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How to use Java Collections.shuffle() on a Scala array?

Asked 2009-08-11T09:11:51.437
11

I have an array that I want to permutate randomly. In Java, there is a method Collections.shuffle() that can shuffle the elements of a List randomly. It can be used on an array too:

String[] array = new String[]{"a", "b", "c"};

// Shuffle the array; works because the list returned by Arrays.asList() is backed by the array
Collections.shuffle(Arrays.asList(array));

I tried using this on a Scala array, but the Scala interpreter responds with a lengthy answer:

scala> val a = Array("a", "b", "c")
a: Array[java.lang.String] = Array(a, b, c)

scala> java.util.Collections.shuffle(java.util.Arrays.asList(a))
<console>:6: warning: I'm seeing an array passed into a Java vararg.
I assume that the elements of this array should be passed as individual arguments to the vararg.
Therefore I follow the array with a `: _*', to mark it as a vararg argument.
If that's not what you want, compile this file with option -Xno-varargs-conversion.
       java.util.Collections.shuffle(java.util.Arrays.asList(a))
                                                             ^
<console>:6: error: type mismatch;
 found   : Array[java.lang.String]
 required: Seq[Array[java.lang.String]]
       java.util.Collections.shuffle(java.util.Arrays.asList(a))
                                                             ^

What exactly is happening here? I don't want to compile my code with a special flag (-Xno-varargs-conversion), if that is the solution at all, just because of this.

So, how do I use Java's Collections.shuffle() on a Scala array?

I wrote my own shuffle method in Scala in the meantime:

// Fisher-Yates shuffle, see: http://en.wikipedia.org/wiki/Fisher–Yates_shuffle
def shuffle[T](array: Array[T]): Array[T] = {
    val rnd = new java.util.Random
    for (n <- Iterator.range(array.length - 1, 0, -1)) {
        val k = rnd.nextInt(n + 1)
        val t = array(k); array(k) =
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1 Answer

6

It seems Scala is doing something different from Java when it comes to varargs. At least, I can't get that array shuffled any way I try. Supposedly, the shuffle on the list would shuffle the array because the list is array-backed. Well, it seems that Scala will create a new array when passing vararg arguments to Java, therefore making the aforementioned example useless.

scala> val b = java.util.Arrays.asList(a: _*)
b: java.util.List[java.lang.String] = [a, b, c]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [a, b, c]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [c, b, a]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [a, c, b]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [b, a, c]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [a, b, c]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(a, b, c) [c, a, b]

It does works with Ints, despite the claim otherwise, though:

scala> val a = Array(1,2,3)
a: Array[Int] = Array(1, 2, 3)

scala> val b = java.util.Arrays.asList(a: _*)
b: java.util.List[Int] = [1, 2, 3]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(1, 2, 3) [2, 3, 1]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(1, 2, 3) [3, 2, 1]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(1, 2, 3) [3, 2, 1]

scala> java.util.Collections.shuffle(b); println(a.toString+" "+b.toString)
Array(1, 2, 3) [1, 2, 3]

On Scala 2.8, there's a simpler way:

scala> scala.util.Random.shuffle(a)
res32: Sequence[Int] = Array(1, 2, 3)

scala> scala.util.Random.shuffle(a)
res33: Sequence[Int]
answered 2009-08-11T14:00:59.067

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