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Find top 3 nodes with maximum relationships


Find top 3 nodes with maximum relationships

By : Robet Habibi
Date : November 22 2020, 02:59 PM
wish helps you The structure of my data base is: , Using ORDER BY and LIMIT should solve your problem:
code :
Match ( n:node ) -[r:give { money: some_int_value } ]-> ( :Org )
RETURN n
ORDER BY count(r) DESC //Order by the number of relations each node has
LIMIT 3 //We only want the top 3 nodes


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neo4j - Find all nodes with relationships to nodes that match a list of property values of x length

neo4j - Find all nodes with relationships to nodes that match a list of property values of x length


By : king kong
Date : March 29 2020, 07:55 AM
seems to work fine You can use the IN operator to test if a value matches a value in a list.
code :
MATCH (n:Persons)-[:KNOWS]-(x:Skill)
WHERE x.name IN ['PHP', 'JavaScript', 'MySql']
RETURN n, COUNT(x) AS nSkills
ORDER BY nSkills DESC;
MATCH (n:Persons)-[:KNOWS]-(x:Skill)
WHERE x.name IN {needed_skills}
RETURN n, COUNT(x) AS nSkills
ORDER BY nSkills DESC;
Neo4j: Query to find the nodes with most relationships, and their connected nodes

Neo4j: Query to find the nodes with most relationships, and their connected nodes


By : M Shahab
Date : March 29 2020, 07:55 AM
To fix the issue you can do Aggregations only have context based on the non-aggregation columns, and with your match, a unique relationship will only occur once in your results.
So your first query is asking for each relationship on a row, and the count of that particular relationship, which is 1.
code :
MATCH (a)-[:WRITES]->(b)
RETURN b, COLLECT(a) as authors
ORDER BY SIZE(authors) DESC LIMIT 10
MATCH (b:Book)
WITH b, SIZE(()-[:WRITES]->(b)) as authorCnt
ORDER BY authorCnt DESC LIMIT 10
MATCH (a)-[:WRITES]->(b)
RETURN b, a
Delete duplicate relationships between two nodes with Cypher where multiple relationships exist between two nodes

Delete duplicate relationships between two nodes with Cypher where multiple relationships exist between two nodes


By : Bonnie Starrett
Date : March 29 2020, 07:55 AM
it should still fix some issue I would like to delete duplicate relationships between two nodes with Cypher where multiple relationships exist between the nodes. As an example, given: , Assuming you have labels on your nodes, this should work:
code :
MATCH (a:A)-[r]->(b:B)
WITH a, type(r) as type, collect(r) as rels, b
WHERE size(rels) > 1
UNWIND tail(rels) as rel
DELETE rel
Find maximum distance given a grid of nodes and a set of source nodes

Find maximum distance given a grid of nodes and a set of source nodes


By : user3549311
Date : March 29 2020, 07:55 AM
I hope this helps . Yes You can solve this problem using a concept called multisource BFS . The concept of multisource BFS is just like BFS but here is the minor difference .
In normal BFS we just push 1 node in the queue initially while in multisource BFS we push all the source nodes .
code :
import java.util.*;
class Pair{

    int x , y , dist;

    Pair(int first , int second){

        this.x = first;
        this.y = second;
        this.dist = 0;
    }

    Pair(int first , int second , int dist){

        this.x = first;
        this.y = second;
        this.dist = dist;
    }
}

public class MultisourceBFS {
    public static void main(String[] args) {
        int[][] sourceNodes = new int[][] {
            {0, 0, 1, 0},
            {1, 0, 0, 0},
            {0, 0, 0, 0},
            {0, 0, 0, 1}
        };

        int maximumDistance = computeMaximumDistance(sourceNodes);
        System.out.println(String.format(
            "The maximum distance in this grid is %d.",
            maximumDistance));
    }

    private static int computeMaximumDistance(int[][] sourceNodes) {
        int m = sourceNodes.length;
        int n = sourceNodes[0].length;

        int maximumDistance = 0;

        int xx[] = {0 , 0 , 1 , -1}; // Normal array to go to up , down , left , right;
        int yy[] = {1 , -1 , 0 , 0}; // Normal array to go to up ,down , left , right

        Queue<Pair> q = new LinkedList<Pair>();
        boolean isVisited[][] = new boolean[m][n]; // An array to check if a cell is visited or not .

        // I am adding all the source nodes to the queue

        for(int i = 0 ; i < m ; ++i)
            for(int j = 0 ; j < n ; ++j)
                if(sourceNodes[i][j]==1){
                    q.add(new Pair(i , j));
                    isVisited[i][j] = true;
                }

        // Now it is going to be normal BFS

        while(!q.isEmpty()){

            Pair node = q.remove();

            for(int k = 0 ; k < 4 ; ++k){

                int new_i = node.x + xx[k];
                int new_j = node.y + yy[k];

                if(new_i >= 0 && new_i < m && new_j >= 0 && new_j < n && isVisited[new_i][new_j]==false){

                    maximumDistance = Math.max(node.dist + 1 , maximumDistance);

                    isVisited[new_i][new_j] = true;

                    q.add(new Pair(new_i , new_j , node.dist + 1));
                }
            }
        }


        return maximumDistance;
    }

}
In Neo4j, can one find all nodes whose relationships are a superset of another node's relationships?

In Neo4j, can one find all nodes whose relationships are a superset of another node's relationships?


By : Biju Philip
Date : March 29 2020, 07:55 AM
I wish this helpful for you Given the following contrived database:
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