Create an array of random numbers in Swift

eloist picture eloist · Jan 25, 2015 · Viewed 15.4k times · Source

I'm just starting to learn Swift.

I'm attempting to create an array of several random numbers, and eventually sort the array. I'm able to create an array of one random number, but what's the best way to iterate this to create an array of several random numbers?

func makeList() {
   var randomNums = arc4random_uniform(20) + 1

    let numList = Array(arrayLiteral: randomNums)

}

makeList()

Answer

Leo Dabus picture Leo Dabus · Jan 25, 2015

In Swift 4.2 there is a new static method for fixed width integers that makes the syntax more user friendly:

func makeList(_ n: Int) -> [Int] {
    return (0..<n).map { _ in .random(in: 1...20) }
}

Edit/update: Swift 5.1 or later

We can also extend Range and ClosedRange and create a method to return n random elements:

extension RangeExpression where Bound: FixedWidthInteger {
    func randomElements(_ n: Int) -> [Bound] {
        precondition(n > 0)
        switch self {
        case let range as Range<Bound>: return (0..<n).map { _ in .random(in: range) }
        case let range as ClosedRange<Bound>: return (0..<n).map { _ in .random(in: range) }
        default: return []
        }
    }
}

extension Range where Bound: FixedWidthInteger {
    var randomElement: Bound { .random(in: self) }
}

extension ClosedRange where Bound: FixedWidthInteger {
    var randomElement: Bound { .random(in: self) }
}

Usage:

let randomElements = (1...20).randomElements(5)  // [17, 16, 2, 15, 12]
randomElements.sorted() // [2, 12, 15, 16, 17]

let randomElement = (1...20).randomElement   // 4 (note that the computed property returns a non-optional instead of the default method which returns an optional)

let randomElements = (0..<2).randomElements(5)  // [1, 0, 1, 1, 1]
let randomElement = (0..<2).randomElement   // 0

Note: for Swift 3, 4 and 4.1 and earlier click here.