Scala Syntactic Sugar for converting to `Option`

Question

When working in Scala, I often want to parse a field of type [A] and convert it to a Option[A], with a single case (for example, "NA" or "") being converted to None, and the other cases being wrapped in some.

Right now, I'm using the following matching syntax.

match {
  case "" => None
  case s: String => Some(s)
}
// converts an empty String to None, and otherwise wraps it in a Some.

Is there any more concise / idiomatic way to write this?

Answer 1

There are a more concise ways. One of:

Option(x).filter(_ != "")
Option(x).filterNot(_ == "")

will do the trick, though it's a bit less efficient since it creates an Option and then may throw it away.

If you do this a lot, you probably want to create an extension method (or just a method, if you don't mind having the method name first):

implicit class ToOptionWithDefault[A](private val underlying: A) extends AnyVal {
  def optNot(not: A) = if (underlying == not) None else Some(underlying)
}

Now you can

scala> 47.toString optNot ""
res1: Option[String] = Some(47)

(And, of course, you can always create a method whose body is your match solution, or an equivalent one with if, so you can reuse it for that particular case.)