I am using Java, and I want to get the property" name "of each vertex of the shortest path between #26:1 and #24.0 . I am using the sql command select dijkstra (#26:1,#24.0,"distance") from V. And I get the result OResultSet. I dont know how the get the rid of each vertex in my java program (I mean OVertex or ORID of each vertex : objects offered by orientdb in my java program) .
try to do it with following code:
String query3 = "SELECT dijkstra (#26:1, #28:1, 'valeur') FROM V";
OResultSet rs3 = db.query(query3);
while(rs3.hasNext()) {
OResult row = rs3.next();
String rid= row.getProperty("#rid");
}
rs3.close();
For more information, you can look for in official java-api
I hope it will help you!
Related
Using networkx 2.0 I try to dynamically add an additional edge attribute by looping through all the edges. The graph is a MultiDiGraph.
According to the tutorial it seems to be possible to add edge attributes the way I do in the code below:
g = nx.read_gpickle("../pickles/" + gname)
yearmonth = gname[:7]
g.name = yearmonth # works
for source, target in g.edges():
g[source][target]['yearmonth'] = yearmonth
This code throws the following error:
TypeError: 'AtlasView' object does not support item assignment
What am I doing wrong?
That should happen if your graph is a nx.MultiGraph. From which case you need an extra index going from 0 to n where n is the number of edges between the two nodes.
Try:
for source, target in g.edges():
g[source][target][0]['yearmonth'] = yearmonth
The tutorial example is intended for a nx.Graph.
I'm looking to find the most liked nodes so basically the degree centrality recipe. This query kind of works but I'd like to return the full vertex (including properties) rather than just the id's.
( I am using Tinkerpop 3.0.1-incubating )
g.V()
.where( inE("likes") )
.group()
.by()
.by( inE("likes").count() )
Result
{
"8240": [
2
],
"8280": [
1
],
"12376": [
1
],
"24704": [
1
],
"40976": [
1
]
}
You're probably looking for the order step, using an anonymous traversal passed to the by() modulator:
g.V().order().by(inE('likes').count(), decr)
Note: this will require iterating over all vertices in Titan v1.0.0 and this query cannot be optimized, it will only work over smaller graphs in OLTP.
To get the 10 most liked:
g.V().order().by(inE('likes').count(), decr).limit(10)
If you want to get the full properties, simply chain .valueMap() or .valueMap(true) (for id and label) on the query.
See also:
http://tinkerpop.apache.org/docs/3.0.1-incubating/#order-step
https://groups.google.com/d/topic/gremlin-users/rt3qRKyAqts/discussion
GraphSON, as it is JSON based, does not support the conversion of complex objects to keys. A key in JSON must be string based and, as in this case, cannot be a map. To deal with this JSON limitation, GraphSON converts complex objects that are to be keys via the Java's toString() or by other methods for certain objects like graph elements (which returns a string representation of the element identifier, explaining why you received the output that you did).
If you want to return properties of elements while using GraphSON, you will have to figure out a workaround. In this specific case, you might do:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V().group().
......1> by(id).
......2> by(union(__(), outE('knows').count()).fold())
==>[1:[v[1],2],2:[0,v[2]],3:[v[3],0],4:[0,v[4]],5:[v[5],0],6:[0,v[6]]]
In this way you get the vertex identifier as the map id and in the value you get the full vertex plus the count. TinkerPop is working on improving this issue, but I don't expect a fast fix.
Currently i have two gremlin queries which will fetch two different values and i am populating in a map.
Scenario : A->B , A->C , A->D
My queries below,
graph.V().has(ID,A).out().label().toList()
Fetch the list of outE labels of A .
Result : List(B,C,D)
graph.traversal().V().has("ID",A).outE("interference").as("x").otherV().has("ID",B).select("x").values("value").headOption()
Given A and B , get the egde property value (A->B)
Return : 10
Is it possible that i can combine both there queries to get a return as Map[(B,10)(C,11)(D,12)]
I am facing some performance issue when i have two queries. Its taking more time
There is probably a better way to do this but I managed to get something with the following traversal:
gremlin> graph.traversal().V().has("ID","A").outE("interference").as("x").otherV().has("ID").label().as("y").select("x").by("value").as("z").select("y", "z").select(values);
==>[B,1]
==>[C,2]
I would wait for more answers though as I suspect there is a better traversal out there.
Below is working in scala
val b = StepLabel[Edge]()
val y = StepLabel[Label]()
val z = StepLabel[Integer]()
graph.traversal().V().has("ID",A).outE("interference").as(b)
.otherV().label().as(y)
.select(b).values("name").as(z)
.select((y,z)).toMap[String,Integer]
This will return Map[String,Int]
I need to get all list of vertices label of all outgoing egdes from a vertex using scala gremlin.
My code looks like below,
val names :ListBuffer[String] = ListBuffer()
val toList: List[Vertex] = graph.V().hasLabel(100).outE().outV().toList()
for(vertex <- toList){
names += vertex.label()
}
Its returning the same label name for all vertex
Eg :
Vertex A is having outE to B,C,D . It returns the label of A.
Output:
ListBuffer(100, 100, 100)
Anything am i missing?
I believe you asking for the wrong vertex in the end. Honestly, I often make the same mistake. Maybe this is the traversal you looking for:
graph.V().hasLabel(100).outE().inV().label().toList()
If you like me and often get confused by inV() and outV() you can use otherV which gets the opposite vertex. Like so:
graph.V().hasLabel(100).outE().otherV().label().toList()
Finally you can even shorten your traversal by not explicitly stating the edge part:
graph.V().hasLabel(100).out().label().toList()
By using out() instead of outE() you don't need to specify you want the vertex, out() gets you the vertex directly.
I'm interested in creating an edge (u,v) between two nodes of the same class in a graph if they share the same day of year and v.year = u.year+1.
Say I have vertices.csv:
id,date
A,2014-01-02
B,2015-01-02
C,2016-01-02
D,2013-06-01
E,2014-06-01
F,2016-06-01
The edge structure I'd like to see would be this:
A --> B --> C
D --> E
F
Let's set the vertex class to be "myVertex" and edge class to be "myEdge"? Is it possible to generate these edges using the SQL interface?
Based on this question, I started trying something like this:
BEGIN
LET source = SELECT FROM myVertex
LET target = SELECT from myVertex
LET edge = CREATE EDGE myEdge
FROM $source
TO (SELECT FROM $target WHERE $source.date.format('MM-dd') = $target.date.format('MM-dd')
AND $source.date.format('yyyy').asInteger() = $target.date.format('yyyy').asInteger()-1)
COMMIT
Unfortunately, this isn't correct. So I got less ambitious and wanted to see if I can create edges just based on matching day-of-year:
BEGIN
LET source = SELECT FROM myVertex
LET target = SELECT from myVertex
LET edge = CREATE EDGE myEdge FROM $source TO (SELECT FROM $target WHERE $source.date.format('MM-dd') = $target.date.format('MM-dd'))
COMMIT
Which still has errors... I'm sure it's something pretty simple to an experienced OrientDB user.
I thought about putting together a JavaScript function like Michela suggested on this question, but I'd prefer to stick to using the SQL commands as much as possible for now.
Help is greatly appreciated.
Other Stack Overflow References
How to print or log on function javascript OrientDB
I tried with OSQL batch but I think that you can't get what you want.
With whis OSQL batch
begin
let a = select #rid, $a1 as toAdd from test let $a1 = (select from test where date.format("MM") == $parent.$current.date.format("MM") and date.format("dd") == $parent.$current.date.format("dd") and #rid<>$parent.$current.#rid and date.format("yyyy") == sum($parent.$current.date.format("yyyy").asInteger(),1))
commit
return $a
I got this
but the problem is that when you create the edge you can not cycle on the table obtained in the previous step.
I think the best solution is to use an JS server-side function.
Hope it helps.