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Property graph (Neo4j) design: Single node with multiple relations or new nodes for each event occurence?


Property graph (Neo4j) design: Single node with multiple relations or new nodes for each event occurence?

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Property graph (Neo4j) design: Single node with multiple relations or new nodes for each event occurence?
Tag : development , By : user104292
Date : November 25 2020, 07:06 PM


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Tag : neo4j , By : SilverRuby
Date : March 29 2020, 07:55 AM
To fix this issue How about something like this: collect answers per group, collect answers per respondent, filter on where all answers for a respondent are present in the answers for a group. I think then the count on respondent should be ok (you'll have to extract the answer texts since the answers are already collected.)
MATCH (g:WeightingGroup)<-[:BelongsTo]-(a)
WITH g, collect(a) as apg
MATCH (r:Respondent)-[:Answered]->(a)
WITH g, apg, r, collect(a) as apr
WHERE ALL(a IN apr WHERE a IN apg)
RETURN g.Weight, EXTRACT(a IN apg | a.Text), count(r) as n

Neo4J: find a sub-graph of arbitrary depth with nodes connected by a given set of relations?


Tag : neo4j , By : Rik
Date : March 29 2020, 07:55 AM
this will help This query will return just the nodes, as you stated in your question:
MATCH (n)-[:FRIEND_OF|COLLEAGUE_WITH*]->(m)
RETURN n, m;
MATCH (n)-[r:FRIEND_OF|COLLEAGUE_WITH*]->(m)
RETURN n, r, m;

neo4j: Replace multiple nodes with same property by one node


Tag : java , By : Stephen Dewar
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Let's say I have a property "name" of nodes in neo4j. Now I want to enforce that there is maximally one node for a given name by identifying all nodes with the same name. More precisely: If there are three nodes where name is "dog", I want them to be replaced by just one node with name "dog", which: , I have made a testcase with following schema:
CREATE (n1:TestX {name:'A', val1:1})
CREATE (n2:TestX {name:'B', val2:2})
CREATE (n3:TestX {name:'B', val3:3})
CREATE (n4:TestX {name:'B', val4:4})
CREATE (n5:TestX {name:'C', val5:5})

MATCH (n6:TestX {name:'A', val1:1}) MATCH (m7:TestX {name:'B', val2:2}) CREATE (n6)-[:TEST]->(m7)
MATCH (n8:TestX {name:'C', val5:5}) MATCH (m10:TestX {name:'B', val3:3}) CREATE (n8)<-[:TEST]-(m10)
//copy all properties
MATCH (n:TestX), (m:TestX) WHERE n.name = m.name AND ID(n)<ID(m) WITH n, m SET n += m;

//copy all outgoing relations
MATCH (n:TestX), (m:TestX)-[r:TEST]->(endnode) WHERE n.name = m.name AND ID(n)<ID(m) WITH n, collect(endnode) as endnodes
FOREACH (x in endnodes | CREATE (n)-[:TEST]->(x));

//copy all incoming relations
MATCH (n:TestX), (m:TestX)<-[r:TEST]-(endnode) WHERE n.name = m.name AND ID(n)<ID(m) WITH n, collect(endnode) as endnodes
FOREACH (x in endnodes | CREATE (n)<-[:TEST]-(x));

//delete duplicates
MATCH (n:TestX), (m:TestX) WHERE n.name = m.name AND ID(n)<ID(m) detach delete m;

Neo4j : How to remove or filter node from set of all nodes with conditions on relations


Tag : neo4j , By : Harvey
Date : March 29 2020, 07:55 AM
it should still fix some issue for the clarification, it is a slightly tricky query. Since it seems like we'll be returning most of the images anyway, it should be okay to start the match with the user and all :Images. We can then use an OPTIONAL MATCH on the :RATES relationship between the user and the image, and r will be null if no such relationship exists. That will let us use WITH with one more predicate where we'll keep only rows where the relationship doesn't exist to the given :Image, or where the user has rated that :Image greater than 5.
MATCH (u:User{userId:"123"}), (i:Image)
OPTIONAL MATCH (u)-[r:RATES]->(i)
WITH i
WHERE r IS NULL OR r.rate > 5
RETURN i

Multiple Nodes graph to single node graph in python


Tag : python , By : bdurbin
Date : March 29 2020, 07:55 AM
I hope this helps . I have a Directed Acyclic graph, something like this: , If you have a list of the node sets you can do something like this -
from itertools import product
s =[[1, 2], [3, 4, 5], [6, 7]]
list(product(*s))
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