How to avoid calling toVector on each Scala for comprehension / yield? - scala

Question

I have several methods that operate on Vector sequences and the following idiom is common when combining data from multiple vectors into a single one with the use of a for comprehension / yield:
(for (i <- 0 until y.length) yield y(i) + 0.5*dy1(i)) toVector
Notice the closing toVector and the enclosing parentheses around the for comprehension. I want to get rid of it because it's ugly, but removing it produces the following error:
type mismatch;
found : scala.collection.immutable.IndexedSeq[Double]
required: Vector[Double]
Is there a better way of achieving what I want that avoids explicitly calling toVector many times to essentially achieve a non-operation (converting and indexed sequence...to an indexed sequence)?

Answer 1

One way to avoid collection casting, e.g. toVector, is to invoke, if possible, only those methods that return the same collection type.
y.zipWithIndex.map{case (yv,idx) => yv + 0.5*dy1(idx)}

Answer 2

for yield on Range which you are using in your example yields a Vector[T] by default.
example,
scala> val squares= for (x <- Range(1, 3)) yield x * x
squares: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 4)
check the type,
scala> squares.isInstanceOf[Vector[Int]]
res14: Boolean = true
Note that Vector[T] also extends IndexedSeq[T].
#SerialVersionUID(-1334388273712300479L)
final class Vector[+A] private[immutable] (private[collection] val startIndex: Int, private[collection] val endIndex: Int, focus: Int)
extends AbstractSeq[A]
with IndexedSeq[A]
with GenericTraversableTemplate[A, Vector]
with IndexedSeqLike[A, Vector[A]]
with VectorPointer[A #uncheckedVariance]
with Serializable
with CustomParallelizable[A, ParVector[A]]
That's why above result is also an instance of IndexedSeq[T],
scala> squares.isInstanceOf[IndexedSeq[Int]]
res15: Boolean = true
You can define the type of your result as IndexedSeq[T] and still achieve what you want with Vector without explicitly calling .toVector
scala> val squares: IndexedSeq[Int] = for (x <- Range(1, 3)) yield x * x
squares: IndexedSeq[Int] = Vector(1, 4)
scala> squares == Vector(1, 4)
res16: Boolean = true
But for yield on Seq[T] gives List[T] by default.
scala> val squares = for (x <- Seq(1, 3)) yield x * x
squares: Seq[Int] = List(1, 9)
Only in that case if you want vector you must .toVector the result.
scala> squares.isInstanceOf[Vector[Int]]
res21: Boolean = false
scala> val squares = (for (x <- Seq(1, 3)) yield x * x).toVector
squares: Vector[Int] = Vector(1, 9)