What's the best way to inverse sort in scala?

Question

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 conveinient way of using sortBy but getting a reverse sort? I would rather not need to use sortWith.

Answer 1

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.

Answer 2

list.sortBy(_.size)(Ordering[Int].reverse)