Convert a List into an Option if it is populated

Question

I have a method that should convert a list to an Option of an object, or None if the list is empty.

def listToOption(myList: List[F         


        

Answer 1

Lee's answer is good, but I think this corresponds to the intention a bit more clearly:

Option(myList).filter(_.nonEmpty).map(Bar)

Answer 2

import scalaz._; import Scalaz._
myList.toNel.map(Bar)

toNel - is "to non-empty list" here, it returns Option[NonEmptyList] for safety:

scala> case class Bar(a: NonEmptyList[Int])
defined class Bar

scala> List(1,2,3).toNel.map(Bar)
res64: Option[Bar] = Some(Bar(NonEmptyList(1, 2, 3)))

scala> List[Int]().toNel.map(Bar)
res65: Option[Bar] = None

Answer 3

myList.headOption.map(_ => Bar(myList))

Answer 4

How about:

Some(myList) collect { case(l@hd::tl) => Bar(l) }

Seems pretty scala-esque to me.

Answer 5

Starting Scala 2.13, Option has a when builder:

Option.when(condition)(result)

which in our case gives:

Option.when(myList.nonEmpty)(Bar(myList))
// val myList = List[Int]()    =>    Option[Bar] = None
// val myList = List(1, 2)     =>    Option[Bar] = Some(Bar(List(1, 2)))

Also note Option.unless which promotes the opposite condition:

Option.unless(myList.isEmpty)(Bar(myList))
// val myList = List[Int]()    =>    Option[Bar] = None
// val myList = List(1, 2)     =>    Option[Bar] = Some(Bar(List(1, 2)))