Is matching with id performant in Neo4J? - neo4j

I'm wondering, when I have read the data of a node and I want to match it in another query, which way will have the best performance? Using id like this:
MATCH (n) WHERE ID(n) = 1234
or using indices of the node:
MATCH (n:Label {SomeIndexProperty: 3456})
Which one is better?

IDs are a technical ID for Neo4j, and those should not be used as a primary key for your application.
Every node (and relationship) has a technical ID, and it's stable over time.
But if you delete a node, for example the node 32, Neo4j will reuse this ID for a new node.
So you can use it in your queries inside the same transaction (there is no problem), otherwise you should know what you are doing.
The only way to retrieve the technical ID, is to use the function ID like you do on your first query : MATCH (n) WHERE ID(n) = 1234 RETURN n.
The ID is not exposed as a node's property, so you can't do MATCH (n {ID:1234}) RETURN n.
You have noticed that if you want to do a WHERE on a strict equality, you can do put the condition directly on the node.
For example :
MATCH (n:Node) WHERE n.name = 'logisima' RETURN n
MATCH (n:Node {name:'logisima'}) RETURN n
Those two queries are identicals, they generate the same query plan, it's just a syntactic sugar.
Is it faster to retrieve a node by its ID or by an indexed property ?
The easier way to know the answer to this question is to profile the two queries.
On the one based on the ID, you will see the box NodeByIdSeek that cost 1 db hit, and on the one with a unique constrainst you will see the box NodeUniqueIndexSeek with 2 db hits.
So searching a node by its ID is faster.

Related

Neo4j count Query

match(m:master_node:Application)-[r]-(k:master_node:Server)-[r1]-(n:master_node)
where (m.name contains '' and (n:master_node:DeploymentUnit or n:master_node:Schema))
return distinct m.name,n.name
Hi,I am trying to get total number of records for the above query.How I change the query using count function to get the record count directly.
Thanks in advance
The following query uses the aggregating funtion COUNT. Distinct pairs of m.name, n.name values are used as the "grouping keys".
MATCH (m:master_node:Application)--(:master_node:Server)--(n:master_node)
WHERE EXISTS(m.name) AND (n:DeploymentUnit OR n:Schema)
RETURN m.name, n.name, COUNT(*) AS cnt
I assume that m.name contains '' in your query was an attempt to test for the existence of m.name. This query uses the EXISTS() function to test that more efficiently.
[UPDATE]
To determine the number of distinct n and m pairs in the DB (instead of the number of times each pair appears in the DB):
MATCH (m:master_node:Application)--(:master_node:Server)--(n:master_node)
WHERE EXISTS(m.name) AND (n:DeploymentUnit OR n:Schema)
WITH DISTINCT m.name AS n1, n.name AS n2
RETURN COUNT(*) AS cnt
Some things to consider for speeding up the query even further:
Remove unnecessary label tests from the MATCH pattern. For example, can we omit the master_node label test from any nodes? In fact, can we omit all label testing for any nodes without affecting the validity of the result? (You will likely need a label on at least one node, though, to avoid scanning all nodes when kicking off the query.)
Can you add a direction to each relationship (to avoid having to traverse relationships in both directions)?
Specify the relationship types in the MATCH pattern. This will filter out unwanted paths earlier. Once you do so, you may also be able to remove some node labels from the pattern as long as you can still get the same result.
Use the PROFILE clause to evaluate the number of DB hits needed by different Cypher queries.
You can find examples of how to use count in the Neo4j docs here
In your case the first example where:
count(*)
Is used to return a count of each returned item should work.

Cypher - Neo4j Query Profiling

I have some questions regarding Neo4j's Query profiling.
Consider below simple Cypher query:
PROFILE
MATCH (n:Consumer {mobileNumber: "yyyyyyyyy"}),
(m:Consumer {mobileNumber: "xxxxxxxxxxx"})
WITH n,m
MATCH (n)-[r:HAS_CONTACT]->(m)
RETURN n,m,r;
and output is:
So according to Neo4j's Documentation:
3.7.2.2. Expand Into
When both the start and end node have already been found, expand-into
is used to find all connecting relationships between the two nodes.
Query.
MATCH (p:Person { name: 'me' })-[:FRIENDS_WITH]->(fof)-->(p) RETURN
> fof
So here in the above query (in my case), first of all, it should find both the StartNode & the EndNode before finding any relationships. But unfortunately, it's just finding the StartNode, and then going to expand all connected :HAS_CONTACT relationships, which results in not using "Expand Into" operator. Why does this work this way? There is only one :HAS_CONTACT relationship between the two nodes. There is a Unique Index constraint on :Consumer{mobileNumber}. Why does the above query expand all 7 relationships?
Another question is about the Filter operator: why does it requires 12 db hits although all nodes/ relationships are already retrieved? Why does this operation require 12 db calls for just 6 rows?
Edited
This is the complete Graph I am querying:
Also I have tested different versions of same above query, but the same Query Profile result is returned:
1
PROFILE
MATCH (n:Consumer{mobileNumber: "yyyyyyyyy"})
MATCH (m:Consumer{mobileNumber: "xxxxxxxxxxx"})
WITH n,m
MATCH (n)-[r:HAS_CONTACT]->(m)
RETURN n,m,r;
2
PROFILE
MATCH (n:Consumer{mobileNumber: "yyyyyyyyy"}), (m:Consumer{mobileNumber: "xxxxxxxxxxx"})
WITH n,m
MATCH (n)-[r:HAS_CONTACT]->(m)
RETURN n,m,r;
3
PROFILE
MATCH (n:Consumer{mobileNumber: "yyyyyyyyy"})
WITH n
MATCH (n)-[r:HAS_CONTACT]->(m:Consumer{mobileNumber: "xxxxxxxxxxx"})
RETURN n,m,r;
The query you are executing and the example provided in the Neo4j documentation for Expand Into are not the same. The example query starts and ends at the same node.
If you want the planner to find both nodes first and see if there is a relationship then you could use shortestPath with a length of 1 to minimize the DB hits.
PROFILE
MATCH (n:Consumer {mobileNumber: "yyyyyyyyy"}),
(m:Consumer {mobileNumber: "xxxxxxxxxxx"})
WITH n,m
MATCH Path=shortestPath((n)-[r:HAS_CONTACT*1]->(m))
RETURN n,m,r;
Why does this do this?
It appears that this behaviour relates to how the query planner performs a database search in response to your cypher query. Cypher provides an interface to search and perform operations in the graph (alternatives include the Java API, etc.), queries are handled by the query planner and then turned into graph operations by neo4j's internals. It make sense that the query planner will find what is likely to be the most efficient way to search the graph (hence why we love neo), and so just because a cypher query is written one way, it won't necessarily search the graph in the way we imagine it will in our head.
The documentation on this seemed a little sparse (or, rather I couldn't find it properly), any links or further explanations would be much appreciated.
Examining your query, I think you're trying to say this:
"Find two nodes each with a :Consumer label, n and m, with contact numbers x and y respectively, using the mobileNumber index. If you find them, try and find a -[:HAS_CONTACT]-> relationship from n to m. If you find the relationship, return both nodes and the relationship, else return nothing."
Running this query in this way requires a cartesian product to be created (i.e., a little table of all combinations of n and m - in this case only one row - but for other queries potentially many more), and then relationships to be searched for between each of these rows.
Rather than doing that, since a MATCH clause must be met in order to continue with the query, neo knows that the two nodes n and m must be connected via the -[:HAS_CONTACT]-> relationship if the query is to return anything. Thus, the most efficient way to run the query (and avoid the cartesian product) is as below, which is what your query can be simplified to.
"Find a node n with the :Consumer label, and value x for the index mobileNumber, which is connected via a -[:HAS_CONTACT]-> relationshop to a node m with the :Consumer label, and value y for its proprerty mobileNumber. Return both nodes and the relationship, else return nothing."
So, rather than perform two index searches, a cartesian product and a set of expand into operations, neo performs only one index search, an expand all, and a filter.
You can see the result of this simplification by the query planner through the presence of AUTOSTRING parameters in your query profile.
How to Change Query to Implement Search as Desired
If you want to change the query so that it must use an expand into relationship, make the requirement for the relationship optional, or use explicitly iterative execution. Both these queries below will produce the initially expected query profiles.
Optional example:
PROFILE
MATCH (n:Consumer{mobileNumber: "xxx"})
MATCH (m:Consumer{mobileNumber: "yyy"})
WITH n,m
OPTIONAL MATCH (n)-[r:HAS_CONTACT]->(m)
RETURN n,m,r;
Iterative example:
PROFILE
MATCH (n1:Consumer{mobileNumber: "xxx"})
MATCH (m:Consumer{mobileNumber: "yyy"})
UNWIND COLLECT(n1) AS n
MATCH (n)-[r:HAS_CONTACT]->(m)
RETURN n,m,r;

Auto-generation of id in Neo4j

Will Neo4j auto-generate unique ids for all the nodes created using 'CREATE' query, as 'id' in Mysql? We created a node with
CREATE (n: user{name:"x", age:10}) return n
If we want to update 'name' property, how to do it in Neo4j?
From the documentation though,
Searching for nodes by id can be done with the id() function in a predicate.
Neo4j reuses its internal ids when nodes and relationships are deleted. This means that applications using, and relying on internal Neo4j ids, are brittle or at risk of making mistakes. It is therefor recommended to rather use application generated ids.
It is a good idea to as they say, use Application generated ids that are stored as properties of the node. You need to set the application generated id while creating the node and then use that key on merge statements
MERGE (n:user{key:<myApplicationKey>}) return n
SET n.name="x", age=10
There is an internal id, you can access it the id() function
CREATE (n:User {name="x", age: 10}) RETURN n, id(n) as nodeId
Later on, you can update it with
MATCH (n) WHERE id(n) = 12345 SET n.name = "y"

Neo4j: Transitive query and node ordering

I am using Neo4j to track relationships in OOP architecture. Let us assume that nodes represent classes and (u) -[:EXTENDS]-> (v) if class u extends class v (i.e. for each node there is at most one outgoing edge of type EXTENDS). I am trying to find out a chain of predecessor classes for a given class (n). I have used the following Cypher query:
start n=node(...)
match (n) -[:EXTENDS*]-> (m)
return m.className
I need to process nodes in such an order that the direct predecessor of class n comes first, its predecessor comes as second etc. It seems that the Neo4j engine returns the nodes in exactly this order (given the above query) - is this something I should rely on or could this behavior suddenly change in some of the future releases?
If I should not rely on this behavior, what Cypher query would allow me to get all predecessor nodes in given ordering? I was thinking about following query:
start n=node(...)
match p = (n) -[:EXTENDS*]-> (m {className: 'Object'})
return p
Which would work quite fine, but I would like to avoid specifying the root class (Object in this case).
It's unlikely to change anytime soon as this is really the nature of graph databases at work.
The query you've written will return ALL possible "paths" of nodes that match that pattern. But given that you've specified that there is at most one :EXTENDS edge from each such node, the order is implied with the direction you've included in the query.
In other words, what's returned won't start "skipping" nodes in a chain.
What it will do, though, is give you all "sub-paths" of a path. That is, assuming you specified you wanted the predecessors for node "a", for the following path...
(a)-[:EXTENDS]->(b)-[:EXTENDS]->(c)
...your query (omitting the property name) will return "a, b, c" and "a, b". If you only want ALL of its predecessors, and you can use Cypher 2.x, consider using the "path" way, something like:
MATCH p = (a)-[:EXTENDS*]->(b)
WITH p
ORDER BY length(p) DESC
LIMIT 1
RETURN extract(x in nodes(p) | p.className)
Also, as a best practice, given that you're looking at paths of indefinite length, you should likely limit the number of hops your query makes to something reasonable, e.g.
MATCH (n) -[:EXTENDS*0..10]-> (m)
Or some such.
HTH

The magic behind Neo4J Cypher query results

I have the following query in Neo4J Cypher 2.0:
MATCH (u:User{uid:'1111111'}), (c1:Concept), (c2:Concept),
c1-[:BY]->u, c2-[:BY]->u, c1-[rel:TO]->c2
WITH c1,c2,rel
MATCH c1-[:AT]->ctx, c2-[:AT]-ctx
WHERE ctx.uid = rel.context
RETURN c1.uid AS source_id, c1.name AS source_name,
c2.uid AS target_id, c2.name AS target_name,
rel.uid AS edge_id,
rel.context AS context_id, ctx.name AS context_name;
What it does is that it looks for all the nodes of the Concept label (c1 and c2) connected to User node u, finds their (c1 to c2) connections to one another (rel), then it tries to find which different contexts (ctx) those concept nodes (c1 and c2) appear in, but only those, whose uid matches the uid of the .context property of the relationships rel (rel.context) and then returns them in a table, where we have the source id and name, the target id and name, the connection id, as well as the .context id property of that relation and the name of the context with that id.
So all works fine, but the question is: WHY?
I mean how does Cypher matches so neatly the right ctx.uid to the right rel.context to know that it should be inserted exactly at the right place of the results table?
Can somebody explain me the magic behind this?
Or am I completely wrong and just getting messy results?
Thank you!
It creates a pattern-graph that represents your combined match patterns. And then it uses indexes to find bound nodes that it starts to apply the pattern graph to and returns a result row for every match found.
While applying the pattern graph it uses your WHERE conditions to filter out paths that you don't want eagerly as early as possible.
If it can't find bound nodes it has to go over all nodes of a label (like :Concept) or over all nodes of the graph (if you haven't specify any label or lookup condition).

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