I am unfamiliar with Docker but a cursory glance at the project suggests
that it may impose environmental conditions that were not anticipated when
I wrote py2neo. I would be happy to review a pull request with any fixes
you are able to provide for this setup.
Nigel
On 23 March 2014 05:43, Alan
Nobody can help me?
--
You received this message because you are subscribed to the Google Groups
Neo4j group.
To unsubscribe from this group and stop receiving emails from it, send an email
to neo4j+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.
You can use findAllPaths with dijkstra too, and then just pull the first 3
paths from the iterator.
Or why do you want to use shortestPath instead? Dijsktra is calculating the
shortest path based on your costProperty.
Am 21.03.2014 um 17:26 schrieb Antonio Grimaldi
Happy I could help.
Please consider blogging about it after you worked out your solution. And don't
hesitate to come back with other questions.
Cheers,
Michael
Am 20.03.2014 um 11:36 schrieb costas.bar...@gmail.com:
Dear Michael,
Thank you for your detailed response and the multiple
Hi, i wonder how to solve the following using cypher
1) using a match i find all paths for a certain relstype (REL1) between all
nodes of certain labeltypes (LABEL1 and LABEL2).. for example
match rel1paths=(n:LABEL1)-[:REL1*]-(m:LABEL2)
lets assume this gives me in total 4 paths
2) now i
Hi Michael,
thanks for your answer.
I tried to use WeightedPath path = dijkstraPath.findAllPaths(startNode,
endNode);
But always one path is returned, except when there are two or more paths
with the same cost.
Instead i would like to found the firts 3 alternative path, with different
cost.
as an idea , not tested..
match p=(n)-[:REL1*]-(m)
where NOT n-[:REL2*]-m
return n,m
or return p
--
You received this message because you are subscribed to the Google Groups
Neo4j group.
To unsubscribe from this group and stop receiving emails from it, send an email
to
hi thx, that one works (just tested it) :
match p=(n:label1)-[:REL1*]-(m:label2)
where NOT (n)-[:REL2*]-(m)
return p
any other syntax ideas ?
ps .. i was thing far more complex withdual match ... with to pass data
to the secondary match ... and some functions to compare the 2 sets ...
hard to say without any idea about the far more complex stuff you're trying
to achieve, and the comparisons you'd like to do ..
--
You received this message because you are subscribed to the Google Groups
Neo4j group.
To unsubscribe from this group and stop receiving emails from it, send an
hi, sure.. actually i would like to build a kind of report that reflects
the amount of paths found for a specific relstype (REL1) between each 2
nodes (actully this means duplicate paths if more than 1 found) , their
length, the shortest based on weight, and whether there is at least also
one
I agree with Tom, but if you really want to do it:
You could do:
match p=(n:label1)-[:REL1*]-(m:label2)
with collect([n,m]) as start_end
match p=(n:label1)-[:REL2*]-(m:label2)
where [n,m] not in start_end
return p
if you want to compare collections
match p=(n:label1)-[:REL1*]-(m:label2)
11 matches
Mail list logo
