Cartesian results

General
1

I successfully ran a simple optimization query for a single project: ‘Project A’ and got 2 set of lists

List 1: Employee label {Employee A, Employee B, Employee C}

List 2: Relationship property Hours {50,60,10}

My objective is to create relationship between Project and Employee and set Hours as a property.

Employee A will_work 50 hours on Project A

Employee B will_work 60 hours on Project A

Employee C will_work 10 hours on Project A

Without success I tried nested FOREACH just to receive a cartesian results (9 options).

2

You can get the Cartesian product between two lists with a double unwind, as follows:

with ["A", "B", "C"] as employees, [50, 60, 10] as hours
unwind employees as employee
unwind hours as hour
return employee, hour

Are you looking to create those three relationships from list1 and list2?

3

Thank you again for taking the time, after rereading my request I realized that I didn't do a good job describing the issue.

My objective is to avoid a cartesian query and unwind the two lists and create relationship.
the desired result are 3 relationships (not 9) :

Employee A will_work 50 hours on Project A
Employee B will_work 60 hours on Project A
Employee C will_work 10 hours on Project A

this is a "test" that I ran without success to set a property (Temp) on Employee label

WITH "Employee A,Employee B,Employee C" as vEmployee, "50,60,10" as vHRS

FOREACH (name IN split(vEmployee, ",")|FOREACH (hrs IN split(vHRS, ",")|

MERGE (n:Employee {ID:name})

SET n.Temp=hrs))

so to summarize I need to unwind two coordinated list and create 3 relationships and set accordingly Hours.
(Employee)-[will_work]->(Project A)
SET Hours=vHRS

4

That can be several approaches to solve this depending on how your are setting up the data. He is an example of a way to approach it. Do you have your entire query that we can look at to modify it? Do you intend to read the data from a spreadsheet?

with ["A", "B", "C"] as employees, [50, 60, 10] as hours
unwind range(0,size(employees)-1) as index
merge(e:Employee{name:employees[index]})
merge(p:Project{name:"A"})
merge(e)-[r:WILL_WORK]->(p)
SET r.hours=hours[index]
return e, p

Screen Shot 2022-10-20 at 1.01.09 PM.png

5

This is amazing , thank you for taking the time.