Combining List, Future and Option in for-comprehension - scalaz

Question

I\'ve got following problem:

val sth: Future[Seq[T, S]] = for {
  x <- whatever: Future[List[T]]
  y <- x: List[T]
  z <- f(y): Future[Option[S]]
           


        

Answer 1

The problem with for comprehension is that it's not some kind of magic monadic "unwrapper", it's just a sequence of map, flatMap and filter.

As you may know map and flatMap operate only on "inner" type, leaving "outer" type of monad unchanged. This means you can't do this:

for {
  x <- whatever: Future[List[T]]
  y <- x: List[T]
} yield y

inside single for. Instead, you can do something like this:

for (x <- whatever: Future[List[T]])
  yield for (y <- x: List[T]) yield y

Which looks kinda ugly.

Back to your case, I's easier to write whole transformation explicitly using map and flatMap, as it gives you greater visibility and control:

whatever.flatMap {
  x: List[T] =>
    Future.sequence(x.map {
      y: T => f(y).map(y -> _)
    }).map(_.collect {
      case (y, Some(n)) => y -> n
    })
}

Also, @trustnoone mentioned, you can't get rid of the Future without explicitly calling Await.

Answer 2

You could do something like what you need using the scalaz ListT monad transformer

 object Test {
   import scalaz._
   import ListT._
   type T = String
   type S = Int
   val whatever: Future[List[T]] = ??? // you get this somewhere
   def f(y: T): Future[Option[S]] = ??? // function that returns future of option

   val sth: Future[List[(T, S)]] = (for {
     y <- listT(whatever) 
     // you cannot mix list and option, but you can convert the option to a list of 1 item
     n <- listT(f(y).map(_.toList)) 
   } yield y -> n).run
 }

N.B.: Since you start with a future, you cannot return a Seq[(T,S)], you can only have a future. You have to call Await.result if you want to block and get the result.