Replace random elements of Array

Question

I have two arrays of same length

 import scala.util.Random
 val length = 10
 val x = 60 // Selection percentage
 val rnd = new Random
 var arrayOne =  Array.fill(length)(rnd .nextInt(100))
 arrayOne: Array[Int] = Array(8, 77, 11, 19, 17, 73, 5, 18, 45, 69)

 val arrayTwo =  Array.fill(length)(rnd .nextInt(100))
 arrayTwo: Array[Int] = Array(96, 21, 85, 70, 28, 5, 31, 56, 27, 76)

I can select first x percent element from arrayTwo and those selected elements can replace first x percent elements of arrayOne in the following way.

arrayOne = arrayTwo.take((length * x / 100).toInt) ++ arrayOne.drop((length * x / 100).toInt)
arrayOne: Array[Int] = Array(96, 21, 85, 70, 28, 5, 5, 18, 45, 69)

Now I want to select random x percent elements from arrayTwo and that selected elements will replace random x percent elements of arrayOne. How can I do this?

Andronicus · Answer 1 · 2020-08-11T08:41:56.603

2

You can exchange every item with a probability x:

val x = 60D
val exchanged = arrayOne.indices
  .map(x => if(math.random > x / 100) arrayOne(x) else arrayTwo(x))

But that way you have no guarantee that (length * x / 100).toInt elements will be from arrayTwo. To achieve that I would go for iterative / recursive algorithm, where I'd pick random index until I have as much as I want.

edited Aug 11 '20 at 08:41

answered Aug 11 '20 at 08:32

Andronicus

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can you share your iterative approach? – Asif Aug 11 '20 at 08:36
I set `x = 60` , and your code is returning **arrayOne** every time. – Asif Aug 11 '20 at 08:40
@Asif yes, because you're making it an `Int` – Andronicus Aug 11 '20 at 08:41
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@Asif iterative approach is easy to find in other languages, in Scala this will not be that different. Here is an example: https://stackoverflow.com/questions/19269545/how-to-get-a-number-of-random-elements-from-an-array – Andronicus Aug 11 '20 at 08:43

esse · Answer 2 · 2020-08-11T09:09:12.943

2

You can do it via Random.shuffle:

scala> val randSet = Random.shuffle(arrayOne.indices.toBuffer).take((arrayOne.length * x / 100).toInt).toSet
randSet: scala.collection.immutable.Set[Int] = HashSet(0, 6, 9, 3, 8, 4)

scala> val randMerged = arrayOne.indices.map(i => if(randSet(i)) arrayTwo(i) else arrayOne(i))
randMerged: IndexedSeq[Int] = Vector(96, 77, 11, 70, 28, 73, 31, 18, 27, 76)

The randSet will take x percent indices randomly.

If you do not care the number's position, there is a simple one:

scala> val size = (arrayOne.length * x / 100).toInt
size: Int = 6

scala> Random.shuffle(arrayTwo).take(size) ++ Random.shuffle(arrayOne).drop(size)
res11: scala.collection.mutable.ArraySeq[Int] = ArraySeq(76, 85, 28, 56, 21, 27, 69, 45, 17, 77)

edited Aug 11 '20 at 09:09

answered Aug 11 '20 at 09:01

esse

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shuffle is expensive operation – Asif Aug 11 '20 at 09:30
1

@Asif It's a O(n) time complexity, you can not do better than O(n) in any way of your problem. – esse Aug 11 '20 at 10:21

Replace random elements of Array

2 Answers2