Easy idiomatic way to define Ordering for a simple case class

Solution 1:

My personal favorite method is to make use of the provided implicit ordering for Tuples, as it is clear, concise, and correct:

case class A(tag: String, load: Int) extends Ordered[A] {
  // Required as of Scala 2.11 for reasons unknown - the companion to Ordered
  // should already be in implicit scope
  import scala.math.Ordered.orderingToOrdered

  def compare(that: A): Int = (this.tag, this.load) compare (that.tag, that.load)
}

This works because the companion to Ordered defines an implicit conversion from Ordering[T] to Ordered[T] which is in scope for any class implementing Ordered. The existence of implicit Orderings for Tuples enables a conversion from TupleN[...] to Ordered[TupleN[...]] provided an implicit Ordering[TN] exists for all elements T1, ..., TN of the tuple, which should always be the case because it makes no sense to sort on a data type with no Ordering.

The implicit ordering for Tuples is your go-to for any sorting scenario involving a composite sort key:

as.sortBy(a => (a.tag, a.load))

As this answer has proven popular I would like to expand on it, noting that a solution resembling the following could under some circumstances be considered enterprise-grade™:

case class Employee(id: Int, firstName: String, lastName: String)

object Employee {
  // Note that because `Ordering[A]` is not contravariant, the declaration
  // must be type-parametrized in the event that you want the implicit
  // ordering to apply to subclasses of `Employee`.
  implicit def orderingByName[A <: Employee]: Ordering[A] =
    Ordering.by(e => (e.lastName, e.firstName))

  val orderingById: Ordering[Employee] = Ordering.by(e => e.id)
}

Given es: SeqLike[Employee], es.sorted() will sort by name, and es.sorted(Employee.orderingById) will sort by id. This has a few benefits:

• The sorts are defined in a single location as visible code artifacts. This is useful if you have complex sorts on many fields.
• Most sorting functionality implemented in the scala library operates using instances of Ordering, so providing an ordering directly eliminates an implicit conversion in most cases.

Solution 2:

object A {
  implicit val ord = Ordering.by(unapply)
}

This has the benefit that it is updated automatically whenever A changes. But, A's fields need to be placed in the order by which the ordering will use them.

Solution 3:

To summarize, there are three ways to do this:

1. For one-off sorting use .sortBy method, as @Shadowlands have showed
2. For reusing of sorting extend case class with Ordered trait, as @Keith said.

3. Define a custom ordering. The benefit of this solution is that you can reuse orderings and have multiple ways to sort instances of the same class:

   case class A(tag:String, load:Int)
   
   object A {
     val lexicographicalOrdering = Ordering.by { foo: A => 
       foo.tag 
     }
   
     val loadOrdering = Ordering.by { foo: A => 
       foo.load 
     }
   }
   
   implicit val ord = A.lexicographicalOrdering 
   val l = List(A("words",1), A("article",2), A("lines",3)).sorted
   // List(A(article,2), A(lines,3), A(words,1))
   
   // now in some other scope
   implicit val ord = A.loadOrdering
   val l = List(A("words",1), A("article",2), A("lines",3)).sorted
   // List(A(words,1), A(article,2), A(lines,3))
   

Answering your question Is there any standard function included into the Scala that can do magic like List((2,1),(1,2)).sorted

There is a set of predefined orderings, e.g. for String, tuples up to 9 arity and so on.

No such thing exists for case classes, since it is not easy thing to roll off, given that field names are not known a-priori (at least without macros magic) and you can't access case class fields in a way other than by name/using product iterator.

Solution 4:

The unapply method of the companion object provides a conversion from your case class to an Option[Tuple], where the Tuple is the tuple corresponding to the first argument list of the case class. In other words:

case class Person(name : String, age : Int, email : String)

def sortPeople(people : List[Person]) = 
    people.sortBy(Person.unapply)

Solution 5:

The sortBy method would be one typical way of doing this, eg (sort on tag field):

scala> l.sortBy(_.tag)foreach(println)
A(article,2)
A(lines,7)
A(words,50)