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.
ベストアンサー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.