Update List with Loop Arguments Scala

Question

I'm trying create new random List based on other List. But this code doesn't work to update the var

import scala.util.Random
import scala.math

val randSymbol = List(1,2,2,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,8)
val _finalSymbol = List(List(0,0,0,0,0),List(0,0,0,0,0),List(0,0,0,0,0))
    var i:Int = 0
    var a:Int = 0
    for (i <- 0 to 2){
      _finalSymbol(i) = new List
      for (a <- 0 to 4){
        var iRandIndex = floor(Random.nextInt() * randSymbol.length).toInt
        var iRandSymbol = randSymbol(iRandIndex)
        _finalSymbol(i)(a) = iRandSymbol
      }
    }

score 3 · Accepted Answer · answered Oct 28 '16 at 14:56

3

Try something like this:

val _finalSymbol = (0 to 2) map { _ => Random.shuffle(randSymbol).take(5) }

And do yourself a favor: buy a scala book. It's not javascript. At all.

answered Oct 28 '16 at 14:56

Dima

39,570
6
44
70

Sure I will buy them, I've tried this, but this one is indexed sequence, and how do I convert this to List[List[Int]] ? – Brandy Geeks Oct 28 '16 at 15:05
So this works for me: `List(_finalSymbol(0),_finalSymbol(1),_finalSymbol(2))`, but can I do something very simple? – Brandy Geeks Oct 28 '16 at 15:10
try `_finalSymbol.toList` – Tzach Zohar Oct 28 '16 at 15:13
1

Why do you care if it is a list or not? You should not. It doesn't matter for most cases you care about. If you want it to be a list for some reason, add `.toList` at the end as @TzachZohar suggested, or replace `(0 to 2)` with `List(1,2,3)` ... But that's just a waste of ticks. You don't need it. – Dima Oct 28 '16 at 15:18

Update List with Loop Arguments Scala

1 Answers1