Scala ArraySeq[Vector[data]]] processing - scala

Question

I have below format of data:
ArraySeq(Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)))
I want
Seq[None, None, None, None, None, Some(1.1), Some(1.1), Some(1.1), Some(1.1), Some(1.1), Some(2.2),Some(2.2),Some(2.2),Some(2.2),Some(2.2), Some(3.3), Some(3.3), Some(3.3), Some(3.3),Some(3.3),Some(4.4), Some(4.4), Some(4.4), Some(4.4), Some(4.4)]

Answer 1

You have multiples solution :
s.transpose.flatten
The transpose method will pair and overlay elements from another collections into a single collection.
And flatten method are used to convert a list of lists into a single list.
A other solution available is use concatenation of foldLeft.
import scala.collection.immutable.ArraySeq;
val s = ArraySeq(Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)),
Vector(None, Some(1.1), Some(2.2), Some(3.3), Some(4.4)))
val result = s.foldLeft(Seq[Seq[Option[Double]]](Seq.empty[Option[Double]], Seq.empty[Option[Double]], Seq.empty[Option[Double]], Seq.empty[Option[Double]], Seq.empty[Option[Double]])) {
case (acc, v) => Seq((acc(0) :+ v(0)), (acc(1) :+ v(1)), (acc(2) :+ v(2)), (acc(3) :+ v(3)), (acc(4) :+ v(4)))
}.foldLeft(Seq.empty[Option[Double]]) {
case (acc, s) => acc ++ s
}
result are equal to => List(None, None, None, None, None, Some(1.1), Some(1.1), Some(1.1), Some(1.1), Some(1.1), Some(2.2), Some(2.2), Some(2.2), Some(2.2), Some(2.2), Some(3.3), Some(3.3), Some(3.3), Some(3.3), Some(3.3), Some(4.4), Some(4.4), Some(4.4), Some(4.4), Some(4.4))