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 Ben Lings' answer, I wrote my own version of sort:

def sort[A : Ordering](coll: Seq[Iterable[A]]) = coll.sorted

which is equivalent to:

def sort[A](coll: Seq[Iterable[A]])(implicit ordering: Ordering[A]) = coll.sorted

Note that ordering is implicitly converted to Ordering[Iterable[A]].

Examples:

scala> def sort[A](coll: Seq[Iterable[A]])(implicit ordering: Ordering[A]) = coll.sorted
sort: [A](coll: Seq[Iterable[A]])(implicit ordering: Ordering[A])Seq[Iterable[A]]

scala> val coll = List(List(1, 3), List(1, 2), List(0), Nil, List(2))
coll: List[List[Int]] = List(List(1, 3), List(1, 2), List(0), List(), List(2))

scala> sort(coll)
res1: Seq[Iterable[Int]] = List(List(), List(0), List(1, 2), List(1, 3), List(2))

It was asked how to supply your own comparison function (say, _ > _ instead of _ < _). It suffices to use Ordering.fromLessThan:

scala> sort(coll)(Ordering.fromLessThan(_ > _))
res4: Seq[Iterable[Int]] = List(List(), List(2), List(1, 3), List(1, 2), List(0))

Ordering.by allows you to map your value into another type for which there is already an Ordering instance. Given that also tuples are ordered, this can be useful for lexicographical comparison of case classes.

To make an example, let's define a wrapper of an Int, apply Ordering.by(_.v), where _.v extracts the underlying value, and show that we obtain the same result:

scala> case class Wrap(v: Int)
defined class Wrap

scala> val coll2 = coll.map(_.map(Wrap(_)))
coll2: List[List[Wrap]] = List(List(Wrap(1), Wrap(3)), List(Wrap(1), Wrap(2)), List(Wrap(0)), List(), List(Wrap(2)))

scala> sort(coll2)(Ordering.by(_.v))
res6: Seq[Iterable[Wrap]] = List(List(), List(Wrap(0)), List(Wrap(1), Wrap(2)), List(Wrap(1), Wrap(3)), List(Wrap(2)))

Finally, let's do the same thing on a case class with more members, reusing the comparators for Tuples:

scala> case class MyPair(a: Int, b: Int)
defined class MyPair

scala> val coll3 = coll.map(_.map(MyPair(_, 0)))
coll3: List[List[MyPair]] = List(List(MyPair(1,0), MyPair(3,0)), List(MyPair(1,0), MyPair(2,0)), List(MyPair(0,0)), List(), List(MyPair(2,0)))

scala> sort(coll3)(Ordering.by(x => (x.a, x.b)))
res7: Seq[Iterable[MyPair]] = List(List(), List(MyPair(0,0)), List(MyPair(1,0), MyPair(2,0)), List(MyPair(1,0), MyPair(3,0)), List(MyPair(2,0)))

EDIT:

My definition of sort above is deprecated in 2.13:

warning: method Iterable in object Ordering is deprecated (since 2.13.0):
Iterables are not guaranteed to have a consistent order; if using a type
with a consistent order (e.g. Seq), use its Ordering (found in the
Ordering.Implicits object)

Use instead:

def sort[A](coll: Seq[Seq[A]])(implicit ordering: Ordering[A]) = {
  import Ordering.Implicits._
  coll.sorted
}