Group 2 case classes

[Michael Larsson]

Hi group.

I have a question about grouping 2 lists of case classes together, and in this example sorting employees into EmployeeDepartment.

1. case class EmployeeDepartment( @Key("_id") id:ObjectId, name:String)

2. case class Employees( @Key("_id") id:ObjectId, name:String, employeeDepartmentName:String)

Basically, gettin a new list or map (don't know what would be better) looking like this:

(EmployeeDepartment(12456, "Marketing department"), List( 123456, "Michael Larsson" ), List( 123456, "etc" ))

Thanks

[Johannes Rudolph]

You can use `groupBy` to group a sequence of values by some
classification [1]. From your description it is not clear how your
model is supposed to work in detail so it's hard to give a more
precise answer. Some questions:

1. Why is the class called "Employees" (plural)?
2. Is the department matched by name? Why? Where do the target values come from?
3. What does your supposed result mean? Is it a valid Scala expression?

Actually, I'd guess if you start with stating your question in a way
that is fully specified you will already have 90% of the answer found
out yourself :)

HTH

Johannes

[1] http://www.scala-lang.org/api/current/index.html#scala.collection.Seq

> --
> You received this message because you are subscribed to the Google Groups
> "scala-user" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to scala-user+...@googlegroups.com.
> For more options, visit https://groups.google.com/groups/opt_out.
>
>



--
Johannes

-----------------------------------------------
Johannes Rudolph
http://virtual-void.net

[Michael Larsson]

Hi Johannes and thanks for your answear.

1. Misstake by me, I see that now. It should be Employee.

2. The Employee class is matched by it's id. I see it says :String but it should be ObjectId. The classes contains more parameters (ie, phone, email etc.) and missed that

3. When looking in to it, basically a Map("Department 1" -> List("Employee 1", "Employee 2"), "Department 2" -> List("Employee 1", "Employee 2"))

I'm I making any sense? :)

[Alec Zorab]

case class Employee(Department: DepartmentId, Name:String) //etc, etc

val employees : List[Employee] = ???

val departments : Map[DepartmentId, List[Employee]] = employees.groupBy(_.Department)

Is that what you want?

[Michael Larsson]

Don't know if this is a "good" way of doing it but it gives me the result I'm looking for:

for(department <- departments) {

      val emp= for(employee <- employees; if(department .id== employee .employeeDepartment)) yield employee 

      println(Map(department  -> emp))

}

[Alec Zorab]

uh. Are you sure?

presuming you meant to return something, not just print a map with only a single entry, that'd give you a Seq[Map[Department, List[Employee]] which probably isn't what you actually meant.

I supect you wanted this:

  case class DepartmentId(Id:String)

  case class Department(Id: DepartmentId)

  case class Employee(Department: DepartmentId, Name:String) //etc, etc

  val departments: List[Department] = ???

  val employees : List[Employee] = ???

  val employeesByDepartmentId: Map[DepartmentId, List[Employee]] = employees.groupBy(_.Department)

  val result : Map[Department, List[Employee]] = departments.map(d => d -> employeesByDepartmentId(d.Id)).toMap

--

[Michael Larsson]

Yes that's what I get of this: (ugly I know)

val departmentAndEmployees:List[Map[EmployeeDepartment, List[Employees]]] = {

      for(department <- departments) yield { Map(department -> (for(employee <- employees; if(department.id== employee.employeeDepartment)) yield employee)) }

    }

However, I'll try the example you gave me. :)

Thanks