Hi Krishna Item # 1: I forgot to mention that the graph that I created is adapted from :
https://www.coursera.org/learn/big-data-graph-analytics/lecture/H9w6T/hands-on-connectivity-analytics-in-neo4j-with-cypher with my changes. -Kamal On Friday, July 28, 2017 at 7:36:09 PM UTC-7, Kamal Murthy wrote: > > Hi Krishna, > > Here is my take on this. > > 1. created a graph: > > > <https://lh3.googleusercontent.com/-eVGGJk4fG6c/WXvxtbNQ1XI/AAAAAAAACPk/xnu2MISmN6QDS0TwPLBuLjWT1zq5bfjigCLcBGAs/s1600/kmahi1.png> > > > > As you can see nodes "H" and "H1" are the two root nodes. > > 2. Here is the query and the result to get root nodes > > match (u:Company)<-[r]-(root) > where not((root)<--()) > return distinct root.name as Node, count(r) as Outdegree; > > Result: H and H1 with count 1 each > > 3. Here is the query and the result for OutDegree nodes for each root node" > > For H: > > > <https://lh3.googleusercontent.com/-STPjaiifKy4/WXvzmIrPY9I/AAAAAAAACPs/9vGzIV3c2nwTmev2wVBdgd6TfW9L-bz3gCLcBGAs/s1600/kmahi2.png> > > For H1: > > > <https://lh3.googleusercontent.com/-3PguhoQ6CNM/WXvzsRAcutI/AAAAAAAACPw/mb1NwtQfRoYZlyQPiC-V47JeHvqm-ij2wCLcBGAs/s1600/kmahi3.png> > > 4.Query to find common nodes > > MATCH (n:Company {name: "H1"})-[:TO*]->(v) > WITH COLLECT (v) as nodes > UNWIND nodes as v1 > > MATCH (n:Company {name: "H"})-[:TO*]->(v1) > WITH COLLECT (v1) as nodes > UNWIND nodes as v2 > > RETURN distinct v2.name; > > > > <https://lh3.googleusercontent.com/-ROB2yAgxoHU/WXv0UWog_KI/AAAAAAAACP0/3eGq5vOz5a8OirLznG5HxVNe8XYBayUAgCLcBGAs/s1600/Screen%2BShot%2B07-28-17%2Bat%2B07.34%2BPM.PNG> > > Hope this is what you are looking for. > > -Kamal > > > > On Thursday, July 27, 2017 at 11:34:38 PM UTC-7, Krishna Mahi wrote: >> >> I want to find out the common node between 2 different queries. I am >> trying hard to find out, but could not think of a solution. My motive is to >> collect top 5 outdegree nodes, collect top 5 root nodes, and return nodes >> that are common between top 5 outdegree and root nodes. I don't know how to >> merge the results, because after using "return" option in first query, no >> further statements would be executed, but without "return" option we cannot >> collect results. (Please do correct me, if I am thinking wrong). Following >> are the queries, >> // for root nodes >> match (u:Port1)<-[r]-(root) >> where not((root)<--()) >> return distinct(root.id) as Node, count(r) as Outdegree >> ORDER BY count(r) desc limit 5 >> // for outdegree nodes >> match (n:Port1)-[r]->() >> return n.id as Node, count(r) as Outdegree >> order by Outdegree DESC >> union >> match (a:Port1)-[r]->(leaf) >> where not((leaf)-->()) >> return leaf.id as Node, 0 as Outdegree limit 5 >> >> How should I combine both results, and get the output of list of nodes >> that are common? Please do help me. Thanks in advance. >> > -- You received this message because you are subscribed to the Google Groups "Neo4j" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
