Converting from Long to List[Int]

Question

I'm trying to convert a Long into a List[Int] where each item in the list is a single digit of the original Long.

scala> val x: Long = 123412341L
x: Long = 123412341

scala> x.toString.split("").toList
res8: List[String] = List("", 1, 2, 3, 4, 1, 2, 3, 4, 1)

scala> val desired = res8.filterNot(a => a == "")
desired: List[String] = List(1, 2, 3, 4, 1, 2, 3, 4, 1)

Using split("") results in a "" list element that I'd prefer not to have in the first place.

I could simply filter it, but is it possible for me to go from 123412341L to List(1, 2, 3, 4, 1, 2, 3, 4, 1) more cleanly?

Answer 1

If you want the integer values of the digits, it can be done like so:

scala> x.toString.map(_.asDigit).toList
res65: List[Int] = List(1, 2, 3, 4, 1, 2, 3, 4, 1)

Note the difference that the .map(_.asDigit) makes:

scala> x.toString.toList
res67: List[Char] = List(1, 2, 3, 4, 1, 2, 3, 4, 1)

scala> res67.map(_.toInt)
res68: List[Int] = List(49, 50, 51, 52, 49, 50, 51, 52, 49)

x.toString.toList is a List of Chars, i.e. List('1','2','3',...). The toString rendering of the list makes it look the same, but the two are quite different -- a '1' character, for example, has integer value 49. Which you should use depends on whether you want the digit characters or the integers they represent.