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\"
...          


        

Answer 1

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 2

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 3

A functional* solution using fold:

def group(s : String) : Seq[String] = {
  s.tail.foldLeft(Seq(s.head.toString)) { case (carry, elem) =>
    if ( carry.last(0) == elem ) {
      carry.init :+ (carry.last + elem)
    }
    else {
      carry :+ elem.toString
    }
  }
}

There is a lot of cost hidden in all those sequence operations performed on strings (via implicit conversion). I guess the real complexity heavily depends on the kind of Seq strings are converted to.

(*) Afaik all/most operations in the collection library depend in iterators, an imho inherently unfunctional concept. But the code looks functional, at least.