Does Scala have any collections that sort by value instead of key?

Question

For example the following code produces the order Stock1, Stock2, Stock3

var tm = TreeMap("Stock1" -> 4.2, "Stock3" -> 3.7, "Stock2" -> 5.9)

What I'm looking for is a collection or technique that can produce the following

Stock2, Stock1, Stock3 which represents the prices in descending order

Answer 1

scala> val sorted1 = tm.toList.sortBy (_._2)                                                           
sorted1: List[(java.lang.String, Double)] = List((Stock3,3.7), (Stock1,4.2), (Stock2,5.9))

scala> val sorted2 = tm.toList.sortBy (_._1)
sorted2: List[(java.lang.String, Double)] = List((Stock1,4.2), (Stock2,5.9), (Stock3,3.7))

Revert sorted1, to get it descending, or sortWith:

scala> val sorted3 = tm.toList.sortWith (_._2 > _._2) 
sorted3: List[(java.lang.String, Double)] = List((Stock2,5.9), (Stock1,4.2), (Stock3,3.7))

Version 4, sort by -x (minus x):

scala> val sorted4 = tm.toList.sortBy (-_._2)   
sorted4: List[(java.lang.String, Double)] = List((Stock2,5.9), (Stock1,4.2), (Stock3,3.7))

Answer 2

I don't think there is a built-in collection that keeps the map sorted by value so that you can read out the sorted values in linear time, but how to sort a scala.collection.Map[java.lang.String, Int] by its values? shows ways to get the entries out sorted by value.