小编典典

在Swift中从数组中获取随机元素

swift

我有一个像这样的数组:

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#,但是我不确定该从哪里开始。我想我应该先对数组进行随机排序,然后再从中选择前3个项目?

我尝试使用以下扩展名将其改组:

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

但随后在“ shuffle()”的位置说“’()’不可转换为’[Int]’”。

为了挑选一些元素,我使用:

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

到目前为止看起来还不错。

如何洗牌?还是有人对此有更好/更优雅的解决方案?


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2020-07-07

共1个答案

小编典典

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"
2020-07-07