You could use zip:
val ms: List[Message] = ???
val as: List[Author] = ???
var sms = for ( (m, a) <- (ms zip as)) yield new SmartMessage(m, a)
If you don’t like for-comprehensions you could use map:
var sms = (ms zip as).map{ case (m, a) => new SmartMessage(m, a)}
Method zip creates collection of pairs. In this case List[(Message, Author)].
You could also use zipped method on Tuple2 (and on Tuple3):
var sms = (ms, as).zipped.map{ (m, a) => new SmartMessage(m, a)}
As you can see you don’t need pattern matching in map in this case.
Extra
List is Seq and Seq preserves order. See scala collections overview.
There are 3 main branches of collections: Seq, Set and Map.
- Seq preserves order of elements.
- Set contains no duplicate elements.
- Map contains mappings from keys to values.
List in scala is linked list, so you should prepend elements to it, not append. See Performance Characteristics of scala collections.