neo4j indegree outdegree union - neo4j

Question

I want to compute Indegree and Outdegree and return a graph that has a connection between top 5 Indegree nodes and top 5 Outdegree nodes. I have written a code as
match (a:Port1)<-[r]-()
return a.id as NodeIn, count(r) as Indegree
order by Indegree DESC LIMIT 5
union
match (n:Port1)-[r]->()
return n.id as NodeOut, count(r) as Outdegree
order by Outdegree DESC LIMIT 5
union
match p=(u:Port1)-[:LinkTo*1..]->(t:Port1)
where u.id in NodeIn and t.id in NodeOut
return p
I get an error as
All sub queries in an UNION must have the same column names (line 4, column 1 (offset: 99)) "union"
What are the changes that I need to do to the code?

Answer 1

There's a few things we can improve.
The matches you're doing isn't the most efficient way to get incoming and outgoing degrees for relationships.
Also, UNION can only be used to combine query results with identical columns. In this case, we won't even need UNION, we can use WITH to pipe results from one part of a query to another, and COLLECT() the nodes you need in between.
Try this query:
match (a:Port1)
with a, size((a)<--()) as Indegree
order by Indegree DESC LIMIT 5
with collect(a) as NodesIn
match (a:Port1)
with NodesIn, a, size((a)-->()) as Outdegree
order by Outdegree DESC LIMIT 5
with NodesIn, collect(a) as NodesOut
unwind NodesIn as NodeIn
unwind NodesOut as NodeOut
// we now have a cartesian product between both lists
match p=(NodeIn)-[:LinkTo*1..]->(NodeOut)
return p
Be aware that this performs two NodeLabelScans of :Port1 nodes, and does a cross product of the top 5 of each, so there are 25 variable length path matches, which can be expenses, as this generates all possible paths from each NodeIn to each NodeOut.
If you only one the shortest connection between each, then you might try replacing your variable length match with a shortestPath() call, which only returns the shortest path found between each two nodes:
...
match p = shortestPath((NodeIn)-[:LinkTo*1..]->(NodeOut))
return p
Also, make sure your desired direction is correct, as you're matching nodes with the highest in degree and getting an outgoing path to nodes with the highest out degree, that seems like it might be backwards to me, but you know your requirements best.