What is the best way to do an inverse sort in scala? I imagine the following is somewhat slow.
list.sortBy(_.size).reverse
Is there a conv
maybe to shorten it a little more:
def Desc[T : Ordering] = implicitly[Ordering[T]].reverse
List("1","22","4444","333").sortBy( _.size )(Desc)
Both sortWith
and sortBy
have a compact syntax:
case class Foo(time:Long, str:String)
val l = List(Foo(1, "hi"), Foo(2, "a"), Foo(3, "X"))
l.sortWith(_.time > _.time) // List(Foo(3,X), Foo(2,a), Foo(1,hi))
l.sortBy(- _.time) // List(Foo(3,X), Foo(2,a), Foo(1,hi))
l.sortBy(_.time) // List(Foo(1,hi), Foo(2,a), Foo(3,X))
I find the one with sortWith
easier to understand.
There may be the obvious way of changing the sign, if you sort by some numeric value
list.sortBy(- _.size)
More generally, sorting may be done by method sorted with an implicit Ordering, which you may make explicit, and Ordering has a reverse (not the list reverse below) You can do
list.sorted(theOrdering.reverse)
If the ordering you want to reverse is the implicit ordering, you can get it by implicitly[Ordering[A]] (A the type you're ordering on) or better Ordering[A]. That would be
list.sorted(Ordering[TheType].reverse)
sortBy is like using Ordering.by, so you can do
list.sorted(Ordering.by(_.size).reverse)
Maybe not the shortest to write (compared to minus) but intent is clear
Update
The last line does not work. To accept the _
in Ordering.by(_.size)
, the compiler needs to know on which type we are ordering, so that it may type the _
. It may seems that would be the type of the element of the list, but this is not so, as the signature of sorted is
def sorted[B >: A](ordering: Ordering[B])
. The ordering may be on A
, but also on any ancestor of A
(you might use byHashCode : Ordering[Any] = Ordering.by(_.hashCode)
). And indeed, the fact that list is covariant forces this signature.
One can do
list.sorted(Ordering.by((_: TheType).size).reverse)
but this is much less pleasant.