Converting List to Map using keys in Scala

Question

I am struggling with finding an elegant FP approach to solving the following problem in Scala:

Say I have a set of candidate keys

val validKeys = Set("key1", "key2", "key3")

And a list that

Starts with a key
has some number of non-keys (> 0) between each key
Does not end with a key

For example:

val myList = List("key3", "foo", "bar", "key1", "baz")

I'd like to transform this list into a map by choosing using valid keys as the key and aggregating non-keys as the value. So, in the example above:

("key3" -> "foo\nbar", "key1" -> "baz")

Thanks in advance.

What should happen if the same key turns up more than once in the list? — The Archetypal Paul
– The Archetypal Paul, Commented Dec 25, 2015 at 9:02

Rok Kralj · Accepted Answer · 2015-12-25 12:18:06Z

2

Short and simple:

def create(a: List[String]): Map[String, String] = a match {
    case Nil => Map()
    case head :: tail =>
        val (vals, rest) = tail.span(!validKeys(_))
        create(rest) + (head -> vals.mkString("\n"))
}

edited Dec 25, 2015 at 12:18

answered Dec 25, 2015 at 11:37

Rok Kralj

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4 Comments

The Archetypal Paul Over a year ago

Heh. I just wrote something identical apart from variable names :) it doesn't collect all values if keys can appear more than once but it's possible that's not something that can happen in the OP's scenario. And it's not tail recursive, but it's easy to make it so.

Rok Kralj Over a year ago

Yours uses appending to string. Strings in Scala are immutable, therefore the whole string is copied on append. It is better to collect all the strings and join them in one go.

The Archetypal Paul Over a year ago

Agreed. The problem with the foldLeft is you don't get a chance to collect all the strings at the end (it could convert to a list of strings then do a map of mkString, but that makes it less clear).

Dan Over a year ago

Thanks, I went with the tail-rec version of this, but otherwise exactly what I was looking for.

The Archetypal Paul · Accepted Answer · 2015-12-25 08:59:21Z

1

Traversing a list from left to right, accumulating a result should suggest foldLeft

 myList.foldLeft((Map[String, String](), "")) {
    case ((m, lk), s) =>
      if (validKeys contains s)
        (m updated (s, ""), s)
      else (m updated (lk, if (m(lk) == "") s else m(lk) + "\n" + s), lk)
  }._1
  // Map(key3 -> foo\nbar, key1 -> baz)

answered Dec 25, 2015 at 8:59

The Archetypal Paul

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Comments

Nyavro · Accepted Answer · 2015-12-25 09:29:08Z

0

As a first approximation solution:

def group(list:List[String]):List[(String, List[String])] = {
  @tailrec
  def grp(list:List[String], key:String, acc:List[String]):List[(String, List[String])] =
    list match {
      case Nil => List((key, acc.reverse))
      case x :: xs if validKeys(x) => (key, acc.reverse)::group(x::xs)
      case x :: xs => grp(xs, key, x::acc)
    }
  list match {
    case Nil => Nil
    case x::xs => grp(xs, x, List())
  }
}

val map = group(myList).toMap

Another option:

list.foldLeft((Map[String, String](), "")) {
  case ((map, key), item) if validKeys(item) => (map, item)
  case ((map, key), item) => 
    (map.updated(key, map.get(key).map(v => v + "\n" + item).getOrElse(item)), key)
}._1

edited Dec 25, 2015 at 9:29

answered Dec 25, 2015 at 8:58

Nyavro

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