Whats the nicest way to convert a Set into a SortedSet by providing an Ordering?

Le 16/07/2010 02:17, Ben Hutchison a écrit :
> I was slightly surprised to find that turning a Set into a SortedSet
> required me to first create an empty SortedSet, then add the Set's
> contents to it
>
> Two other options that came to mind are shown below. One was to use
> what Groovy calls the "spread operator" (does it have another name?)
> :_*, to indicate to SortedSet.apply() that I want the contents of the
> allMatches set to be added to the SortedSet.
>
> But it seems use of the spread operator is restricted to Seq. Why not Iterable?
>
> Also, I kinda hoped for some kind of sortedBy method that would turn a
> Set into a SortedSet in a single call. Impractical?
>
> Have I missed a nicer way to do it?
>
> val allMatches: Set[Match] = Set()
> val matchesSortedByDate =
> scala.collection.SortedSet.empty[Match](Ordering.by(_.date)) ++
> allMatches
>
> //type mismatch; found : scala.collection.immutable.Set[Match]
> required: Seq[?]
> //val matchesSortedByDate =
> scala.collection.SortedSet(allMatches:_*)(Ordering.by(_.date))
Maybe:
val matchesSortedByDate = SortedSet(allMatches.iterator.toList
:_*)(Ordering.by(_.date))
> //No such method
> //val matchesSortedByDate = allMatches.sortedBy(Ordering.by(_.date))
>
>
> -Ben