How to convert data in a Set[Tuple2] into a case class

Question

One of my methods return a List[(String,Option[Int])]. I convert it into a Set calling toSet[(String,Option[Int])] method of List

question 1 - would this give me a Set[(String,Option[Int])]?

then I have a case class which can map the data of the Tuple(String,Option[Int])

case class TagCount

case class TagCount (tag:String,count:Int)

Is it possible that using the toSet method of the List, I actually get a Set[TagCount] instead of `Set[(String,Option[Int])]

I notice that toSet can accept type argument B >: (String,Option[Int]) but I can't figure out how to make TagCount >: (String,Option[Int])

def toSet[B >: A]: Set[B]

user_5205022 · Answer

Assuming tuple matches the case class field types exactly

case class TagCount(tag: String, count: Option[Int])
val s: Set[(String, Option[Int])] = Set("a" -> Some(42), "b" -> Some(11), "c" -> None)

we could get Set[TagCount] using tupled method like so

s.map(TagCount.tupled)

which outputs

res1: scala.collection.immutable.Set[TagCount] = Set(TagCount(a,Some(42)), TagCount(b,Some(11)), TagCount(c,None))

user_3347384 · Answer

You can try:

def foo(in: List[(String, Option[Int])]): Set[TagCount] =
  in.toSet.collect {
    case (key, Some(c)) => TagCount(key, c)
  }