How can I get a random number in Kotlin? - random

Question

A generic method that can return a random integer between 2 parameters like ruby does with rand(0..n).
Any suggestion?

Answer 1

My suggestion would be an extension function on IntRange to create randoms like this: (0..10).random()
TL;DR Kotlin >= 1.3, one Random for all platforms
As of 1.3, Kotlin comes with its own multi-platform Random generator. It is described in this KEEP. The extension described below is now part of the Kotlin standard library, simply use it like this:
val rnds = (0..10).random() // generated random from 0 to 10 included
Kotlin < 1.3
Before 1.3, on the JVM we use Random or even ThreadLocalRandom if we're on JDK > 1.6.
fun IntRange.random() =
Random().nextInt((endInclusive + 1) - start) + start
Used like this:
// will return an `Int` between 0 and 10 (incl.)
(0..10).random()
If you wanted the function only to return 1, 2, ..., 9 (10 not included), use a range constructed with until:
(0 until 10).random()
If you're working with JDK > 1.6, use ThreadLocalRandom.current() instead of Random().
KotlinJs and other variations
For kotlinjs and other use cases which don't allow the usage of java.util.Random, see this alternative.
Also, see this answer for variations of my suggestion. It also includes an extension function for random Chars.

Answer 2

Generate a random integer between from(inclusive) and to(exclusive)
import java.util.Random
val random = Random()
fun rand(from: Int, to: Int) : Int {
return random.nextInt(to - from) + from
}

Answer 3

As of kotlin 1.2, you could write:
(1..3).shuffled().last()
Just be aware it's big O(n), but for a small list (especially of unique values) it's alright :D

Answer 4

In Kotlin SDK >=1.3 you can do it like
import kotlin.random.Random
val number = Random.nextInt(limit)

Answer 5

You can create an extension function similar to java.util.Random.nextInt(int) but one that takes an IntRange instead of an Int for its bound:
fun Random.nextInt(range: IntRange): Int {
return range.start + nextInt(range.last - range.start)
}
You can now use this with any Random instance:
val random = Random()
println(random.nextInt(5..9)) // prints 5, 6, 7, 8, or 9
If you don't want to have to manage your own Random instance then you can define a convenience method using, for example, ThreadLocalRandom.current():
fun rand(range: IntRange): Int {
return ThreadLocalRandom.current().nextInt(range)
}
Now you can get a random integer as you would in Ruby without having to first declare a Random instance yourself:
rand(5..9) // returns 5, 6, 7, 8, or 9

Answer 6

Possible Variation to my other answer for random chars
In order to get random Chars, you can define an extension function like this
fun ClosedRange<Char>.random(): Char =
(Random().nextInt(endInclusive.toInt() + 1 - start.toInt()) + start.toInt()).toChar()
// will return a `Char` between A and Z (incl.)
('A'..'Z').random()
If you're working with JDK > 1.6, use ThreadLocalRandom.current() instead of Random().
For kotlinjs and other use cases which don't allow the usage of java.util.Random, this answer will help.
Kotlin >= 1.3 multiplatform support for Random
As of 1.3, Kotlin comes with its own multiplatform Random generator. It is described in this KEEP. You can now directly use the extension as part of the Kotlin standard library without defining it:
('a'..'b').random()

Answer 7

Building off of #s1m0nw1 excellent answer I made the following changes.
(0..n) implies inclusive in Kotlin
(0 until n) implies exclusive in Kotlin
Use a singleton object for the Random instance (optional)
Code:
private object RandomRangeSingleton : Random()
fun ClosedRange<Int>.random() = RandomRangeSingleton.nextInt((endInclusive + 1) - start) + start
Test Case:
fun testRandom() {
Assert.assertTrue(
(0..1000).all {
(0..5).contains((0..5).random())
}
)
Assert.assertTrue(
(0..1000).all {
(0..4).contains((0 until 5).random())
}
)
Assert.assertFalse(
(0..1000).all {
(0..4).contains((0..5).random())
}
)
}

Answer 8

Examples random in the range [1, 10]
val random1 = (0..10).shuffled().last()
or utilizing Java Random
val random2 = Random().nextInt(10) + 1

Answer 9

No need to use custom extension functions anymore. IntRange has a random() extension function out-of-the-box now.
val randomNumber = (0..10).random()

Answer 10

Kotlin >= 1.3, multiplatform support for Random
As of 1.3, the standard library provided multi-platform support for randoms, see this answer.
Kotlin < 1.3 on JavaScript
If you are working with Kotlin JavaScript and don't have access to java.util.Random, the following will work:
fun IntRange.random() = (Math.random() * ((endInclusive + 1) - start) + start).toInt()
Used like this:
// will return an `Int` between 0 and 10 (incl.)
(0..10).random()

Answer 11

Another way of implementing s1m0nw1's answer would be to access it through a variable. Not that its any more efficient but it saves you from having to type ().
val ClosedRange<Int>.random: Int
get() = Random().nextInt((endInclusive + 1) - start) + start
And now it can be accessed as such
(1..10).random

Answer 12

If the numbers you want to choose from are not consecutive, you can use random().
Usage:
val list = listOf(3, 1, 4, 5)
val number = list.random()
Returns one of the numbers which are in the list.

Answer 13

Using a top-level function, you can achieve exactly the same call syntax as in Ruby (as you wish):
fun rand(s: Int, e: Int) = Random.nextInt(s, e + 1)
Usage:
rand(1, 3) // returns either 1, 2 or 3

Answer 14

There is no standard method that does this but you can easily create your own using either Math.random() or the class java.util.Random. Here is an example using the Math.random() method:
fun random(n: Int) = (Math.random() * n).toInt()
fun random(from: Int, to: Int) = (Math.random() * (to - from) + from).toInt()
fun random(pair: Pair<Int, Int>) = random(pair.first, pair.second)
fun main(args: Array<String>) {
val n = 10
val rand1 = random(n)
val rand2 = random(5, n)
val rand3 = random(5 to n)
println(List(10) { random(n) })
println(List(10) { random(5 to n) })
}
This is a sample output:
[9, 8, 1, 7, 5, 6, 9, 8, 1, 9]
[5, 8, 9, 7, 6, 6, 8, 6, 7, 9]

Answer 15

Kotlin standard lib doesn't provide Random Number Generator API. If you aren't in a multiplatform project, it's better to use the platform api (all the others answers of the question talk about this solution).
But if you are in a multiplatform context, the best solution is to implement random by yourself in pure kotlin for share the same random number generator between platforms. It's more simple for dev and testing.
To answer to this problem in my personal project, i implement a pure Kotlin Linear Congruential Generator. LCG is the algorithm used by java.util.Random. Follow this link if you want to use it :
https://gist.github.com/11e5ddb567786af0ed1ae4d7f57441d4
My implementation purpose nextInt(range: IntRange) for you ;).
Take care about my purpose, LCG is good for most of the use cases (simulation, games, etc...) but is not suitable for cryptographically usage because of the predictability of this method.

Answer 16

First, you need a RNG. In Kotlin you currently need to use the platform specific ones (there isn't a Kotlin built in one). For the JVM it's java.util.Random. You'll need to create an instance of it and then call random.nextInt(n).

Answer 17

To get a random Int number in Kotlin use the following method:
import java.util.concurrent.ThreadLocalRandom
fun randomInt(rangeFirstNum:Int, rangeLastNum:Int) {
val randomInteger = ThreadLocalRandom.current().nextInt(rangeFirstNum,rangeLastNum)
println(randomInteger)
}
fun main() {
randomInt(1,10)
}
// Result – random Int numbers from 1 to 9
Hope this helps.

Answer 18

Below in Kotlin worked well for me:
(fromNumber.rangeTo(toNumber)).random()
Range of the numbers starts with variable fromNumber and ends with variable toNumber. fromNumber and toNumber will also be included in the random numbers generated out of this.

Answer 19

You could create an extension function:
infix fun ClosedRange<Float>.step(step: Float): Iterable<Float> {
require(start.isFinite())
require(endInclusive.isFinite())
require(step.round() > 0.0) { "Step must be positive, was: $step." }
require(start != endInclusive) { "Start and endInclusive must not be the same"}
if (endInclusive > start) {
return generateSequence(start) { previous ->
if (previous == Float.POSITIVE_INFINITY) return#generateSequence null
val next = previous + step.round()
if (next > endInclusive) null else next.round()
}.asIterable()
}
return generateSequence(start) { previous ->
if (previous == Float.NEGATIVE_INFINITY) return#generateSequence null
val next = previous - step.round()
if (next < endInclusive) null else next.round()
}.asIterable()
}
Round Float value:
fun Float.round(decimals: Int = DIGITS): Float {
var multiplier = 1.0f
repeat(decimals) { multiplier *= 10 }
return round(this * multiplier) / multiplier
}
Method's usage:
(0.0f .. 1.0f).step(.1f).forEach { System.out.println("value: $it") }
Output:
value: 0.0 value: 0.1 value: 0.2 value: 0.3 value: 0.4 value: 0.5
value: 0.6 value: 0.7 value: 0.8 value: 0.9 value: 1.0

Answer 20

Full source code. Can control whether duplicates are allowed.
import kotlin.math.min
abstract class Random {
companion object {
fun string(length: Int, isUnique: Boolean = false): String {
if (0 == length) return ""
val alphabet: List<Char> = ('a'..'z') + ('A'..'Z') + ('0'..'9') // Add your set here.
if (isUnique) {
val limit = min(length, alphabet.count())
val set = mutableSetOf<Char>()
do { set.add(alphabet.random()) } while (set.count() != limit)
return set.joinToString("")
}
return List(length) { alphabet.random() }.joinToString("")
}
fun alphabet(length: Int, isUnique: Boolean = false): String {
if (0 == length) return ""
val alphabet = ('A'..'Z')
if (isUnique) {
val limit = min(length, alphabet.count())
val set = mutableSetOf<Char>()
do { set.add(alphabet.random()) } while (set.count() != limit)
return set.joinToString("")
}
return List(length) { alphabet.random() }.joinToString("")
}
}
}

Answer 21

Whenever there is a situation where you want to generate key or mac address which is hexadecimal number having digits based on user demand, and that too using android and kotlin, then you my below code helps you:
private fun getRandomHexString(random: SecureRandom, numOfCharsToBePresentInTheHexString: Int): String {
val sb = StringBuilder()
while (sb.length < numOfCharsToBePresentInTheHexString) {
val randomNumber = random.nextInt()
val number = String.format("%08X", randomNumber)
sb.append(number)
}
return sb.toString()
}

Answer 22

Here is a straightforward solution in Kotlin, which also works on KMM:
fun IntRange.rand(): Int =
Random(Clock.System.now().toEpochMilliseconds()).nextInt(first, last)
Seed is needed for the different random number on each run. You can also do the same for the LongRange.

Answer 23

Gets the next random Int from the random number generator.
Random.nextInt()

Answer 24

to be super duper ))
fun rnd_int(min: Int, max: Int): Int {
var max = max
max -= min
return (Math.random() * ++max).toInt() + min
}