在 Swift 中从数组中获取随机元素

Question

我有一个类似的数组：

var names: String = [ "Peter", "Steve", "Max", "Sandra", "Roman", "Julia" ]

我想从该数组中获取 3 个随机元素。我来自 C#，但很快我不确定从哪里开始。例如，我认为我应该先对数组进行洗牌，然后从中挑选前 3 个项目？

我尝试使用以下扩展名对其进行洗牌：

extension Array
{
    mutating func shuffle()
    {
        for _ in 0..<10
        {
            sort { (_,_) in arc4random() < arc4random() }
        }
    }
}

但它会在“shuffle()”的位置显示“'()' is not convertible to '[Int]'”。

为了挑选我使用的一些元素：

var randomPicks = names[0..<4];

目前看起来不错。

如何洗牌？或者有人对此有更好/更优雅的解决方案吗？

Answer 1

Xcode 11 • Swift 5.1

extension Collection {
    func choose(_ n: Int) -> ArraySlice<Element> { shuffled().prefix(n) }
}

游乐场测试

var alphabet = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
let shuffledAlphabet = alphabet.shuffled()  // "O", "X", "L", "D", "N", "K", "R", "E", "S", "Z", "I", "T", "H", "C", "U", "B", "W", "M", "Q", "Y", "V", "A", "G", "P", "F", "J"]
let letter = alphabet.randomElement()  // "D"
var numbers = Array(0...9)
let shuffledNumbers = numbers.shuffled()
shuffledNumbers                              // [8, 9, 3, 6, 0, 1, 4, 2, 5, 7]
numbers            // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
numbers.shuffle() // mutate it  [6, 0, 2, 3, 9, 1, 5, 7, 4, 8]
numbers            // [6, 0, 2, 3, 9, 1, 5, 7, 4, 8]
let pick3numbers = numbers.choose(3)  // [8, 9, 2]

---

extension RangeReplaceableCollection {
    /// Returns a new Collection shuffled
    var shuffled: Self { .init(shuffled()) }
    /// Shuffles this Collection in place
    @discardableResult
    mutating func shuffledInPlace() -> Self  {
        self = shuffled
        return self
    }
    func choose(_ n: Int) -> SubSequence { shuffled.prefix(n) }
}

---

var alphabetString = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
let shuffledAlphabetString = alphabetString.shuffled  // "DRGXNSJLFQHPUZTBKVMYAWEICO"
let character = alphabetString.randomElement()  // "K"
alphabetString.shuffledInPlace() // mutate it  "WYQVBLGZKPFUJTHOXERADMCINS"
alphabetString            // "WYQVBLGZKPFUJTHOXERADMCINS"
let pick3Characters = alphabetString.choose(3)  // "VYA"

Answer 2

或者有人对此有更好/更优雅的解决方案吗？

我愿意。在算法上比公认的答案更好，它为完全随机播放执行 count-1 arc4random_uniform 操作，我们可以简单地在 n arc4random_uniform 操作中选择 n 个值。 p>

实际上，我有 两种 方法比公认的答案做得更好：

更好的解决方案

extension Array {
    /// Picks `n` random elements (straightforward approach)
    subscript (randomPick n: Int) -> [Element] {
        var indices = [Int](0..<count)
        var randoms = [Int]()
        for _ in 0..<n {
            randoms.append(indices.remove(at: Int(arc4random_uniform(UInt32(indices.count)))))
        }
        return randoms.map { self[$0] }
    }
}

最佳解决方案

以下解决方案比前一个解决方案快两倍。

适用于 Swift 3.0 和 3.1

extension Array {
    /// Picks `n` random elements (partial Fisher-Yates shuffle approach)
    subscript (randomPick n: Int) -> [Element] {
        var copy = self
        for i in stride(from: count - 1, to: count - n - 1, by: -1) {
            let j = Int(arc4random_uniform(UInt32(i + 1)))
            if j != i {
                swap(&copy[i], &copy[j])
            }
        }
        return Array(copy.suffix(n))
    }
}

适用于 Swift 3.2 和 4.x

extension Array {
    /// Picks `n` random elements (partial Fisher-Yates shuffle approach)
    subscript (randomPick n: Int) -> [Element] {
        var copy = self
        for i in stride(from: count - 1, to: count - n - 1, by: -1) {
            copy.swapAt(i, Int(arc4random_uniform(UInt32(i + 1))))
        }
        return Array(copy.suffix(n))
    }
}

用法：

let digits = Array(0...9)  // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
let pick3digits = digits[randomPick: 3]  // [8, 9, 0]

Answer 3

Swift 4.1 及更低版本

let playlist = ["Nothing Else Matters", "Stairway to Heaven", "I Want to Break Free", "Yesterday"]
let index = Int(arc4random_uniform(UInt32(playlist.count)))
let song = playlist[index]

Swift 4.2 及以上版本

if let song = playlist.randomElement() {
  print(song)
} else {
  print("Empty playlist.")
}

Answer 4

您可以在 Array 上定义扩展：

extension Array {
    func pick(_ n: Int) -> [Element] {
        guard count >= n else {
            fatalError("The count has to be at least \(n)")
        }
        guard n >= 0 else {
            fatalError("The number of elements to be picked must be positive")
        }

        let shuffledIndices = indices.shuffled().prefix(upTo: n)
        return shuffledIndices.map {self[$0]}
    }
}

[ "Peter", "Steve", "Max", "Sandra", "Roman", "Julia" ].pick(3)

---

如果初始数组可能有重复，并且您希望值的唯一性：

extension Array where Element: Hashable {
    func pickUniqueInValue(_ n: Int) -> [Element] {
        let set: Set<Element> = Set(self)
        guard set.count >= n else {
            fatalError("The array has to have at least \(n) unique values")
        }
        guard n >= 0 else {
            fatalError("The number of elements to be picked must be positive")
        }

        return Array(set.prefix(upTo: set.index(set.startIndex, offsetBy: n)))
    }
}

[ "Peter", "Steve", "Max", "Sandra", "Roman", "Julia" ].pickUniqueInValue(3)

Answer 5

您也可以使用 arc4random() 从数组中选择三个元素。像这样的：

extension Array {
    func getRandomElements() -> (T, T, T) {
        return (self[Int(arc4random()) % Int(count)],
                self[Int(arc4random()) % Int(count)],
                self[Int(arc4random()) % Int(count)])
    }
}

let names = ["Peter", "Steve", "Max", "Sandra", "Roman", "Julia"]
names.getRandomElements()

这只是一个示例，您还可以在函数中包含逻辑来为每个函数获取不同的名称。