How do I sort a collection of Lists in lexicographic order in Scala?

Question

If A has the Ordered[A] trait, I\'d like to be able to have code that works like this

val collection: List[List[A]] = ... // constr         


        

Answer 1

Inspired by Daniel's comment, here is a recursive version:

implicit def toOrdered[A <% Ordered[A]](list1: List[A]): Ordered[List[A]] = { 
  @scala.annotation.tailrec
  def c(list1:List[A], list2:List[A]): Int = {
    (list1, list2) match {
      case (Nil, Nil) => 0
      case (x::xs, Nil) => 1
      case (Nil, y::ys) => -1
      case (x::xs, y::ys) => (x compare y) match {
        case 0 => c(xs, ys)
        case i => i
      }
    }
  }
  new Ordered[List[A]] {
    def compare(list2: List[A]): Int = c(list1, list2)
  }
}

With respect to the comment: I used to think it's more a matter of taste. Sometimes it's easier to verify correctness on a recursive function, and certainly your version is short enough that there is no compelling reason to prefer recursive.

I was intrigued by the performance implications though. So I tried to benchmark it: see http://gist.github.com/468435. I was surprised to see that the recursive version is faster (assuming I did the benchmark correctly). The results still hold true for list of about length 10.