How to find if a Scala String is parseable as a Double or not?
For Scala 2.13+ see Xavier's answer below. Apparently there's a toDoubleOption
method now.
For older versions:
def parseDouble(s: String) = try { Some(s.toDouble) } catch { case _ => None }
Fancy version (edit: don't do this except for amusement value; I was a callow youth years ago when I used to write such monstrosities):
case class ParseOp[T](op: String => T)
implicit val popDouble = ParseOp[Double](_.toDouble)
implicit val popInt = ParseOp[Int](_.toInt)
// etc.
def parse[T: ParseOp](s: String) = try { Some(implicitly[ParseOp[T]].op(s)) }
catch {case _ => None}
scala> parse[Double]("1.23")
res13: Option[Double] = Some(1.23)
scala> parse[Int]("1.23")
res14: Option[Int] = None
scala> parse[Int]("1")
res15: Option[Int] = Some(1)
Scalaz provides an extension method parseDouble
on String
s, which gives a value of type Validation[NumberFormatException, Double]
.
scala> "34.5".parseDouble
res34: scalaz.Validation[NumberFormatException,Double] = Success(34.5)
scala> "34.bad".parseDouble
res35: scalaz.Validation[NumberFormatException,Double] = Failure(java.lang.NumberFormatException: For input string: "34.bad")
You can convert it to Option
if so required.
scala> "34.bad".parseDouble.toOption
res36: Option[Double] = None