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[Foo]): Option[Bar] = {
  if(myList.nonEmpty) Some(Bar(myList))
  else None
}

case class Bar(fooList: List[Foo]) {}

For some reason, my solution feels rather inelegant, and not the Scala way. It seems I should be able to use a method on List to do this sort of thing, but I can't wrap my head around it.

Is there a more Scala-like way to do this?

I first thought that you wanted the head of the list when present but then I re-read your question I really do not understand why you would want to do this. It's hard finding a idiomatic way to do stuff when it's not a common operation. It would be nice if you could clarify what you are going to use this functionality for. — Johan S, Jan 30 '15 at 20:47
From a wider perspective, it was already clear to me that there are some design problems in this class I'm working with, but I chose not to refactor the class and instead try to write something elegant given the weird restraints. My `if-else` solution merely felt lacking, and I thought there could be a better way to do it. — Cory Klein, Jan 30 '15 at 20:52
I've been doing Scala for about 9 months, and I truly find it to be awesome, but it's both a blessing and a curse that you can do stuff exactly how you want to do it. I would have refactored the crap out of that code. Well thanks for the clarification, and GLHF coding Scala! — Johan S, Jan 30 '15 at 20:57

score 31 · Accepted Answer · answered Jan 31 '15 at 07:16

31

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

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

answered Jan 31 '15 at 07:16

Alexey Romanov

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score 17 · Answer 2 · answered Jan 23 '19 at 13:16

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)))

Lee · Answer 3 · 2015-01-30T21:00:07.157

14

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

edited Jan 30 '15 at 21:00

answered Jan 30 '15 at 20:47

Lee

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3

The force is strong with this one. – Cory Klein Jan 30 '15 at 20:48
@CoryKlein - I'm not sure this is actually better than an explicit match since it seems a bit messy to ignore the value. – Lee Jan 30 '15 at 20:57
1

But the sheer brevity of it! – Cory Klein Jan 30 '15 at 21:40

dk14 · Answer 4 · 2015-01-30T21:09:35.030

3

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

edited Jan 30 '15 at 21:09

answered Jan 30 '15 at 21:03

dk14

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score 3 · Answer 5 · answered Jan 30 '15 at 21:49

3

How about:

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

Seems pretty scala-esque to me.

answered Jan 30 '15 at 21:49

Ben Reich

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You could also use `case l if l.nonEmpty`. – Alexey Romanov Jan 31 '15 at 07:17

Convert a List into an Option if it is populated

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