List of HLists to HList of Lists? - scala

Question

Basically I want to transpose my object. Is there a simple way to do this? If my HList is large and I don't want to fold.
This is in service of unzipping a List of a high-arity tuples.

Answer 1

I don't know of an existing solution, but it seemed to be an interesting task, so I tried my hand at it.
Assuming we have a list of HLists, like
val myList: List[Int :: String :: Boolean :: HNil] = List(
1 :: "sample1" :: true :: HNil,
2 :: "sample2" :: false :: HNil,
3 :: "sample3" :: true :: HNil
)
and we want a HList of Lists, like
val expected: List[Int] :: List[String] :: List[Boolean] :: HNil =
List(3, 2, 1) ::
List("sample3", "sample2", "sample") ::
List(true, false, true) ::
HNil
we need to create one typeclass. I named it EmptyOf. It has a single method that creates an HList containing empty lists.
trait EmptyOf[T <: HList, AsList <: HList] {
def empty: AsList
}
Here is the implementation for HNil:
implicit val emptyOfHNil: EmptyOf[HNil, HNil] = () => HNil
And now for HCons:
implicit def emptyOfHCons[H, T <: HList, TEmpty <: HList](
implicit ev: EmptyOf[T, TEmpty]
): EmptyOf[H :: T, List[H] :: TEmpty] = () => List.empty[H] :: ev.empty()
So, for EmptyOf[T :: U :: HNil, List[T] :: List[U] :: HNil] our instance's empty will return Nil :: Nil :: HNil.
With these, we can implement transpose by folding on the list of HLists and using a Poly2 to merge the HLists.
object concat extends Poly2 {
implicit def prepend[S] = at[List[S], S]((list, s) => s :: list)
}
def transpose[T <: HList, U <: HList](lists: List[T])
(implicit emptyOf: EmptyOf[T, U],
zip: ZipWith.Aux[U, T, concat.type, U]): U =
lists.foldLeft(emptyOf.empty())(_.zipWith(_)(concat))
Now transpose(myList) should be equal to expected (and of the correct type).