How to get a random element from a Set in Scala

问题:

For any given set, for instance,

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

how to get a random element from fruits ?

Many Thanks.

回答1:

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

回答2:

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
}

回答3:

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

回答4:

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

回答5:

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.

回答6:

   import Scala.util.Random

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

   val sz =fruits.size

   val num = Random.nextInt(sz)

   fruits(num)

回答7:

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

回答8:

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

import util.Random

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