I am trying to apply filter on my Neo4j Graph DB which has 733922 nodes and 303378 relationships, the DB size is 913.56 MiB. I want to fetch all nodes which has related labels from a specific set, it works if I give upto three related nodes but takes indefinite time to process queries beyond that. If I remove order by then it works for upto five related labels. Is this the most optimised query or am I doing something wrong here? I have attached the PROFILE output for one related label.
MATCH p=(node1:label1{value:'a2cb2c487d9da6941f0f9c1692ed1a5a', source: 'a0b116e2-d9f5-40d4-97ff-6ca6f2ec6c9b'})-[r]-> (:a),(:b),(:c),(:d),(:e),(:f),(:g),(:h),(:i),(:j)
RETURN node1,r
ORDER BY node1.created DESC
LIMIT 25
The following
(:b),(:c),(:d),(:e),(:f),(:g),(:h),(:i),(:j)
creates a Cartesian product between all nodes with labels b,c,d etc. which will quickly grow.
If you want any of the labels a..b use following:
MATCH p=(node1:label1{value:'a2cb2c487d9da6941f0f9c1692ed1a5a', source: 'a0b116e2-d9f5-40d4-97ff-6ca6f2ec6c9b'})-[r]-> (related)
WHERE related:a OR related:b OR ....
If you want to pass labels as parameter labels you could use following in the WHERE clause
WHERE any(x in labels(n) WHERE x in $labels)
Related
I have the following graph:
I would look to get all contractors and subcontractors and clients, starting from David.
So I thought of a query likes this:
MATCH (a:contractor)-[*0..1]->(b)-[w:works_for]->(c:client) return a,b,c
This would return:
(0:contractor {name:"David"}) (0:contractor {name:"David"}) (56:client {name:"Sarah"})
(0:contractor {name:"David"}) (1:subcontractor {name:"John"}) (56:client {name:"Sarah"})
Which returns the desired result. The issue here is performance.
If the DB contains millions of records and I leave (b) without a label, the query will take forever. If I add a label to (b) such as (b:subcontractor) I won't hit millions of rows but I will only get results with subcontractors:
(0:contractor {name:"David"}) (1:subcontractor {name:"John"}) (56:client {name:"Sarah"})
Is there a more efficient way to do this?
link to graph example: https://console.neo4j.org/r/pry01l
There are some things to consider with your query.
The relationship type is not specified- is it the case that the only relationships from contractor nodes are works_for and hired? If not, you should constrain the relationship types being matched in your query. For example
MATCH (a:contractor)-[:works_for|:hired*0..1]->(b)-[w:works_for]->(c:client)
RETURN a,b,c
The fact that (b) is unlabelled does not mean that every node in the graph will be matched. It will be reached either as a result of traversing the works_for or hired relationships if specified, or any relationship from :contractor, or via the works_for relationship.
If you do want to label it, and you have a hierarchy of types, you can assign multiple labels to nodes and just use the most general one in your query. For example, you could have a label such as ExternalStaff as the generic label, and then further add Contractor or SubContractor to distinguish individual nodes. Then you can do something like
MATCH (a:contractor)-[:works_for|:hired*0..1]->(b:ExternalStaff)-[w:works_for]->(c:client)
RETURN a,b,c
Depends really on your use cases.
Let's assume this use case;
We have few nodes (labeled Big) and each having a simple integer ID property.
Each Big node has a relation with millions of (labeled Small) nodes.
such as :
(Small)-[:BELONGS_TO]->(Big)
How can I phrase a Cypher query to represent the following in natural language:
For each Big node in the range of ids between 4-7, get me 10 of Small nodes that belongs to it.
The supposed result would give 2 Big nodes, 20 Small nodes, and 20 Relations
The needed result would be represented by this graph:
2 Big nodes, each with a subset of 10 of Small nodes that belongs to them
What I've tried but failed (it only shows 1 big node (id=5) along with 10 of its related Small nodes, but doesn't show the second node (id=6):
MATCH (s:Small)-[:BELONGS_TO]->(b:Big)
Where 4<b.bigID<7
return b,s limit 10
I guess I need a more complex compound query.
Hope I could phrase my question in an understandable way!
As stdob-- says, you can't use limit here, at least not in this way, as it limits the entire result set.
While the aggregation solution will return you the right answer, you'll still pay the cost for the expansion to those millions of nodes. You need a solution that will lazily get the first ten for each.
Using APOC Procedures, you can use apoc.cypher.run() to effectively perform a subquery. The query will be run per-row, so if you limit the rows first, you can call this and use LIMIT within the subquery, and it will properly limit to 10 results per row, lazily expanding so you don't pay for an expansion to millions of nodes.
MATCH (b:Big)
WHERE 4 < b.bigID < 7
CALL apoc.cypher.run('
MATCH (s:Small)-[:BELONGS_TO]->(b)
RETURN s LIMIT 10',
{b:b}) YIELD value
RETURN b, value.s
Your query does not work because the limit applies to the entire previous flow.
You need to use aggregation function collect:
MATCH (s:Small)-[:BELONGS_TO]->(b:Big) Where 4<b.bigID<7
With b,
collect(distinct s)[..10] as smalls
return b,
smalls
To keep things simple, as part of the ETL on my time-series data, I added a sequence number property to each row corresponding to 0..370365 (370,366 nodes, 5,555,490 properties - not that big). I later added a second property and named it "outeseq" (original) and "ineseq" (second) to see if an outright equivalence to base the relationship on might speed things up a bit.
I can get both of the following queries to run properly on up to ~30k nodes (LIMIT 30000) but past that, its just an endless wait. My JVM has 16g max (if it can even use it on a windows box):
MATCH (a:BOOK),(b:BOOK)
WHERE a.outeseq=b.outeseq-1
MERGE (a)-[s:FORWARD_SEQ]->(b)
RETURN s;
or
MATCH (a:BOOK),(b:BOOK)
WHERE a.outeseq=b.ineseq
MERGE (a)-[s:FORWARD_SEQ]->(b)
RETURN s;
I also added these in hopes of speeding things up:
CREATE CONSTRAINT ON (a:BOOK)
ASSERT a.outeseq IS UNIQUE
CREATE CONSTRAINT ON (b:BOOK)
ASSERT b.ineseq IS UNIQUE
I can't get the relationships created for the entire data set! Help!
Alternatively, I can also get bits of the relationships built with parameters, but haven't figured out how to parameterize the sequence over all of the node-to-node sequential relationships, at least not in a semantically general enough way to do this.
I profiled the query, but did't see any reason for it to "blow-up".
Another question: I would like each relationship to have a property to represent the difference in the time-stamps of each node or delta-t. Is there a way to take the difference between the two values in two sequential nodes, and assign it to the relationship?....for all of the relationships at the same time?
The last Q, if you have the time - I'd really like to use the raw data and just chain the directed relationships from one nodes'stamp to the next nearest node with the minimum delta, but didn't run right at this for fear that it cause scanning of all the nodes in order to build each relationship.
Before anyone suggests that I look to KDB or other db's for time series, let me say I have a very specific reason to want to use a DAG representation.
It seems like this should be so easy...it probably is and I'm blind. Thanks!
Creating Relationships
Since your queries work on 30k nodes, I'd suggest to run them page by page over all the nodes. It seems feasible because outeseq and ineseq are unique and numeric so you can sort nodes by that properties and run query against one slice at time.
MATCH (a:BOOK),(b:BOOK)
WHERE a.outeseq = b.outeseq-1
WITH a, b ORDER BY a.outeseq SKIP {offset} LIMIT 30000
MERGE (a)-[s:FORWARD_SEQ]->(b)
RETURN s;
It will take about 13 times to run the query changing {offset} to cover all the data. It would be nice to write a script on any language which has a neo4j client.
Updating Relationship's Properties
You can assign timestamp delta to relationships using SET clause following the MATCH. Assuming that a timestamp is a long:
MATCH (a:BOOK)-[s:FORWARD_SEQ]->(b:BOOK)
SET s.delta = abs(b.timestamp - a.timestamp);
Chaining Nodes With Minimal Delta
When relationships have the delta property inside, the graph becomes a weighted graph. So we can apply this approach to calculate the shortest path using deltas. Then we just save the length of the shortest path (summ of deltas) into the relation between the first and the last node.
MATCH p=(a:BOOK)-[:FORWARD_SEQ*1..]->(b:BOOK)
WITH p AS shortestPath, a, b,
reduce(weight=0, r in relationships(p) : weight+r.delta) AS totalDelta
ORDER BY totalDelta ASC
LIMIT 1
MERGE (a)-[nearest:NEAREST {delta: totalDelta}]->(b)
RETURN nearest;
Disclaimer: queries above are not supposed to be totally working, they just hint possible approaches to the problem.
When writing a query to add relationships to existing nodes, it keeps me warning with this message:
"This query builds a cartesian product between disconnected patterns.
If a part of a query contains multiple disconnected patterns, this will build a cartesian product between all those parts. This may produce a large amount of data and slow down query processing. While occasionally intended, it may often be possible to reformulate the query that avoids the use of this cross product, perhaps by adding a relationship between the different parts or by using OPTIONAL MATCH (identifier is: (e))"
If I run the query, it creates no relationships.
The query is:
match
(a{name:"Angela"}),
(b{name:"Carlo"}),
(c{name:"Andrea"}),
(d{name:"Patrizia"}),
(e{name:"Paolo"}),
(f{name:"Roberta"}),
(g{name:"Marco"}),
(h{name:"Susanna"}),
(i{name:"Laura"}),
(l{name:"Giuseppe"})
create
(a)-[:mother]->(b),
(a)-[:grandmother]->(c), (e)-[:grandfather]->(c), (i)-[:grandfather]->(c), (l)-[:grandmother]->(c),
(b)-[:father]->(c),
(e)-[:father]->(b),
(l)-[:father]->(d),
(i)-[:mother]->(d),
(d)-[:mother]->(c),
(c)-[:boyfriend]->(f),
(g)-[:brother]->(f),
(g)-[:brother]->(h),
(f)-[:sister]->(g), (f)-[:sister]->(h)
Can anyone help me?
PS: if I run the same query, but with just one or two relationships (and less nodes in the match clause), it creates the relationships correctly.
What is wrong here?
First of all, as I mentionned in my comments, you don't have any Labels, it's a really bad practice because Labels are useful to match properties in a certains dataset (if you match "name" property, you don't want to match it on a node who doesn't have a name, Labels are here for that.
The second problem is that your query doesn't know how many nodes it will get before it does. It means that if you have 500 000 nodes having name : "Angela" and 500 000 nodes having name : "Carlo", you will create one relation from each Angela node, going on each Carlo, that's quite a big query (500 000 * 500 000 relations to create if my maths aren't bad). Cypher is giving you a warning for that.
Cypher will still tell you this warning because you aren't using Unique properties to match your nodes, even with Labels, you will still have the warning.
Solution?
Use unique properties to create and match your nodes, so you avoid cartesian product.
Always use labels, Neo4j without labels is like using one giant table in SQL to store all of your data.
If you want to know how your query will run, use PROFILE before your query, here is the profile plan for your query:
Does every single one of those name strings exist? If not then you're not going to get any results because it's all one big match. You could try changing it to a MERGE.
But Supamiu is right, you really should have a label (say Person) and an index on :Person(name).
I have connected Neo4j with Qlikview and I'm firing the cypher query. Now I have two graphs with the same labels so when I make a report in Qlikview I'm finding the aggregate count value of a particular label retrieved from both the graphs. So in order to get the data from only one graph can I fire the query with graph name say 'emp' prefixed with the label as below.
SQL MATCH a=(emp.cname)-[:has]->(emp.clusters) RETURN cname.name as cName, clusters.name as ClustName; where emp is the name of the graph from which I want to get the data and cname and clusters are the labels.
Thanks in advance...
There is technically only one graph. It seems that you have create two sub-graphs in that space and need a way of disguising them. That is typically done by creating an anchor node that at least one of the sub-graph nodes is related to or to add a unique label to each set of nodes to distinguish them. If that is not an option then you could run a preliminary query to determine the two start nodes (distinguished by some property/label or that are not connected in any way) and using one or the other for the query of interest.