Combining Scala Futures and Collections for Understanding

Question

I try to use a for expression to iterate through a list, and then do the conversion for each item using a utility that returns Future. In short, it does not compile, and I would like to understand why. I read this question , which is similar, and was of great help, but what I'm trying to do is even simpler, which is even more confusing why it does not work. I am trying to do something like:

import scala.concurrent.Future import scala.concurrent.ExecutionContext.Implicits.global val numberList = List(1, 2, 3) def squareInTheFuture(number: Int): Future[Int] = Future { number * number} val allTheSquares = for { number <- numberList square <- squareInTheFuture(number) } yield { square } 

And I get:

error: type of discrepancy; found: scala.concurrent.Future [Int] required: scala.collection.GenTraversableOnce [?] square <- squareInTheFuture (number) ^

Can someone help me understand why this is not working and what is the best alternative?

Answer 1

flatMap requires that the constructors of the types numberList and squareInTheFuture(number) be the same (modulo any implicit conversions that the collection library does). This is not true. Instead, this is a workaround:

 val allSquaresInTheFuture: Future[List[Int]] = Future.traverse(numberList)(squareInTheFuture) 

Answer 2

Future companion object has a traverse method that does exactly what you want:

 val allTheSquares: Future[List[Int]] = Future.traverse(numberList)(squareInTheFuture) 

This will start all calculations asynchronously and return a future that will be completed after all these futures are completed.

Answer 3

@Right. In addition, if you are trying to do parallel computation:

  val numberList = List(1, 2, 3) val allTheSquares = numberList.par.map(x => x * x)(breakOut) 

If you really want Future :

 val allTheSquares: Future[List[Int]] = Future.traverse(numberList)(squareInTheFuture) 

Answer 4

Your understanding is the same as

 val allTheSquares = numberList.flatMap(number => squareInTheFuture(number)) 

flatMap requires the argument function to return the value of GenTraversableOnce[Int] , however yours returns a Future[Int] , therefore, a mismatch.

Answer 5

Change & lt; - on = works. In this case, flatmap / map is not called for the Future object.

  import scala.concurrent.Future import scala.concurrent.ExecutionContext.Implicits.global val numberList = List(1, 2, 3) def squareInTheFuture(number: Int): Future[Int] = Future { number * number} val allTheSquares = for { number <- numberList square = squareInTheFuture(number) } yield { square } 

Reading this helped me understand how profitability works.