How to fold a Scala iterator and get a lazily evaluated sequence as result?

mvallebr :

I have an iterator of strings, where each string can be either "H" (header) or "D" (detail). I want to split this iterator into blocks, where each block starts with one header and can have 0 to many detail.

I know how to solve this problem loading everything into memory. For example, the code bellow:

Seq("H","D","D","D","H","D","H","H","D","D","H","D").toIterator
  .foldLeft(List[List[String]]())((acc, x) => x match {
    case "H" => List(x) :: acc
    case "D" => (x :: acc.head) :: acc.tail })
  .map(_.reverse)
  .reverse

returns 5 blocks - List(List(H, D, D, D), List(H, D), List(H), List(H, D, D), List(H, D)) - which is what I want.

However, instead of List[List[String]] in the result, I want either Iterator[List[String]] or some other structure that allows me to evaluate the result lazily and do not load the entire input into memory if the entire iterator in consumed, I want to load into memory only the block being consumed at a time (e.g.: when I call iterator.next).

How can I modify the code above to achieve the result I want?

EDIT: I need this in Scala 2.11 specifically, as the environment I use sticks to it. Glad to also accept answers for other versions though.

Scalway :

Here is the simplest implementation I could find (It's generic and lazy):

/** takes 'it' and groups consecutive elements 
 *  until next item that satisfy 'startGroup' predicate occures. 
 *  It returns Iterator[List[T]] and is lazy 
 *  (keeps in memory only last group, not whole 'it'). 
*/
def groupUsing[T](it:Iterator[T])(startGroup:T => Boolean):Iterator[List[T]] = {
  val sc = it.scanLeft(List.empty[T]) {
    (a,b) => if (startGroup(b)) b::Nil else b::a
  }

  (sc ++ Iterator(Nil)).sliding(2,1).collect { 
    case Seq(a,b) if a.length >= b.length => a.reverse
  }
}

use it like that:

val exampleIt = Seq("H1","D1","D2","D3","H2","D4","H3","H4","D5","D6","H5","D7").toIterator
groupUsing(exampleIt)(_.startsWith("H"))
// H1 D1 D2 D3 / H2 D4 / H3 / H4 D5 D6 / H5 D7

here is specyfication:

X | GIVEN            | EXPECTED     |
O |                  |              | empty iterator
O | H                | H            | single header
O | D                | D            | single item (not header)
O | HD               | HD           |
O | HH               | H,H          | only headers
O | HHD              | H,HD         |
O | HDDDHD           | HDDD,HD      |
O | DDH              | DD,H         | heading D's have no Header as you can see.
O | HDDDHDHDD        | HDDD,HD,HDD  |

scalafiddle with tests and additional comments: https://scalafiddle.io/sf/q8xbQ9N/11

(if answer is helpful up-vote please. I spent a little too much time on it :))

SECOND IMPLEMENTATION:

You have propose version that does not use sliding. Here it is, but it has its own problems listed below.

def groupUsing2[T >: Null](it:Iterator[T])(startGroup:T => Boolean):Iterator[List[T]] = {
  type TT = (List[T], List[T])
  val empty:TT = (Nil, Nil)
  //We need this ugly `++ Iterator(null)` to close last group.
  val sc = (it ++ Iterator(null)).scanLeft(empty) {
    (a,b) => if (b == null || startGroup(b)) (b::Nil, a._1) else (b::a._1, Nil)
  }

  sc.collect { 
    case (_, a) if a.nonEmpty => a.reverse
  }
}

Traits:

• (-) It works only for T>:Null types. We just need to add element that will close last collection on the end (null is perfect but it limits our type).
• (~) it should create same amount of trsh as previous version. We just create tuples in first step instead of second one.
• (+) it does not check length of List (and this is big gain to be honest).
• (+) In core it is Ivan Kurchenko answer but without extra boxing.

Here is scalafiddle: https://scalafiddle.io/sf/q8xbQ9N/11