Scala map function to remove fields - scala

Question

I have a list of Person objects with many fields and I can easily do:
list.map(person => person.getName)
In order to generate another collection with all the peoples names.
How can you use the map function to create a new collection with all the fields of the Person class, BUT their name though?
In other words, how can you create a new collection out of a given collection which will contain all the elements of your initial collection with some of their fields removed?

Answer 1

You can use unapply method of your case class to extract the members as tuple then remove the things that you don't want from the tuple.
case class Person(name: String, Age: Int, country: String)
// defined class Person
val personList = List(
Person("person_1", 20, "country_1"),
Person("person_2", 30, "country_2")
)
// personList: List[Person] = List(Person(person_1,20,country_1), Person(person_2,30,country_2))
val tupleList = personList.flatMap(person => Person.unapply(person))
// tupleList: List[(String, Int, String)] = List((person_1,20,country_1), (person_2,30,country_2))
val wantedTupleList = tupleList.map({ case (name, age, country) => (age, country) })
// wantedTupleList: List[(Int, String)] = List((20,country_1), (30,country_2))
// the above is more easy to understand but will cause two parses of list
// better is to do it in one parse only, like following
val yourList = personList.flatMap(person => {
Person.unapply(person) match {
case (name, age, country) => (age, country)
}
})
// yourList: List[(Int, String)] = List((20,country_1), (30,country_2))