How to get a random element from a Set in Scala

Question

For any given set, for instance,

val fruits = Set("apple", "grape", "pear", "banana")

how to get a random element from fruits ?

Many Thanks.

Rok Kralj · Answer 1 · 2017-04-11T08:41:01.710

22

So, every answer posted before has complexity O(n) in terms of space, since they create a copy a whole collection in some way. Here is a solution without any additional copying (therefore it is "constant space"):

def random[T](s: Set[T]): T = {
  val n = util.Random.nextInt(s.size)
  s.iterator.drop(n).next
}

edited Apr 11 '17 at 08:41

answered Nov 13 '15 at 18:14

Rok Kralj

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This is a better answer for the reason you put forward. It could be done in a one liner like this: val s = Set(1,2,3); s.drop(random.nextInt(s.size).head – justinhj Feb 12 '16 at 19:59
9

@justinhj: Nope, your solution is O(n) space again, since `drop` on `Set` causes a set with the remaining elements to be created. You *have* to use iterators, because they are non-strict. – Rok Kralj Feb 12 '16 at 21:52
Good solution. Its worth pointing out that if the set is empty `it.next` will throw. – Antony Perkov Nov 09 '16 at 05:22
1

@AntonyPerkov which makes sense because ... what does it mean to sample from an empty set? – Jus12 Feb 08 '17 at 12:10
1

This is still O(n) time, though. – Brian McCutchon Apr 28 '17 at 15:34
Yes, I also look for a O(1) time and O(1) space version – pihentagy Aug 29 '17 at 16:05
You will have to create your own Set implementation for that. and the time will probably be O(log n) if you want to have other operations efficient. – Rok Kralj Aug 29 '17 at 19:09

Govind Singh · Accepted Answer · 2014-07-31T10:21:21.607

21

convert into Vector and get random element from it

scala> val fruits = Set("apple", "grape", "pear", "banana")
fruits: scala.collection.immutable.Set[String] = Set(apple, grape, pear, banana)

scala> import scala.util.Random
import scala.util.Random

scala> val rnd=new Random
rnd: scala.util.Random = scala.util.Random@31a9253

scala> fruits.toVector(rnd.nextInt(fruits.size))
res8: String = apple

edited Jul 31 '14 at 10:21

answered Jul 31 '14 at 08:21

Govind Singh

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Indexing is `O(n)` on List. Why not use `Array` or `Vector`? – Kigyo Jul 31 '14 at 10:18
Can't scala.util.Random.nextInt occasionally return negative numbers? – Tim Barrass Jan 30 '18 at 09:07
2

@TimBarrass `nextInt()` will return a negative number half of the time, but `nextInt(n: Int)`, which is used here, starts at 0 and ends at `n` (exclusive): https://www.scala-lang.org/api/current/scala/util/Random.html#nextInt(n:Int):Int – László van den Hoek Sep 21 '18 at 10:01
Or convert to `IndexedSeq`, and code to interfaces. – Abhijit Sarkar Nov 18 '18 at 11:54

László van den Hoek · Answer 3 · 2020-05-25T12:59:15.427

Drawing inspiration from the other answers to this question, I've come up with:

private def randomItem[T](items: Traversable[T]): Option[T] = {
  val i = Random.nextInt(items.size)
  items.view(i, i + 1).headOption
}

This doesn't copy anything, doesn't fail if the Set (or other type of Traversable) is empty, and it's clear at a glance what it does. If you're certain that the Set is not empty, you could use .head instead of headOption, returning a T.

score 2 · Answer 4 · edited May 19 '18 at 18:35

2

Solution1

Random way ( import scala.util.Random )

scala>  fruits.toList(Random.nextInt(fruits.size))
res0: java.lang.String = banana

Solution2

Math way (no imports)

scala> fruits.toList((math.random*fruits.size).toInt)
res1: String = banana

edited May 19 '18 at 18:35

matanster

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answered Aug 02 '14 at 07:00

AmeyChavan

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Chained List is not a good option to pick one element at random. Chose array or vector instead – Boris Sep 21 '17 at 08:03

score 2 · Answer 5 · answered May 05 '16 at 23:43

You can directly access an element of a Set with slice. I used this when I was working with a set that was changing in size, so converting it to a Vector every time seemed like overkill.

val roll = new Random ()

val n = roll nextInt (fruits size)
fruits slice (n, n + 1) last

score 1 · Answer 6 · edited Jul 31 '14 at 09:13

1

   import Scala.util.Random

   val fruits = Set("apple", "grape", "pear", "banana").toVector

   val sz =fruits.size

   val num = Random.nextInt(sz)

   fruits(num)

edited Jul 31 '14 at 09:13

Govind Singh

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answered Jul 31 '14 at 08:23

Ashalynd

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Yep, Set should be converted to container that supports accessing elements by sequential number. Vector is quicker than List. – Ashalynd Jul 31 '14 at 08:30
@dividebyzero previously it was set – Govind Singh Jul 07 '15 at 06:00

elm · Answer 7 · 2016-02-16T08:09:12.127

Not converting the Set to an ordered collection but using zipWithIndex we can attribute an index to each item in the collection,

fruits.zipWithIndex
Set((apple,0), (grape,1), (pear,2), (banana,3))

Thus for val rnd = util.Random.nextInt(fruits.size),

fruits.zipWithIndex.find( _._2 == rnd)
Option[(String, Int)] = Some((banana,3))

Given an empty set,

Set[String]().zipWithIndex.find( _._2 == 3)
Option[(String, Int)] = None

score 0 · Answer 8 · answered Jul 03 '19 at 21:32

0

If you don't mind an O(n) solution:

import util.Random

// val fruits = Set("apple", "grape", "pear", "banana")
Random.shuffle(fruits).head
// "pear"

answered Jul 03 '19 at 21:32

Xavier Guihot

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How to get a random element from a Set in Scala

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