How to split a line given a list of positions in Scala

Question

How could you write a functional implementation for split(positions:List[Int], str:String):List[String] , which is similar to splitAt but splits a given string into a list of strings by a specified list of positions?

for instance

• split(List(1, 2), "abc") returns List("a", "b", "c")
• split(List(1), "abc") returns List("a", "bc")
• split(List(), "abc") returns List("abc")

Answer 1

 def lsplit(pos: List[Int], str: String): List[String] = { val (rest, result) = pos.foldRight((str, List[String]())) { case (curr, (s, res)) => val (rest, split) = s.splitAt(curr) (rest, split :: res) } rest :: result } 

Answer 2

Something like that:

 def lsplit(pos: List[Int], s: String): List[String] = pos match { case x :: rest => s.substring(0,x) :: lsplit(rest.map(_ - x), s.substring(x)) case Nil => List(s) } 

(A fair warning: it is not tail recursive, so it will hit the stack for large lists, inefficient due to reassigning indices and strings of substrings. You can solve these things by adding additional arguments and / or an internal method that does recursions.)

Answer 3

What about....

 def lSplit( indices : List[Int], s : String) = (indices zip (indices.tail)) map { case (a,b) => s.substring(a,b) } scala> lSplit( List(0,4,6,8), "20131103") List[String] = List(2013, 11, 03)