I have nodes with label A. Some of them are connected with the relationship TEST (see Figure A).
I want to MATCH the groups of connected nodes, create a new node B for each group and create a relationship from each member of the group to the new node B (see Figure B). I know that the groups are small, never more then 3 steps of TEST relationships.
How can I MATCH the A nodes and return connected groups? Is there a graph algorithm implemented in APOC?
I found the answer, maybe it's still helpful for someone:
There are several algorithms for community detection in the graph algorithm package ()https://neo4j.com/docs/graph-algorithms/current/. In this case, we look for connected components: https://neo4j.com/docs/graph-algorithms/current/algorithms/connected-components/
The algorithm can find connected components and store an ID for the component on the nodes:
CALL algo.unionFind('A', 'TEST', {write:true, partitionProperty:"partition"})
YIELD nodes, setCount, loadMillis, computeMillis, writeMillis;
With this new property it's simple to MATCH all nodes belonging to a particular group:
MATCH (a:A)
WITH a.partition AS p, a
RETURN p, count(a)
algo.unionFind seems to be deprecated. The replacement is:
CALL gds.wcc.write('A', {nodeLabels: 'TEST', writeProperty: 'partition' });
Here is the syntax from the docs
CALL gds.wcc.write(
graphName: String,
configuration: Map
)
YIELD
componentCount: Integer,
nodePropertiesWritten: Integer,
preProcessingMillis: Integer,
computeMillis: Integer,
writeMillis: Integer,
postProcessingMillis: Integer,
componentDistribution: Map,
configuration: Map
See https://neo4j.com/docs/graph-data-science/current/algorithms/wcc/ for details
Related
Given that I'm very new to Neo4j. I have a schema which looks like the below image:
Here Has nodes are different for example Passport, Merchant, Driving License, etc. and also these nodes are describing the customer node (looking for future scope of filtering customers based on these nodes).
SIMILAR is a self-relation meaning there exists a customer with ID:1 is related to another customer with ID:2 with a score of 2800.
I have the following questions:
Is this a good schema given the condition of the future scope I mentioned above, or getting all the properties in a single customer node is viable? (Different nodes may have array of items as well, for example: ()-[:HAS]->(Phone) having {active: "+91-1231241", historic_phone_numbers: ["+91-121213", "+91-1231421"]})
I want to get the customer along with describing nodes in relation to other customers. For that, I tried the below query (w/o number of relation more than 1):
// With number_of_relation > 1
MATCH (searched:Customer)-[r:SIMILAR]->(matched:Customer)
WHERE r.score > 2700
WITH searched, COLLECT(matched.customer_id) AS MatchedList, count(r) as cnt
WHERE cnt > 1
UNWIND MatchedList AS matchedCustomer
MATCH (person:Customer {customer_id: matchedCustomer})-[:HAS|:LIVES_IN|:IS_EMPLOYED_BY]->(related)
RETURN searched, person, related
Result what I got is below, notice one customer node not having its describing nodes:
// without number_of_relation > 1
// second attempt - for a sample customer_id
MATCH (matched)<-[r:SIMILAR]-(c)-[:HAS|:LIVES_IN|:IS_EMPLOYED_BY]->(b)
WHERE size(keys(b)) > 0
AND c.customer_id = "1b093559-a39b-4f95-889b-a215cac698dc"
AND r.score > 2700
RETURN b AS props, c AS src_cust, r AS relation, matched
Result I got are below, notice related nodes are not having their describing nodes:
If I had two describing nodes with some property (some may have a list) upon which I wanted to query and build the expected graph specified in point 2 above, how can I do that?
I want the database to find a similar customer given the describing nodes. Example: A customer {name: "Dave"} has phone {active_number: "+91-12345"} is similar to customer {name: "Mike"} has phone {active_number: "+91-12345"}. How can get started with this?
If something is unclear, please ask. I can explain with examples.
[EDITED]
Yes, the schema seems fine, except that you should not use the same HAS relationship type between different node label pairs.
The main problem with your first query is that its top MATCH clause uses a directional relationship pattern, ()-->(), which does not allow all Customer nodes to have a chance to be the searched node (because some nodes may only be at the tail end of SIMILAR relationships). This tweaked query should work better:
MATCH (searched:Customer)-[r:SIMILAR]-(matched:Customer)
WHERE r.score > 2700
WITH searched, COLLECT(matched) AS matchedList
WHERE SIZE(matchedList) > 1
UNWIND matchedList AS person
MATCH (person)-[:HAS|LIVES_IN|IS_EMPLOYED_BY]->(pDesc)
WITH searched, person, COLLECT(pDesc) AS personDescribers
MATCH (searched)-[:HAS|LIVES_IN|IS_EMPLOYED_BY]->(sDesc)
RETURN searched, person, personDescribers, COLLECT(sDesc) AS searchedDescribers
It's not clear what you want are trying to do.
To get all Customers who have the same phone number:
MATCH (c:Customer)-[:HAS_PHONE]-(p:Phone)
WHERE p.activeNumber = '+91-12345'
WITH p.activeNumber AS phoneNumber, COLLECT(c) AS customers
WHERE SIZE(customers) > 1
RETURN phoneNumber, customers
I am new at Neo4j but not to graphs and I have a specific problem I did not manage to solve with Cypher.
With this type of data:
I would like to be able in a single query to follow some incoming and some outgoing flow.
Example:
Starting on "source"
Follow all "A" relationships in the outgoing way
Follow all "B" relationships in the incoming way
My problem is that Cypher only allows one single direction to be specified in the relationship pattern.
So I could do (source)-[:A|:B*]->() or (source)<-[:A|:B*]-().
But I have no possibility to tell Cypher that I want to follow -[:A]-> and <-[:B]-.
By the way, I know that I could do -[:A|:B]- but this won't solve my problem because I don't want to follow -[:B]-> and <-[:A]-.
Thanks in advance for your help :)
Alternatively to #Gabor Szarnyas answer, I think you can achieve your goal using the APOC procedure apoc.path.expand.
Using this sample data set:
CREATE (:Source)-[:A]->()-[:A]->()<-[:B]-()-[:A]->()
And calling apoc.path.expand:
match (source:Source)
call apoc.path.expand(source,"A>|<B","",0,5) yield path
return path
You will get this path as output:
The apoc.path.expand call starts from the source node following -[:A]-> and <-[:B]- relationships.
Remember to install APOC procedures according to the version of Neo4j you are using. Take a look in the version compatibility matrix.
To express this in a single query would require a regular path query, which has been proposed to and accepted to openCypher, but it is not yet implemented.
I see two possible workarounds. I recreated your example with this command with a Source label for the source node:
CREATE (:Source)-[:A]->()-[:A]->()<-[:B]-()-[:A]->()
(1) Insert additional relationships that have the same direction:
MATCH (s)-[:B]->(t)
CREATE (s)<-[:B2]-(t)
And use this relationship for traversal:
MATCH p=(source)-[:A|:B2*]->()
RETURN p
(2) As you mentioned:
By the way, I know that I could do -[:A|:B]- but this won't solve my problem because I don't want to follow -[:B]-> and <-[:A]-.
You could use this approach to first get potential path candidates and manually check the directions of the relationships afterwards. Of course, this is an expensive operation but you only have to calculate it on the candidates, a possibly small data set.
MATCH p=(source:Source)-[:A|:B*]-()
WITH p, nodes(p) AS nodes, relationships(p) AS rels
WHERE all(i IN range(0, size(rels) - 1) WHERE
CASE type(rels[i])
WHEN 'A' THEN startNode(rels[i]) = nodes[i]
ELSE /* B */ startNode(rels[i]) = nodes[i+1]
END)
RETURN p
Let's break down how this works:
We store path candidates in p and use the nodes and relationships functions to extract the lists of nodes/relationships from it.
We define a range of indexes for the relationships (e.g. from 0, 1, 2 if there are 3 relationships).
To determine the direction of relationships, we use the startNode function. For example, if there is a relationship r between nodes n1 to n2, the paths will like <n1, r, n2>. If r was traversed to in the outgoing direction, the startNode(r) will return n1, if it was traverse in the incoming direction, startNode(r) will return n2. The type of the relationship is checked with the type function and a simple CASE expression is used to differentiate between types.
The WHERE clause uses the all predicate function to check whether all :A and :B relationships had the appropriate directions.
I have database with entities person (name,age) and project (name).
can I query the database in cypher that specifies me it is person or project?
for example consider I have these two instances for each :
Node (name = Alice, age= 20)
Node (name = Bob, age = 31)
Node (name = project1)
Node (name = project2)
-I want to know, is there any way that I just say project1 and it tells me that this is a project.
-or I query Alice and it says me this is a person?
Thanks
So your use case is to search things by name, and those things can be of several types instead of a single type.
Just to note, in general, this is not what Neo4j is built for. Typically in Neo4j queries you know the type of the thing you're searching for, and you're exploring relationships between that thing (or things) to figure out associations or data derived from that.
That said, there are ways to do this, though it's worth going through the rest of your use cases and seeing if Neo4j is really the best tool for what you're trying to do
Whenever you're querying by a property, you either want a unique constraint on the label/property, or an index on the label/property. Note that you need a combination of a label and a property for this; you cannot blindly ask for a node with a property without specifying a label and get good performance, as it will have to do a scan of all nodes in your database (there are some older manual indexes in Neo4j, but I'm not sure if these will continue to be supported; the schema indexes are recommended by the developers).
There is a workaround to this, as Neo4j allows multiple labels on the same node. If you only expect to query certain types by name (for example, only projects and people), you might create a :Named label, and set that label on all :Project and :Person nodes (and any other labels where it should apply). You can then create an index on :Named.name. That way your query would be something like:
MATCH (n:Named)
WHERE n.name = 'blah'
WITH LABELS(n) as types
WITH FILTER(type in types WHERE type <> 'Named') as labels
RETURN labels
Keep in mind that you haven't specified if a name should be unique among node types, so it could be possible for a :Person or a :Project or multiple :Persons to have the same name, unsure how that affects what should happen on your end. If every named thing ought to have a unique name, you should create a unique constraint on :Named.name (though again, it's on you to ensure that every node you create that ought to be :Named has the :Named label applied on creation).
You should use node labels (like Person and Project) to represent node "types".
For example, to create a person and a project:
CREATE (:Person {name: 'Alice', age: 20})
CREATE (:Project {name: 'project1'})
To find the project(s) named 'Fred':
MATCH (p:Project {name: 'Fred'})
RETURN p;
To get a collection of the labels of node n, you can invoke the LABELS(n) function. You can then look in that collection to see if the label you are looking for is in there. For example, if your Cypher query somehow obtains a node n, then this snippet would return n if and only if it has the Person label:
.
.
.
WHERE 'Person' IN LABELS(n)
RETURN n;
[UPDATED]
If you want to find all nodes with the name property value of "Fred":
MATCH (n {name: 'Fred'})
...
If you want to find all relationships with the name property value of "Fred":
MATCH ()-[r {name: 'Fred'})-()
...
If you want to match both in a single query, you have many ways to do that, depending on your exact use case. For example, if you want a cartesian product of the matching nodes and relationships:
OPTIONAL MATCH (n {name: 'Fred'})
OPTIONAL MATCH ()-[r {name: 'Fred'})-()
...
In a graph where the following nodes
A,B,C,D
have a relationship with each nodes successor
(A->B)
and
(B->C)
etc.
How do i make a query that starts with A and gives me all nodes (and relationships) from that and outwards.
I do not know the end node (C).
All i know is to start from A, and traverse the whole connected graph (with conditions on relationship and node type)
I think, you need to use this pattern:
(n)-[*]->(m) - variable length path of any number of relationships from n to m. (see Refcard)
A sample query would be:
MATCH path = (a:A)-[*]->()
RETURN path
Have also a look at the path functions in the refcard to expand your cypher query (I don't know what exact conditions you'll need to apply).
To get all the nodes / relationships starting at a node:
MATCH (a:A {id: "id"})-[r*]-(b)
RETURN a, r, b
This will return all the graphs originating with node A / Label A where id = "id".
One caveat - if this graph is large the query will take a long time to run.
I've built a simple graph of one node type and two relationship types: IS and ISNOT. IS relationships means that the node pair belongs to the same group, and obviouslly ISNOT represents the not belonging rel.
When I have to get the groups of related nodes I run the following query:
"MATCH (a:Item)-[r1:IS*1..20]-(b:Item) RETURN a,b"
So this returns a lot of a is b results and I added some code to group them after.
What I'd like is to group them modifying the query above, but given my rookie level I haven't yet figured it out. What I'd like is to get one row per group like:
(node1, node3, node5)
(node2,node4,node6)
(node7,node8)
I assume what you call groups are nodes present in a path where all these nodes are connected with a :IS relationship.
I think this query is what you want :
MATCH p=(a:Item)-[r1:IS*1..20]-(b:Item)
RETURN nodes(p) as nodes
Where p is a path describing your pattern, then you return all the nodes present in the path in a collection.
Note that, a simple graph (http://console.neo4j.org/r/ukblc0) :
(item1)-[:IS]-(item2)-[:IS]-(item3)
will return already 6 paths, because you use undericted relationships in the pattern, so there are two possible paths between item1 and item2 for eg.