How to run sequence over List[F[G[A]]] to get F[G[List[A]]]

Question

Is it possible to transform a List[F[G[A]]] into F[G[List[A]]] nicely?

I can do this in Scalaz the following way:

val x: List[Future[Option[Int]]] = ???
val transformed: Future[Option[List[Int]]] = x.sequenceU.map(_.sequenceU)

I'm just wondering if there's a nicer way to do this rather than .sequenceU.map(_.sequenceU) Perhaps using a monad transformer? I did attempt this, without much luck.

Answer 1

Monad transformers are the way to go if you want to avoid the nested sequencing. In this case you want an OptionT[Future, A] (which is equivalent to Future[Option[A]]):

import scalaz._, Scalaz._
import scala.concurrent.ExecutionContext.Implicits.global
import scala.concurrent.Future

val xs = List(OptionT(Future(some(1))), OptionT(Future(some(2))))
val ys = OptionT(Future(none[Int])) :: xs

val sequencedXs: Future[Option[List[Int]]] = xs.sequenceU.run
val sequencedYs: Future[Option[List[Int]]] = ys.sequenceU.run

And then:

scala> sequencedXs.foreach(println)
Some(List(1, 2))

scala> sequencedYs.foreach(println)
None

As expected.