I got 2 node types, let's say A and B, and a relationship with a property, let's call it 'a_has_b' with the property 'value'
First I want to count the number of relationships a specific node of type A has.
MATCH (a:A)-[r:a_has_b]->(b:B)
WHERE a.id='123'
RETURN COUNT(r) as count
I also want to get the top n B's ordered by the property from the relationship
MATCH (a:A)-[r:a_has_b]->(b:B)
WHERE a.id='123'
RETURN r, b
ORDER BY r.value
LIMIT 3
Now, it's clearly I am doing the same thing twice, changing the return value.
How can I combine them together to get both needed results?
You can combine collect and range:
MATCH (a:A)-[r:a_has_b]->(b:B)
WHERE a.id='123'
WITH a,
r,
b
ORDER BY r.value
RETURN a,
COUNT(r) AS count,
COLLECT([r,b])[0..3] AS rels
Related
I have a graph where a pair of nodes can have several relationships between them.
I would like to count this relationships between each pair of nodes, and set it as a parameter of each relationship.
I tried something like:
MATCH (s:LabeledExperience)-[r:NextExp]->(e:LabeledExperience)
with s, e, r, length(r) as cnt
MATCH (s2:LabeledExperience{name:s.name})-[r2:NextExp{name:r.name}]->(e2:LabeledExperience{name: e.name})
SET r2.weight = cnt
But this set the weight always to one.
I also tried:
MATCH ()-[r:NextExp]->()
with r, length(r) as cnt
MATCH ()-[r2:NextExp{name:r.name}]->()
SET r2.weight = cnt
But this takes too much time since there are more than 90k relationships and there is no index (since it is not possible to have them on edges).
They are always set to 1 because of the way you are counting.
When you group by s, e, r that is always going to result in a single row. But if you collect(r) for every s, e then you will get a collection of all of the :NextExp relationships between those two nodes.
Also, length() is for measuring the length (number of nodes) in a matched path and should not work directly on a relationship.
Match the relationship and put them in a collection for each pair of nodes. Iterate over each rel in the collection and set the size of the collection of rels.
MATCH (s:LabeledExperience)-[r:NextExp]->(e:LabeledExperience)
WITH s, e, collect(r) AS rels
UNWIND rels AS rel
SET rel.weight = size(rels)
I have a graph with one node type 'nodeName' and one relationship type 'relName'. Each node pair has 0-1 'relName' relationships with each other but each node can be connected to many nodes.
Given an initial list of nodes (I'll refer to this list as the query subset) I want to:
Find all the nodes that connect to the query subset
I'm currently doing this (which may be overly convoluted):
MATCH (a: nodeName)-[r:relName]-()
WHERE (a.name IN ['query list'])
WITH a
MATCH (b: nodeName)-[r2:relName]-()
WHERE NOT (b.name IN ['query list'])
WITH a, b
MATCH (a)--(b)
RETURN DISTINCT b
Then for each connected node (b) I want to return the SUM of the weights that connect to the query subset
For example. If node b1 has 4 edges that connect to nodes in the query subset I would like to RETURN SUM(r2.weight) AS totalWeight for b2. I actually need a list of all the b nodes ordered by totalWeight.
No. 2 is where I'm stuck. I've been reading the docs about FOREACH and reduce() but I'm not sure how to apply them here.
Speed is important as I have 30,000 nodes and 1.5M edges if you have any suggestions regarding this please throw them into the mix.
Many thanks
Matt
Why do you need so many Match statements? You can specify a nodes and b nodes in single Match statement and select only those who have a relationship between them.
After that just return b nodes and sum of the weights. b nodes will automatically be acting as a group by if it is returned along with aggregation function such as sum.
MATCH (a:nodeName)-[r:relName]-(b:nodeName)
WHERE (a.name IN ['query list']) AND NOT((b.name IN ['query list']))
RETURN b.name, sum(r.weight) as weightSum order by weightSum
I think we can simplify that query a bit.
MATCH (a: nodeName)
WHERE (a.name IN ['query list'])
WITH collect(a) as subset
UNWIND subset as a
MATCH (a)-[r:relName]-(b)
WHERE NOT b in subset
RETURN b, sum(r.weight) as totalWeight
ORDER BY totalWeight ASC
Since sum() is an aggregating function, it will make the non-aggregation variables the grouping key (in this case b), so the sum is per b node, then we order them (switch to DESC if needed).
So as a complication to this question, I basically want to do
MATCH (n:TEST) OPTIONAL MATCH (n)-[r]->() RETURN DISTINCT n, r
And I want to return n and r as one column with no repeat values. However, running
MATCH (n:TEST) OPTIONAL MATCH (n)-[r]->() UNWIND n+r AS x RETURN DISTINCT x
gives a "Type mismatch: expected List but was Relationship (line 1, column 47)" error. And this query
MATCH (n:TEST) RETURN DISTINCT n UNION MATCH ()-[n]->() RETURN DISTINCT n
Puts nodes and relationships in the same column, but the context from the first match is lost in the second half.
So how can I return all matched nodes and relationships as one minimal list?
UPDATE:
This is the final modified version of the answer query I am using
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r {properties:properties(r), id:id(r), type:type(r), startNode:id(startNode(r)), endNode:id(endNode(r))})} as n
There are a couple ways to handle this, depending on if you want to hold these within lists, or within maps, or if you want a map projection of a node to include its relationships.
If you're using Neo4j 3.1 or newer, then map projection is probably the easiest approach. Using this, we can output the properties of a node and include its relationships as a collected property:
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r)} as n
Here's what you might do if you wanted each row to be its own pairing of a node and a single one of its relationships as a list:
...
RETURN [n, r] as pair
And as a map:
...
RETURN {node:n, rel:r} as pair
EDIT
As far as returning more data from each relationship, if you check the Code results tab, you'll see that the id, relationship type, and start and end node ids are included, and accessible from your back-end code.
However, if you want to explicitly return this data, then we just need to include it in the query, using another map projection for each relationship:
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r {.*, id:id(r), type:type(r), startNode:startNode(r), endNode:endNode(r)})} as n
How can I craft a query that will return only one relation of a certain type between two nodes?
For example:
MATCH (a)-[r:InteractsWith*..5]->(b) RETURN a,r,b
Because (a) may have interacted with (b) many times, the result will contain many relations between the two. However, the relations are not identical. They have different properties because they occurred at different points in time.
But what if you're only interested in the fact that they have interacted at least once?
Instead of the result as it appears currently I'd like to receive a result that has either:
Only one random relation from the set of relations between (a) and (b)
Only those relations that fit to some criteria (e.g. "newest" or one of each type, ...)
One approach I have thought of is creating new relations of the type "hasEverInteractedWith". But there should be another way, right?
Use shortestPath() to get the quickest single result.
MATCH (a)-[:InteractsWith*..5]->(b)
WITH DISTINCT a, b
MATCH p = shortestPath((a)-[:InteractsWith*..5]->(b))
RETURN a, b, RELATIONSHIPS(p) AS r
If you want to get a specific one, you'll have to get all of the r and then filter them down, which will be slower (but provide more context).
MATCH (a)-[r:InteractsWith*..5]->(b)
WITH a, b, COLLECT(r) AS rs
RETURN a, b, REDUCE(s = HEAD(rs), r IN TAIL(rs)|CASE WHEN s.date > r.date THEN s ELSE r END)
I want to display the users whose sum of amount of transactions is greater than 5000
How do I display the relationship [:TRANS_AMOUNT] too.
My query
MATCH(c)-[r:TRANS_AMOUNT]->(e)
WITH sum(toInt(e.totalAmount))as l,c
WHERE l>5000
RETURN c,l;
The above query groups the sum by customer and checks if sum amount is greater than 5000. How do I display the relationships where this happens too?
Add the relationship to the WITH statement and return it:
MATCH (c)-[r:TRANS_AMOUNT]->(e)
WITH sum(toInt(e.totalAmount))as l, c, r
WHERE l>5000
RETURN c, l, r
You can also aggregate the relationships in order to have one row per user in the result:
MATCH (c)-[r:TRANS_AMOUNT]->(e)
WITH sum(toInt(e.totalAmount))as l, c, collect(r) as rels
WHERE l>5000
RETURN c, l, rels