How to convert a List[Any] in Option[List[String]] - scala

Question

Given a List[Any], I want to convert it to a Option[List[String]]
def convert(ls: List[Any]) : Option[List[String]] = {
if (ls.forall(_.getClass == classOf[String]))
Some(ls.map(_.asInstanceOf[String]))
else
None
}
Is there a better way ?

Answer 1

Like:
scala> val bag = List("a", 1, 2.0, "b")
bag: List[Any] = List(a, 1, 2.0, b)
scala> def unbagged(vs: List[Any]): Option[List[String]] = Option(vs collect { case s: String => s}) filter (_.nonEmpty)
unbagged: (vs: List[Any])Option[List[String]]
scala> unbagged(bag)
res0: Option[List[String]] = Some(List(a, b))
scala> unbagged(List(1, 3.14))
res1: Option[List[String]] = None
or, addressing the use case:
scala> def strung(vs: List[Any]): Option[List[String]] = (Option(vs) filter (_ forall { case _: String => true case _ => false })).asInstanceOf[Option[List[String]]]
strung: (vs: List[Any])Option[List[String]]
scala> strung(bag)
res3: Option[List[String]] = None
scala> strung(List("a","b","c"))
res4: Option[List[String]] = Some(List(a, b, c))

Answer 2

There are already quite a few answers, but I think they're all cleverer than needed. The initial proposal in the question is not that bad, except I would replace the getClass test by isInstanceOf:
def convert(ls: List[Any]): Option[List[String]] = {
if (ls.forall(_.isInstanceOf[String]))
Some(ls.map(_.asInstanceOf[String]))
else
None
}
It's functional, copies the list only once. Yes, the list is traversed twice, but typically that it still going to be faster than throwing an exception (which is usually slow -- if you really want to go that route, at least use a ControlThrowable, which does not record the stack trace when constructed).
Besides, as #som-snytt pointed out quietly in a comment, due to erasure, you don't even need to cast all the elements inside the list. You may just as well cast the list, which, after having checked that all elements are Strings, is just as safe as any other casts:
def convert(ls: List[Any]): Option[List[String]] = {
if (ls.forall(_.isInstanceOf[String]))
Some(ls.asInstanceOf[List[String]])
else
None
}
This is simply the most efficient version, because there is no list copying at all.

Answer 3

There is a toString method, which can make a String from any object. So if it's not a requirement that all elements of your original List should actually be String elements, you can do this:
import scala.util.Try
def convert(l: List[Any]) : Option[List[String]] = {
Try(l.map(_.toString)).toOption
}
Try will return Some(x) if it succeeds and obtains a value x, or None otherwise.
If the conversion should succeed only if all elements are Strings, then we can do the conversion inside Try (at the first failure, Try will fail and hence we'll get None)
import scala.util.Try
def convert(l: List[Any]) : Option[List[String]] = {
Try(l.map(_.asInstanceOf[String])).toOption
}

Answer 4

I would recommend to use pattern matching:
def convert(l: List[Any]) : Option[List[String]] = {
Try(list.collect{
case s : String => s
case x : Any => throw new Exception()
}).toOption
}

Answer 5

Code is a bit ugly but it works.
It doesnt use classOf but it uses pattern matching:
scala> val l1 = List("a", 1, 12.0)
l1: List[Any] = List(a, 1, 12.0)
scala> val l2 = List[Any]("a", "b", "c")
l2: List[Any] = List(a, b, c)
scala> def convert(list: List[Any]) = {
| list.foldLeft(Some(List()): Option[List[String]]) { (x, y) =>
| x match {
| case Some(l) =>
| y match {
| case elem: String => Some(l ::: List(elem))
| case _ => None
| }
| case None => None
| }
| }
| }
convert: (list: List[Any])Option[List[String]]
scala> convert(l1)
res12: Option[List[String]] = None
scala> convert(l2)
res13: Option[List[String]] = Some(List(a, b, c))
scala>

Answer 6

There is a straightforward solution using scalaz :
def convert(ls: List[Any]) : Option[List[String]] =
ls.map { a => if (a.isInstanceOf[String]) Some(a.asInstanceOf[String]) else None}.sequence