Creating Seq after waiting for all results from map/foreach in Scala - scala

Question

I am trying to loop over inputs and process them to produce scores.
Just for the first input, I want to do some processing that takes a while.
The function ends up returning just the values from the 'else' part. The 'if' part is done executing after the function returns the value.
I am new to Scala and understand the behavior but not sure how to fix it.
I've tried inputs.zipWithIndex.map instead of foreach but the result is the same.
def getscores(
inputs: inputs
): Future[Seq[scoreInfo]] = {
var scores: Seq[scoreInfo] = Seq()
inputs.zipWithIndex.foreach {
case (f, i) => {
if (i == 0) {
// long operation that returns Future[Option[scoreInfo]]
getgeoscore(f).foreach(gso => {
gso.foreach(score => {
scores = scores.:+(score)
})
})
} else {
scores = scores.:+(
scoreInfo(
id = "",
score = 5
)
)
}
}
}
Future {
scores
}
}

Answer 1

For what you need, I would drop the mutable variable and replace foreach with map to obtain an immutable list of Futures and recover to handle exceptions, followed by a sequence like below:
def getScores(inputs: Inputs): Future[List[ScoreInfo]] = Future.sequence(
inputs.zipWithIndex.map{ case (input, idx) =>
if (idx == 0)
getGeoScore(input).map(_.getOrElse(defaultScore)).recover{ case e => errorHandling(e) }
else
Future.successful(ScoreInfo("", 5))
})
To capture/print the result, one way is to use onComplete:
getScores(inputs).onComplete(println)

Answer 2

The part your missing is understanding a tricky element of concurrency, and that is that the order of execution when using multiple futures is not guaranteed.
If your block here is long running, it will take a while before appending the score to scores
// long operation that returns Future[Option[scoreInfo]]
getgeoscore(f).foreach(gso => {
gso.foreach(score => {
// stick a println("here") in here to see what happens, for demonstration purposes only
scores = scores.:+(score)
})
})
Since that executes concurrently, your getscores function will also simultaneously continue its work iterating over the rest of inputs in your zipWithindex. This iteration, especially since it's trivial work, likely finishes well before the long-running getgeoscore(f) completes the execution of the Future it scheduled, and the code will exit the function, moving on to whatever code is next after you called getscores
val futureScores: Future[Seq[scoreInfo]] = getScores(inputs)
futureScores.onComplete{
case Success(scoreInfoSeq) => println(s"Here's the scores: ${scoreInfoSeq.mkString(",")}"
}
//a this point the call to getgeoscore(f) could still be running and finish later, but you will never know
doSomeOtherWork()
Now to clean this up, since you can run a zipWithIndex on your inputs parameter, I assume you mean it's something like a inputs:Seq[Input]. If all you want to do is operate on the first input, then use the head function to only retrieve the first option, so getgeoscores(inputs.head) , you don't need the rest of the code you have there.
Also, as a note, if using Scala, get out of the habit of using mutable vars, especially if you're working with concurrency. Scala is built around supporting immutability, so if you find yourself wanting to use a var , try using a val and look up how to work with the Scala's collection library to make it work.

Answer 3

In general, that is when you have several concurrent futures, I would say Leo's answer describes the right way to do it. However, you want only the first element transformed by a long running operation. So you can use the future return by the respective function and append the other elements when the long running call returns by mapping the future result:
def getscores(inputs: Inputs): Future[Seq[ScoreInfo]] =
getgeoscore(inputs.head)
.map { optInfo =>
optInfo ++ inputs.tail.map(_ => scoreInfo(id = "", score = 5))
}
So you neither need zipWithIndex nor do you need an additional future or join the results of several futures with sequence. Mapping the future just gives you a new future with the result transformed by the function passed to .map().