Find top 3 nodes with maximum relationships - neo4j

The structure of my data base is:
( :node ) -[:give { money: some_int_value } ]-> ( :Org )
One node can have multiple relations.
I need to find top 3 nodes with the most number of relations :give with their property money holding: vx <= money <= vy

Using ORDER BY and LIMIT should solve your problem:
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

Instead of using the label 'node', maybe use something more descriptive like Person for the label so the datamodel is more clear:
MATCH (p:Person)-[r:give]->(o:Org)
WITH count(r) AS num, sum(r.money) AS total, p
RETURN p, num, total ORDER BY num DESC LIMIT 3;
I'm not sure what you mean by "their property money holding: vx <= money <= vy". If you could clarify I can update my answer accordingly. You can calculate the total of the money properties using the sum() function.
Edit
To only include relationships with money property with value greater than 10 and less 25:
MATCH (p:Person)-[r:give]->(o:Org)
WHERE r.money >= 10 AND r.money <= 25
WITH count(r) AS num, sum(r.money) AS total, p
RETURN p, num, total ORDER BY num DESC LIMIT 3;

Related

Neo4j - Counting rows with where condition

I'm trying to count the amount of rows that Neo4j will return but the count (or the query) is very slow.
Version 1 (70 sec):
MATCH (person:Person)-[:HAS_ORDER]->(order:Order)
WHERE order.timestamp >= 1632434400 AND size((order)<-[:HAS_ORDER]-(:OrderLine)-[:HAS_PRODUCT]->(:Product)) <= 20
WITH order
MATCH (order)<-[:HAS_ORDER]-(:OrderLine)-[:HAS_PRODUCT]->(product:Product)
RETURN COUNT(product);
Version 2 (68 sec.):
MATCH (person:Person)-[:HAS_ORDER]->(order:Order)
WITH size((order)<-[:HAS_ORDER]-(:OrderLine)-[:HAS_PRODUCT]->(:Product)) AS amount
WHERE order.timestamp >= 1632434400 AND amount <= 20
RETURN SUM(amount)
Using Neo4j 4.4 community with about 800000 orders and about 17000000 order lines.
Is there a more efficient way to count the rows?
These are the indexes:
CREATE INDEX idx_order_torder_id FOR (n:Order) ON (n.order_id);
CREATE INDEX idx_order_timestamp FOR (n:Order) ON (n.timestamp);
CREATE INDEX idx_person_person_id FOR (n:Person) ON (n.person_id);
CREATE INDEX idx_product_product_id FOR (n:Product) ON (n.product_id);
The amount of rows are equal to 4269011.
The EXPLAIN plan:
Please try below, hope it will give faster results
MATCH (person:Person)-[:HAS_ORDER]->(order:Order)
WHERE order.timestamp >= 1632434400
WITH order.order_id AS orderid
MATCH (o:Order { order_id: orderid })<-[:HAS_ORDER]-(:OrderLine)-[:HAS_PRODUCT]->(product:Product)
WITH COUNT(product) as productCount
WHERE productCount <= 20
RETURN productCount;
Because every order line has one product, i can skip the counting of the relation order lines to products:
MATCH (order:Order)
WHERE order.timestamp >= 1632434400
WITH order
MATCH (order)<-[:HAS_ORDER]-(orderLine:OrderLine)
WITH COUNT(orderLine) as productCount
WHERE productCount <= 20
RETURN SUM(productCount);
This query took 0m17.342s
But i managed to snoop some seconds with the following query:
MATCH (order:Order)
WHERE order.timestamp >= 1632434400
WITH order, size((order)<-[:HAS_ORDER]-(:OrderLine)) AS amount
WHERE amount <= 20
RETURN SUM(amount);
This query took 0m15.675s

Finding right path in Cypher Neo4j

I'm working with Flight Analyzer database (https://neo4j.com/graphgist/flight-analyzer).
We have there few nodes and relationships types.
Nodes:
Airport
(SEA:Airport { name:'SEA' })
Flight
(f0:Flight { date:'11/30/2015 04:24:12', duration:218, distance:1721, airline:'19977' })
Ticket
(t1f0:Ticket { class:'economy', price:1344.75 })
Relationships
Destination
(f0)-[:DESTINATION]->(ORD)
Origin
(f0)-[:ORIGIN]->(SEA)
Assign
(t1f0)-[:ASSIGN]->(f0)
Now I need to find some path and I have problem with that connection ORIGIN - FLIGHT - DESTINATION.
I need to find all airports that are connected to LAX airport with sum of ticket prices < 3000.
I tried
MATCH path = (origin:Airport { name:"LAX" })<-[r:ORIGIN|DESTINATION*..5]->(destination:Airport)
WHERE REDUCE(s = 0, n IN [x IN NODES(path) WHERE 'Flight' IN LABELS(x)] |
s + [(n)<-[:ASSIGN]-(ticket) | ticket.price][0]
) < 3000
RETURN path
but in this solution LAX can be ORIGIN and DESTINATION too. I only want to chose paths that always have the same order aiport1 <- origin - flight1 - destination -> airport2 <- origin - flight2 - destination -> aiport etc..
I need to include departure and arrive time so
flight1 date + duration < flight2 date then flight2 date + duration < flight3 date etc...
[UPDATED]
This query should check that:
matched paths have alternating ORIGIN/DESTINATION relationships, and
every departing flight lands at least 30 minutes before the next departing flight (if any), and
the sum of the ticket prices of the Flight nodes (which are every other node starting at the second one) < 3000
MATCH p = (origin:Airport {name: 'LAX'})-[:ORIGIN|DESTINATION*..5]-(destination:Airport)
WHERE
ALL(i IN RANGE(0, LENGTH(p)-1) WHERE
TYPE(RELATIONSHIPS(p)[i]) = ['ORIGIN', 'DESTINATION'][i] AND
(i%4 <> 1 OR (i + 2) > LENGTH(p) OR
(apoc.date.parse(NODES(p)[i].date,'m','MM/dd/yyyy hh:mm:ss') + NODES(p)[i].duration + 30) < apoc.date.parse(NODES(p)[i+2].date,'m','MM/dd/yyyy hh:mm:ss'))
) AND
REDUCE(s = 0, n IN [k IN RANGE(1, LENGTH(p), 2) | NODES(p)[k]] |
s + [(n)<-[:ASSIGN]-(ticket) | ticket.price][0]
) < 3000
RETURN p
The query uses the apoc.date.parse function to convert each date into the number of epoch minutes, so that a duration (assumed to also be in minutes) can be added to it.
I believe, you should create new relationships like flyto from an airport to an airport with ticket price and ticket class. it can be useful.
then you can find flights easier.
match
(a:Airport )<-[:ORIGIN]-(f:Flight)-[:DESTINATION ]->(b:Airport ),
(f)-[:ASSIGN]-(t:Ticket)
CREATE (a)-[r:FLY_TO {price:t.price,Class:t.class} ]->(b)

How to aggregate scores

I have this cypher query:
CALL db.index.fulltext.queryNodes("names","John Snow") YIELD node, score
WITH node, score MATCH (node)-[c:ACTIVE]->() WHERE c.is_active = 'True'
RETURN DISTINCT node, score ORDER BY score DESC LIMIT 10
I would like to filter results based on score percentile, so probably I need percentileDisc() as pd aggregation and then WHERE score > pd clause. How do I apply it here? percentileDisc(score, 0.5) always gives score itself.
This should work:
CALL db.index.fulltext.queryNodes("names","John Snow") YIELD node, score
WHERE EXISTS ((node)-[:ACTIVE {is_active: 'True'}]->())
WITH COLLECT({node: node, score: score}) AS data, percentileDisc(score, 0.5) AS p
UNWIND data AS d
WITH p, d
WHERE d.score > p
RETURN p, d.node AS node, d.score AS score
ORDER BY score DESC LIMIT 10

Conditional Order By Clause In Cypher Query Neo4j

Hi I want to sort my graph results by two filters..
My Cypher Query looks like this..
MATCH (n:User{user_id:304020})-[r:know]->(m:User) with m MATCH (m)-[s:like|create|share]->(o{is_active:1})
with m, s, o, (toInt(timestamp()/1000)-toInt(o.created_on))/86400 as days,
(toInt(timestamp()/1000)-toInt(o.created_on))/3600 as hours,
(1- round(o.impression_count_all/20)/50) as low_boost
with m,s,o,days,low_boost,hours,
CASE
WHEN days > 30 THEN 0.05
WHEN days >=20 AND days <=30 THEN 0.1
WHEN days >=10 AND days <=20 THEN 0.2
WHEN days >=5 AND days <=10 THEN 0.4
WHEN days >=2 AND days <=5 THEN 0.5
WHEN days =1 THEN 0.6
WHEN days < 1 THEN
CASE
WHEN hours <= 2 THEN 1
WHEN hours > 2 AND hours <= 8 THEN 0.9
WHEN hours > 8 AND hours <= 16 THEN 0.8
WHEN hours > 16 AND hours < 23 THEN 0.75
WHEN hours >= 23 AND hours <= 24 THEN 0.7
END
END as rs,
CASE
WHEN low_boost > 0 THEN low_boost
WHEN low_boost <= 0 THEN 0
END as lb
where has(o.trending_score_all) and has(o.impression_count_all) and not(o.is_featured=2)
RETURN distinct o.story_id as story_id,
(o.trending_score_all*4) as ts, (o.trending_score_all + rs + lb) as final_score,
count(s) as rel_count,max(s.activity_id) as id, toInt(o.created_on) as created_on
ORDER BY (CASE WHEN ts > 3 THEN final_score desc, rel_count desc ELSE ts) END) DESC
skip 0 limit 10;
Now If ts > 3 ,I want to sort results by both final_score and rel_count ELSE srt only by ts..
Please modify order by..
Does this much simplified query (which uses a single argument for ORDER BY) work for you?
MATCH (u:User)-[r:like]->(s:Story)
WITH s, (s.trending_score_all*4) AS ts
RETURN DISTINCT s.story_id, ts, TOINT(s.impression_count_72)
ORDER BY (CASE WHEN ts > 3 THEN ts ELSE TOINT(s.impression_count_72) END) DESC
LIMIT 10;
[EDITED]
If you need to sort by a varying number of values (depending on the situation) you have to use a workaround, as Cypher does not support that directly.
For example, suppose when (ts > 3) you wanted to order by ts DESC and then by s.story_id ASC. In this situation, you could change the above ORDER BY clause to this:
ORDER BY
(CASE WHEN ts > 3 THEN ts ELSE TOINT(s.impression_count_72) END) DESC,
(CASE WHEN ts > 3 THEN s.story_id ELSE NULL END) ASC
By using NULL (or any literal value) in this way, you can have any of the sub-sorts effectively do nothing.
1) You use pagination (skip and limit)
2) If I understand what you need, then add "else" to sort:
UNWIND RANGE(1,100) as i
WITH i, rand()*5 as x, toInt(rand()*10) as y
RETURN i, x, y, CASE WHEN x>3 THEN 1 ELSE 0 END as for_sort
ORDER BY
CASE
WHEN for_sort=1 THEN x ELSE y END DESC,
CASE
WHEN for_sort=1 THEN y ELSE x END DESC

Cypher: find a path which takes the maximum valued step each time

I am trying to write a cypher query that finds a path between nodes a and b such that each step has the maximum timestamp value out of all available alternatives that is less than 15.
Here is my query so far, it does everything except for select the maximum possible timestamp at each step. How do I express this condition?
MATCH path=(a:NODE)-[rs:PARENT*]->(b:NODE)
WHERE a.name = 'SOME_VALUE' and b.name = 'SOME_OTHER_VALUE' AND ALL (r IN rs
WHERE r.timestamp < 15)
RETURN path
This is just awful sudo code but I think it expresses what I am looking for
MATCH path=(a:NODE)-[rs:PARENT*]->(b:NODE)
WHERE a.name = 'SOME_VALUE' and b.name = 'SOME_OTHER_VALUE' AND ALL (r IN rs
WHERE r.timestamp < 15 AND r.timestamp = max(allPossibleRsForThisStep))
RETURN path
Can this kind of query be written in cypher?
It won't be fast in cypher, it's possible to compute all maximum values first and then do what you want to do by compare the max value in a list with the current value.
Something like this (not sure if it works)
WITH range(1,10) as max_vals // a list with 10 values (actual values are not that important)
MATCH (a:NODE)-[rs:PARENT*..10]->(b:NODE)
WHERE a.name = 'SOME_VALUE' and b.name = 'SOME_OTHER_VALUE'
WITH a,b,
map(idx in range(0,size(rs)) |
max_vals[idx] = case when max_vals[idx]<rs[idx].timestamp then rs[idx].timestamp else max_vals[idx] end ), max_vals
MATCH path=(a)-[rs:PARENT*..10]->(b)
AND ALL (idx in range(0,size(rs) WHERE rs[idx].timestamp < 15 AND rs[idx].timestamp = max_vals[idx])
RETURN path

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