CODEWITHSUNDEEP
arraysscalafor-loopfunctional-programmingscala-collections
Max Wong
Max Wong

Reputation: 750

Tile a small Array in a large Array multiple times in scala

I want to tile a small array multiple times in a large array. I'm looking for an "official" way of doing this. A naive solution follows:

val arr = Array[Int](1, 2, 3)
val array = {
  val arrBuf = ArrayBuffer[Int]()
  for (_ <- 1 until 10) {
    arrBuf ++= arr
  }
  arrBuf.toArray
}

Upvotes: 0

Views: 149

Answers (2)

Luis Miguel Mej&#237;a Su&#225;rez
Luis Miguel Mej&#237;a Su&#225;rez

Reputation: 22895

If you do not know why Arrays are good for performance (meaning you do not really need raw performance in this case) I would recommend you do not use them, and rather stick with List or Vector instead.

Arrays are not proper Scala collections, they are just plain JVM arrays. Meaning, they are mutable, very efficient (especially for unboxed primitives), fixed in memory size, and very restricted. They behave like normal scala collections because of implicit conversions and extension methods. But, due to their mutability and invariance, you really should avoid them unless you have good reasons for using them.
The proposed solution by Andronicus is not ideal for arrays (but it would be a very good solution for any real collection) because given arrays have fixed memory size, this fattening will end in constant reallocations and memory copying under the hood.

Anyways, here is a slight variation to such solution, using lists instead; which is a little bit more efficient.

implicit class ListOps[A](private val list: List[A]) extends AnyVal {
  def times[B >: A](n: Int): List[B] =
    Iterator.fill(n)(list).flatten.toList
}

List(1, 2, 3).times(3)
// res: List[Int] = List(1, 2, 3, 1, 2, 3, 1, 2, 3)

And here is also an efficient version using the new ArraySeq introduced in 2.13; which is an immutable Array.
(Note, you can do this using plain Arrays too)

implicit class ArraySeqOps[A](private val arr: ArraySeq[A]) extends AnyVal {
  def times[B >: A](n: Int): ArraySeq[B] =
    ArraySeq.tabulate(n * arr.lenght) { i => arr(i % arr.length) }
}

ArraySeq(1, 2, 3).times(3)
// res: ArraySeq[Int] = ArraySeq(1, 2, 3, 1, 2, 3, 1, 2, 3)

Upvotes: 2

Andronicus
Andronicus

Reputation: 26066

You can use Array.fill:

Array.fill(10)(Array(1, 2, 3)).flatten

Upvotes: 1

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