hi Michael/Peter
Peter: yes, that about sums it up. Webadmin console is fine, but REST not so
much.
Michael: my final set is a little over 100K nodes, and while I am still
playing around as part of a technical proof of concept, I think I will need
this data more or less in its entirety.
I
Hi there,
I wanted to look into more efficient ways to send back big responses,
but had not time. Basically, throwing in more memory is what you have
to do right now, or get Cypher or Gremlin to return more effective
representations like only the properties you need, which cuts down the
amount of
hi again
I've been trying that query out (slightly modified now to return all nodes
of a different kind, but attached to a node in the first set), on a data set
of around 100K, and I'm getting an OutOfMemoryError:
{
message : GC overhead limit exceeded,
exception : java.lang.OutOfMemoryError:
Hi,
I've been trying that query out (slightly modified now to return all nodes
of a different kind, but attached to a node in the first set), on a data set
of around 100K, and I'm getting an OutOfMemoryError:
{
message : GC overhead limit exceeded,
exception : java.lang.OutOfMemoryError:
hi, thanks for the quick reply. I need to add some more info (some I left
out, some I discovered later)
This only happens when I use that gremlin over REST.
Executing the gremlin in the webadmin console is pretty quick, and I don't
see any memory spikes or anything.
What I did (after posting the
furthermore, I tried a simple query, that returns about the same number of
rows (still around 100K), and increased the timeout of my REST call (to
several minutes) - and got my OutOfMemoryError again :)
latest query is just g.v(293).in('R_Bought'). Running it through the
webadmin console is
Hi Kevin,
That is unfortunate. Peter is the man to track down such things in Neo4j REST.
Hopefully it's something obvious.
Glad the REPL and WebAdmin is speedy.
Good luck,
Marko.
http://markorodriguez.com
On Nov 10, 2011, at 10:42 PM, Kevin Versfeld kevin.versf...@gmail.com wrote:
Kevin,
So the webadmin console is displaying the 100k nodes ok, men but rest
barfs?
On Nov 11, 2011 6:42 AM, Kevin Versfeld kevin.versf...@gmail.com wrote:
furthermore, I tried a simple query, that returns about the same number of
rows (still around 100K), and increased the timeout of my REST
Kevin,
how large is the dataset that is build up?
I'm not sure but the REST Plugin and the Webadmin use different implementations
where webadmin streams the results with ajax and the REST plugin builds up a
full string result.
Do you need all of the data? Or would it be possible to page it?
Perfect, thanks! (Seems so obvious now... *sigh*)
At least I now have the easier query done, and can get on to the tricky
ones *nervous*
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hi again
I suspect I'm missing something small, but when I run the following in the
webadmin console, I don't get any output?
m = [:]; g.v(162).in('R_PartOf').loop(1){m.put(it.object, it.loops);
true}.cap.next().sort{a, b - a.value = b.value}.keySet []
(m is empty too...)
Thanks for the
hi again
I've played around a bit more, and this is the query that currently works
best for me... would anyone mind reviewing it for me please? Feels a
little like I'm getting the right answer, but the wrong way
gremlin m = [:];
gremlin g.v(162).in('R_PartOf').loop(1){m.put(it.object,
Hey,
I've played around a bit more, and this is the query that currently works
best for me... would anyone mind reviewing it for me please? Feels a
little like I'm getting the right answer, but the wrong way
gremlin m = [:];
gremlin g.v(162).in('R_PartOf').loop(1){m.put(it.object,
Great! Thanks Marko, this has been very helpful.
I did realise my mistake on the sort: when I tried it the way you suggested
originally, I also had a few other things wrong, and obviously did not
apply a methodical approach to solving it.
One last question: I need to just return the nodes, at the
Marko,
And I'm saying nasty with a South African accent so you know its dirty.
Thanks to working with Romiko, I'm getting an eerily clear mental image of that.
-- Tatham
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Hi,
I would do it like this:
m = [:]
g.v(162).in('R_PartOf').loop(1){m.put(it.object, it.loops); true} -1
m.sort{a, b - a.value = b.value}.keySet as List
In short, fill up a Map (m) with key being the vertex and value being the
number of hops (or times through the loop). Then sort the map by
Can you do it on one line? ;)
Great stuff.
Cheers,
/peter neubauer
GTalk: neubauer.peter
Skype peter.neubauer
Phone +46 704 106975
LinkedIn http://www.linkedin.com/in/neubauer
Twitter http://twitter.com/peterneubauer
http://www.neo4j.org - NOSQL for the
Sure:
m = [:]; g.v(162).in('R_PartOf').loop(1){m.put(it.object, it.loops);
true}.cap.next().sort{a, b - a.value = b.value}.keySet []
Marko.
http://markorodriguez.com
On Nov 4, 2011, at 12:17 PM, Peter Neubauer wrote:
Can you do it on one line? ;)
Great stuff.
Cheers,
/peter
Thanks! I will still need to try this out, but the idea (in your first
response) kind of feels more or less like what I knew I needed. I have so
many questions now based on the responses so far - it feels like I'm just
scratching the surface!
Firstly, could you explain the differences between map,
Hi,
Thanks! I will still need to try this out, but the idea (in your first
response) kind of feels more or less like what I knew I needed. I have so
many questions now based on the responses so far - it feels like I'm just
scratching the surface!
Once you get it, you can get nasty with
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