29

如何将下面的函数转换为swift 3？当前出现Binary operator '..<' cannot be applied to operands of type 'Int' and 'Self.IndexDistance'错误。

extension MutableCollection where Index == Int {
  /// Shuffle the elements of `self` in-place.
  mutating func shuffleInPlace() {
    // empty and single-element collections don't shuffle
    if count < 2 { return }

    for i in 0..<count - 1 { //error takes place here
      let j = Int(arc4random_uniform(UInt32(count - i))) + i
      guard i != j else { continue }
      swap(&self[i], &self[j])
    }
  }
}

参考：https ://stackoverflow.com/a/24029847/5222077

4

4 回答 4

78

count返回一个IndexDistance描述两个集合索引之间距离的类型。IndexDistance必须是 a SignedInteger，但不必是 anInt并且可以不同于Index. 因此无法创建范围0..<count - 1。

一个解决方案是使用startIndexandendIndex而不是0and count：

extension MutableCollection where Index == Int {
    /// Shuffle the elements of `self` in-place.
    mutating func shuffle() {
        // empty and single-element collections don't shuffle
        if count < 2 { return }

        for i in startIndex ..< endIndex - 1 {
            let j = Int(arc4random_uniform(UInt32(endIndex - i))) + i
            if i != j {
                swap(&self[i], &self[j])
            }
        }
    }
}

另一个优点是这也适用于数组切片 （其中第一个元素的索引不一定为零）。

请注意，根据新的“Swift API 设计指南”， shuffle()是变异 shuffle 方法的“正确”名称，以及shuffled()返回数组的非变异对应物：

extension Collection {
    /// Return a copy of `self` with its elements shuffled
    func shuffled() -> [Iterator.Element] {
        var list = Array(self)
        list.shuffle()
        return list
    }
}

更新：一个（更通用的）Swift 3 版本已添加到 How do I shuffle an array in Swift? 同时。

对于Swift 4 (Xcode 9) ，必须通过调用集合的方法来替换对swap() 函数的调用。也不再需要swapAt()对类型的限制：Index

extension MutableCollection {
    /// Shuffle the elements of `self` in-place.
    mutating func shuffle() {
        for i in indices.dropLast() {
            let diff = distance(from: i, to: endIndex)
            let j = index(i, offsetBy: numericCast(arc4random_uniform(numericCast(diff))))
            swapAt(i, j)
        }
    }
}

有关. _MutableCollection.swapAt(_:_:) _swapAt

从Swift 4.2（Xcode 10，目前处于测试阶段）开始，实现了 SE-0202 Random Unification， shuffle()并且shuffled()是 Swift 标准库的一部分。

于 2016-06-15T19:11:33.287 回答
10

Gamekit 中有一个fisher-yates shuffle：

import GameKit
let unshuffledArray = [1,2,3,4]
let shuffledArray = GKRandomSource.sharedRandom().arrayByShufflingObjects(in: unshuffledArray)
print(shuffledArray)

您还可以传入并存储一个随机种子，这样每次您提供相同的种子时，您都会获得相同的伪随机洗牌值序列，以防您需要重新创建模拟。

import GameKit
let unshuffledArray = [1,2,3,4]
let randomSource = GKLinearCongruentialRandomSource(seed: 1)
let shuffledArray = randomSource.arrayByShufflingObjects(in: unshuffledArray)
//Always [1,4,2,3]
print(shuffledArray)
于 2016-10-30T14:57:34.673 回答
8

我建议简单地改组数组，而不是尝试将其扩展到一般的集合：

extension Array {
    mutating func shuffle () {
        for i in (0..<self.count).reversed() {
            let ix1 = i
            let ix2 = Int(arc4random_uniform(UInt32(i+1)))
            (self[ix1], self[ix2]) = (self[ix2], self[ix1])
        }
    }
}
于 2016-06-15T19:02:18.543 回答
0

您可以为此使用 GameplayKit 框架中的 NSArray 扩展：

import GameplayKit

extension Collection {
    func shuffled() -> [Iterator.Element] {
        let shuffledArray = (self as? NSArray)?.shuffled()
        let outputArray = shuffledArray as? [Iterator.Element]
        return outputArray ?? []
    }
    mutating func shuffle() {
        if let selfShuffled = self.shuffled() as? Self {
            self = selfShuffled
        }
    }
}

// Usage example:

var numbers = [1,2,3,4,5]
numbers.shuffle()

print(numbers) // output example: [2, 3, 5, 4, 1]

print([10, "hi", 9.0].shuffled()) // output example: [hi, 10, 9]
于 2017-06-24T20:47:26.313 回答