Un-optioning an optioned Option

Question

Say I have a val s: Option[Option[String]]. It can thus have the following values:

Some(Some(\"foo\")) Some(None) None

Answer 1

You might use flatMap like the following:

val options = List(Some(Some(1)), Some(None), None)
options map (_ flatMap (a => a))

This will map the List[Option[Option[Int]]] to a List[Option[Int]].
If you just have an Option you can use it as following:

val option = Some(Some(2))
val unzippedOption = option flatMap (b => b)

This will flatten your Option[Option[Int]] to Option[Int].

Answer 2

You could use scalaz join to do this, as this is one of the monadic operations:

doubleOpt.join

Here it is in the REPL:

scala> import scalaz._; import Scalaz._
import scalaz._
import Scalaz._

scala> some(some("X")).join
res0: Option[java.lang.String] = Some(X)

scala> some(none[String]).join
res1: Option[String] = None

scala> none[Option[String]].join
res3: Option[String] = None

It's available to anything with a typeclass instance for a Monad.

Answer 3

It's a shame that flatten doesn't exist. It should.

Flatten does exist now.

As before,

s getOrElse None

(in addition to the other answers) will also do the same thing.

Answer 4

I think the conversion to the Iterable is just fine. Use these steps to go from Option[Option[String] to a single Option[String]

s.flatten.headOption 

(which returns Option[String])

Answer 5

s.flatten

followed by a bunch of characters to get me up to the minimum that stackoverflow allows

Answer 6

Well, I actually don't understand how come it could be just None (the third case). If it can really be also just None, then I would vote for Rex Kerr's answer, otherwise just .get would be enough:

scala> Some(Some("foo")).get
res0: Some[java.lang.String] = Some(foo)

scala> Some(None).get
res1: None.type = None