For a given Array[Byte] such as
val in = Array(104, 101, 108, 108, 111, 10, 119, 111, 114, 108, 100, 10)
how to split it by value 10 so that
val out = in.arr_split(10)
would deliver
Array( Array(104, 101, 108, 108, 111),
Array(119, 111, 114, 108, 100))
Assume in general many occurrences of splitting elements, for instance many 10.
If possible, a parallel solution is desired.
Many Thanks.
Something like this should work:
def split(l: Array[Int], i:Int):Array[Array[Int]] = {
l match {
case Array() => Array()
case _ =>
val (h, t) = l.span(a => a != i)
Array(h) ++ split(t.drop(1), i)
}
}
val in = Array(104, 101, 108, 108, 111, 10, 119, 111, 114, 108, 100, 10)
val out = split(in, 10)
// res: Array[Array[Int]] = Array(Array(104, 101, 108, 108, 111), Array(119, 111, 114, 108, 100))
scalaz-stream solution. Instead of Array I use Vector here.
val in = Vector(104, 101, 108, 108, 111, 10, 119, 111, 114, 108, 100, 10)
val P = scalaz.stream.Process
implicit val eq = scalaz.Equal.equal[Int]((l, r) => l == r)
println(P.emitSeq[Task, Int](in).splitOn(10).filter(!_.isEmpty).runLog.run)
Output:
Vector(Vector(104, 101, 108, 108, 111), Vector(119, 111, 114, 108, 100))
Pimped version
implicit def toDivide[A, B <% TraversableLike[A, B]](a : B) = new {
private def divide(x : B, condition: (A) => Boolean) : Iterable[B] = {
if (x.size > 0)
x.span(condition) match {
case (e, f) => if (e.size > 0) Iterable(e) ++ divide(f.drop(1),condition) else Iterable(f)
}
else
Iterable()
}
def divide(condition: (A) => Boolean): Iterable[B] = divide(a, condition)
}
