What's Scala's idiomatic way to split a List by separator?

Question

If I have a List of type String,

scala> val items = List("Apple","Banana","Orange","Tomato","Grapes","BREAK","Salt","Pepper","BREAK","Fish","Chicken","Beef")
items: List[java.lang.String] = List(Apple, Banana, Orange, Tomato, Grapes, BREAK, Salt, Pepper, BREAK, Fish, Chicken, Beef)

how can I split it into n separate lists based on a certain string/pattern ("BREAK", in this case).

I've thought about finding the position of "BREAK" with indexOf, and split up the list that way, or using a similar approach with takeWhile (i => i != "BREAK") but I'm wondering if there's a better way?

If it helps, I know there will only ever be 3 sets of items in the items list (thus 2 "BREAK" markers).

Answer 1

def splitBySeparator[T](l: List[T], sep: T): List[List[T]] = {
  l.span( _ != sep ) match {
    case (hd, _ :: tl) => hd :: splitBySeparator(tl, sep)
    case (hd, _) => List(hd)
  }
}

val items = List("Apple","Banana","Orange","Tomato","Grapes","BREAK","Salt","Pepper","BREAK","Fish","Chicken","Beef")
splitBySeparator(items, "BREAK")

Result:

res1: List[List[String]] = List(List(Apple, Banana, Orange, Tomato, Grapes), List(Salt, Pepper), List(Fish, Chicken, Beef))

UPDATE: The above version, while concise and effective, has two problems: it does not handle well the edge cases (like List("BREAK") or List("BREAK", "Apple", "BREAK"), and is not tail recursive. So here is another (imperative) version that fixes this:

import collection.mutable.ListBuffer
def splitBySeparator[T](l: Seq[T], sep: T): Seq[Seq[T]] = {
  val b = ListBuffer(ListBuffer[T]())
  l foreach { e =>
    if ( e == sep ) {
      if  ( !b.last.isEmpty ) b += ListBuffer[T]()
    }
    else b.last += e
  }
  b.map(_.toSeq)
}

It internally uses a ListBuffer, much like the implementation of List.span that I used in the first version of splitBySeparator.

Answer 2

Another option:

val l = Seq(1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5, 9, 1, 2, 3, 4, 5)

l.foldLeft(Seq(Seq.empty[Int])) {
  (acc, i) =>
    if (i == 9) acc :+ Seq.empty
    else acc.init :+ (acc.last :+ i)
}

// produces:
List(List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5))