can help out ...!:)
Jason Plurad
Apr 6, 2017, 12:56:57 PM4/6/17
to Gremlin-users
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Mustafa Tinwala
Apr 11, 2017, 12:35:07 AM4/11/17
to Gremlin-users
Thanks a lot jason...!
:)
you have solve my doubt :)
but as i say it takes this way
gremlin> s=System.currentTimeMillis()
==>1491540194192
gremlin> clockWithResult{g.V(fromNode).repeat(out("childs")).times(12).emit().count().next()}
:)
you have solve my doubt :)
but as i say it takes this way
gremlin> s=System.currentTimeMillis()
==>1491540194192
gremlin> clockWithResult{g.V(fromNode).repeat(out("childs")).times(12).emit().count().next()}
==>142.589291250000002
==>8190
gremlin> (System.currentTimeMillis()-s)/1000
==>38.075
How ...?
So how to tackle this stuation but wait yeah
i know this that clockWithResult state the mean time between the iteration.
so Now this is my problem
can anyone knows the solution for clockWithResult how to
Can anyone help me for this...! so that i can run my query fast with no iteration or mean time added
btw thank you guys ....! : )
==>8190
gremlin> (System.currentTimeMillis()-s)/1000
==>38.075
How ...?
So how to tackle this stuation but wait yeah
i know this that clockWithResult state the mean time between the iteration.
so Now this is my problem
can anyone knows the solution for clockWithResult how to
Can anyone help me for this...! so that i can run my query fast with no iteration or mean time added
btw thank you guys ....! : )
i will be happy if you have any answer :)
Stephen Mallette
Apr 11, 2017, 6:17:37 AM4/11/17
to Gremlin-users
TinkerPop's clockWithResults works how it works. If you need your own implementation that doesn't do iterations or doesn't calculate the mean the way that we do, just write your won function.
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