scala collections : map a list and carry some state?

Question

I seem to run into this problem all the time. I want to modify some of the elements in a list, but I need to keep some state as I do it, so map doesn\'t work.

Here i

Answer 1

The tricky part is that the "d" and "f" elements get no modification.

This is what I came up with. It's a bit more concise, code wise, but does involve multiple traversals.

val l1: List[String] = List("a","b","c","d","e","f","b","c","e","b","a")

l1.reverse.tails.foldLeft(List[String]()){
  case (res, Nil) => res
  case (res, hd::tl) =>
    val count = tl.count(_ == hd)
    if (count > 0) s"$hd${count+1}" +: res
    else if (res.contains(hd+2)) (hd+1) +: res
    else hd +: res
}
//res0: List[String] = List(a1, b1, c1, d, e1, f, b2, c2, e2, b3, a2)

By using tails, each element, hd, is able to see the future, tl, and the past, res.

Answer 2

A simple but slow version

l1.zipWithIndex.map{ case (elem, i) =>
  if (l1.count(_ == elem) == 1) {
    elem
  } else {
    val n = {l1.take(i+1).count(_ == elem)}
    s"$elem$n"
  }
}

The next version is longer, less pretty, and not functional, but should be faster in the unlikely case that you are processing a very long list.

def makeUniq(in: Seq[String]): Seq[String] = {
  // Count occurrence of each element
  val m = mutable.Map.empty[String, Int]

  for (elem <- in) {
    m.update(elem, m.getOrElseUpdate(elem, 0) + 1)
  }

  // Remove elements with a single occurrence
  val dupes = m.filter(_._2 > 1)

  // Apply numbering to duplicate elements
  in.reverse.map(e => {
    val idx = dupes.get(e) match {
      case Some(i) =>
        dupes.update(e, i - 1)
        i.toString
      case _ =>
        ""
    }

    s"$e$idx"
    }).reverse
  }

The code is easier if you wanted to apply a count to every element rather than just the non-unique ones.

def makeUniq(in: Seq[String]): Seq[String] = {
  val m = mutable.Map.empty[String, Int]

  in.map{ e =>
    val i = m.getOrElseUpdate(e, 0) + 1
    m.update(e, i)

    s"$e$i"
  }
}