Splitting string into groups

Question

I'm trying to 'group' a string into segments, I guess this example would explain it more succintly

scala> val str: String = "aaaabbcddeeeeeeffg"
... (do something)
res0: List("aaaa","bb","c","dd","eeeee","ff","g")

I can thnk of a few ways to do this in an imperative style (with vars and stepping through the string to find groups) but I was wondering if any better functional solution could be attained? I've been looking through the Scala API but there doesn't seem to be something that fits my needs.

Any help would be appreciated

Answer 1

You can split the string recursively with span:

def s(x : String) : List[String] = if(x.size == 0) Nil else {
    val (l,r) = x.span(_ == x(0))
    l :: s(r) 
}

Tail recursive:

@annotation.tailrec def s(x : String, y : List[String] = Nil) : List[String] = {
    if(x.size == 0) y.reverse 
    else {
        val (l,r) = x.span(_ == x(0))
        s(r, l :: y)
    }
}

Answer 2

Seems that all other answers are very concentrated on collection operations. But pure string + regex solution is much simpler:

str split """(?<=(\w))(?!\1)""" toList

In this regex I use positive lookbehind and negative lookahead for the captured char

Answer 3

def group(s: String): List[String] = s match {
  case "" => Nil
  case s  => s.takeWhile(_==s.head) :: group(s.dropWhile(_==s.head))
}

---

Edit: Tail recursive version:

def group(s: String, result: List[String] = Nil): List[String] = s match {
  case "" => result reverse
  case s  => group(s.dropWhile(_==s.head), s.takeWhile(_==s.head) :: result)
}

can be used just like the other because the second parameter has a default value and thus doesnt have to be supplied.