I am developing a contact tracing framework using Neo4j. There are 2 types of nodes, namely Person and Location. There exists a relationship VISITED between a Person and a Location, which has properties startTS and endTS. Example:
Now suppose person 1 is infected. I need to find all the persons who have been in contact with this person. For each person identified, I need to find all other persons who have been in contact with that person. This process is repeated until an identified person has not met anyone. Here is a working code:
MATCH path = (infected:Person {id:'1'})-[*]-(otherPerson:Person)
WITH relationships(path) as rels, otherPerson
WHERE all(i in range(1, size(rels)-1)
WHERE i % 2 = 0
OR (rels[i].endTS >= rels[i-1].startTS AND rels[i].startTS <= rels[i-1].endTS)
)
RETURN otherPerson
The problem is that the process is taking way too much time to complete with large datasets. Can the above query be optimised? Thank you for your help.
For this one, unfortunately, there are some limitations on our syntax for filtering these more complex conditions during expansion. We can cover post-expansion filtering, but you'd want an upper bound otherwise this won't perform well on a more complex graph.
To get what you need today (filtering during-expansion instead of after), you would need to implement a custom procedure in Java leveraging our traversal API, and then call the procedure in your Cypher query.
Advanced syntax that can cover these cases has already been proposed for GQL, and we definitely want that in Cypher. It's on our backlog.
Related
I am using neo4j for the first time, and its fun using such an interactive database, but currently i got stuck in a problem, i have a data of people(uid,first name,last name, skills) , i also have a relationship [:has_skill]
my result frame looks like - p1 has a skill s (Robert has skill java)
I need to find out how many people have common skills, so i tried the following cypher query
match (p1:People)-[:has_skill]->(s:Skill)<-[:has_skill]-(p2:People)
where p1.people_uid="49981" and p2.people_uid="34564"
return p1.first_name+' '+p1.last_name as Person1, p2.first_name+' '+p2.last_name as Person2,s.skill_name,s.skillid,count(s)
i am getting p1 as different persons, but due to high skill set, the p2 person is getting repeated, and also the skill is not changing, i tried to delete every node and relationship where skill count of a person is greater then 6 to get good results, but cannot delete it, i am getting "invalid use of aggregating function"
This is my attempt to delete
match (p1:People)-[:has_skill]->(s:Skill)
where count(s)>6
detach delete p1,s
Please if anyone could guide or correct me where i am going wrong, your help would be highly appreciable . Thanks in advance.
Make sure when using count or other aggregating functions, they are within a WITH clause or a RETURN clause - seems to be a design decision that Neo Technology made when creating Neo4j - see some of the following links for similar cases to yours:
How to count the number of relationships in Neo4j
Neo4j aggregate function
I need to count the number of connection between two nodes with a certain property
Also - see the WITH clause documentation here and the RETURN clause documentation here, in particular, this part of the WITH documentation:
Another use is to filter on aggregated values. WITH is used to introduce aggregates which can then be used in predicates in WHERE. These aggregate expressions create new bindings in the results. WITH can also, like RETURN, alias expressions that are introduced into the results using the aliases as the binding name.
In your case, you are going to want your aggregate function to be used within a WITH clause because you need to use WHERE afterwards to filter only those persons with more than 6 skills. You can use the following query to see which persons have more than 6 skills:
match (p1:People)-[r:has_skill]->(s:Skill)
with p1,count(s) as rels, collect (s) as skills
where rels > 6
return p1,rels,skills
After confirming that the result set is correct, you can use the following query to delete the persons who have more than 6 skills along with all the skill nodes that these persons are related to:
MATCH(p1:People)-[r:has_skill]->(s:Skill)
WITH p1,count(s) as rels, collect (s) as skills
WHERE rels > 6
FOREACH(s in skills | DETACH DELETE s)
DETACH DELETE p1
I am new to Neo4j and currently playing with this tree structure:
The numbers in the yellow boxes are a property named order on the relationship CHILD_OF.
My goal was
a) to manage the sorting order of nodes at the same level through this property rather than through directed relationships (like e.g. LEFT, RIGHT or IS_NEXT_SIBLING, etc.).
b) being able to use plain integers instead of complete paths for the order property (i.e. not maintaining sth. like 0001.0001.0002).
I can't however find the right hint on how or if it is possible to recursively query the graph so that it keeps returning the nodes depth-first but for the sorting at each level consider the order property on the relationship.
I expect that if it is possible it might include matching the complete path iterating over it with the collection utilities of Cypher, but I am not even close enough to post some good starting point.
Question
What I'd expect from answers to this question is not necessarily a solution, but a hint on whether this is a bad approach that would perform badly anyways. In terms of Cypher I am interested if there is a practical solution to this.
I have a general idea on how I would tackle it as a Neo4j server plugin with the Java traversal or core api (which doesn't mean that it would perform well, but that's another topic), so this question really targets the design and Cypher aspect.
This might work:
match path = (n:Root {id:9})-[:CHILD_OF*]->(m)
WITH path, extract(r in rels(path) | r.order) as orders
ORDER BY orders
if it complains about sorting arrays then computing a number where each digit (or two digits) are your order and order by that number
match path = (n:Root {id:9})-[:CHILD_OF*]->(m)
WITH path, reduce(a=1, r in rels(path) | a*10+r.order) as orders
ORDER BY orders
Suppose I have a large knowledge base with many relationship types, e.g., hasChild, livesIn, locatedIn, capitalOf, largestCityOf...
The number of capicalOf relationships is relatively small (say, one hundred) compared to that of all nodes and other types of relationships.
I want to fetch any capital which is also the largest city in their country by the following query:
MATCH city-[:capitalOf]->country, city-[:largestCityOf]->country RETURN city
Apparently it would be wise to take the capitalOf type as clue, scan all 100 relationship with this type and refine by [:largestCityOf]. However the current execution plan engine of neo4j would do an AllNodesScan and Expand. Why not consider add an "RelationshipByTypeScan" operator into the current query optimization engine, like what NodeByLabelScan does?
I know that I can transform relationship types to relationship properties, index it using the legacy index and manually indicate
START r=relationship:rels(rtype = "capitalOf")
to tell neo4j how to make it efficient. But for a more complicated pattern query with many relationship types but no node id/label/property to start from, it is clearly a duty of the optimization engine to decide which relationship type to start with.
I saw many questions asking the same problem but getting answers like "negative... a query TYPICALLY starts from nodes... ". I just want to use the above typical scenario to ask why once more.
Thanks!
A relationship is local to its start and end node - there is no global relationship dictionary. An operation like "give me globally all relationships of type x" is therefore an expensive operation - you need to go through all nodes and collect matching relationships.
There are 2 ways to deal with this:
1) use a manual index on relationships as you've sketched
2) assign labels to your nodes. Assume all the country nodes have a Country label. Your can rewrite your query:
MATCH (city)-[:capitalOf]->(country:Country), (city)-[:largestCityOf]->(country) RETURN city
The AllNodesScan is now a NodeByLabelScan. The query grabs all countries and matches to the cities. Since every country does have one capital and one largest city this is efficient and scales independently of the rest of your graph.
If you put all relationships into one index and try to grab to ~100 capitalOf relationships that operation scales logarithmically with the total number of relationships in your graph.
I'm trying to query Book nodes for recommendation by Cypher.
I want to recommend A:Book and C:Book for A:User.
i'm sorry I need some graph to explain this question, but I could't up graph image because my lepletion lacks for upload function.
I wrote query below.
match (u1:User{uid:'1003'})-->(o1:Order)-->(b1:Book)<--(o2:Order)
<--(u2:User)-->(o3:Order)-->(b2:Book)
return b2
This query return all Books(A,B,C,D) dispite cypher's Uniqueness.
I expect to only return A:Book and C:Book.
Is this behavior Neo4j' specification?
How do I get expected return? Thanks, everyone.
environment:
Neo4j ver.v2.0.0-RC1
Using Neo4j Server with REST API
Without the sample graph its hard to say why you get something back when you expected something else. You can share a sample graph by including a create statement that would generate said graph, or by creating it in Neo4j console and putting the link in your question. Here is an example of the latter: console.neo4j.org/r/fnnz6b
In the meantime, you probably want to declare the type of the relationships in your pattern. If a :User has more than one type of outgoing relationships you will be excluding those other paths based on the labels of the nodes on the other end, which is much less efficient than to only traverse the right relationships to begin with.
To my mind its not clear whether (u:User)-->(o:Order)-->(b:Book) means that a user has one or more orders, and each order consists of one or more books; or if it means only that a user ordered a book. If you can share a sample, hopefully that will be clear too.
Edit:
Great, so looking at the graph: You get B and D back because others who bought B also bought D, and others who bought D also bought B, which is your criterion for recommendation. You can add a filter in the WHERE clause to exclude those books that the user has already bought, something like
WHERE NOT (u1)-[:BUY]->()-[:CONTAINS]->(b2)
This will give you A, C, C back, since there are two matching paths to C. It's probably not important to get two result items for C, so you can either limit the return to give only distinct values
RETURN DISTINCT(b2)
or group the return values by counting the matching paths for each result as a 'recommendation score'
RETURN b2, COUNT(b2) as score
Also, if each order only [CONTAINS] one book, you could try modelling without order, just (:User)-[:BOUGHT]->(:Book).
I'm new to NEO4J and I need help on a specific problem. Or an answer if it's even possible.
SETUP:
We have 2 distinct type of nodes: users (A,B,C,D) and Products (1,2,3,4,5,6,7,8)
Next we have 2 distinct type of relationships between users and products where a users WANTS a Product and where a product is OWNED BY a user.
1,2 is owned by A
3,4 is owned by B
5,6 is owned by C
7,8 is owned by D
Now
B wants 1
C wants 3
D wants 5
So for now, I have no problems and I created the graph data with no difficulty. My questions starts here. We have a circle, when A wants product 8.
A-[:WANTS]->8-[:OWNEDBY]->D-[:WANTS]->5-[:OWNEDBY]->C-[:WANTS]->3-[:OWNEDBY]->B-[:WANTS]->1-[:OWNEDBY]->A
So we have a distinct pattern, U-[:WANTS]->P-[:OWNEDBY]->U
Now what I want to do is to find the paths toward the start node (initiating user that wants a product) following that pattern.
How do I define this using Cypher? Or do I need another way?
Thanks upfront.
i got a feeling this can be hacked with reduce and counting every even (every second) relationship:
MATCH p=A-[:OWNEDBY|WANTS*..20]->X
WITH r in relationships(p)
RETURN type(r),count(r) as cnt,
WHERE cnt=10;
or maybe counting all paths where the number of rels is even:
MATCH p=A-[:OWNEDBY|WANTS*..]->X
RETURN p,reduce(total, r in relationships(p): total + 1) as tt
WHERE tt%2=0;
but you graph must have the strict pattern, where all incoming relationship from the set of ownedby and wants must be different from all outgoin relationships from the same set. in other words, this pattern can not exist: A-[:WANTS]->B-[:WANTS]->C or A-[:OWNEDBY]->B-[:OWNEDBY]->C
the query is probably wrong in syntax, but the logic can be implemented in cypher whn you will play more with it.
or use gremlin, I think I saw somewhere a gremlin query, where you could define a pattern and than loop n-times via that pattern further till the end node.
I've played around with this and created http://console.neo4j.org/?id=qq9v1 showing the sample graph. I've found the following cypher statement solving the issue:
start a=node(1)
match p=( a-[:WANTS]->()-[:WANTS|OWNEDBY*]-()-[:OWNEDBY]->a )
return p
order by length(p) desc
limit 1
There is just one glitch: the intermediate part [:WANTS|OWNEDBY*] does not mandate alternating WANT and OWNEDBY chains. As #ulkas stated, this should not be an issue if you take care during data modelling.
You might also look into http://api.neo4j.org/current/org/neo4j/graphalgo/GraphAlgoFactory.html to apply graph algorithms from Java code. You might use unmangaged extensions to provide REST access to that.