快速sorting3

如何将下面的函数转换为swift 3 ？ 目前得到一个Binary operator '..<' cannot be applied to operands of type 'Int' and 'Self.IndexDistance'错误的Binary operator '..<' cannot be applied to operands of type 'Int' and 'Self.IndexDistance' 。

 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

Answer 1

count返回一个IndexDistance ，它是描述两个集合索引之间距离的types。 IndexDistance需要是一个SignedInteger ，但不一定是一个Int ，可以不同于Index 。 因此无法创build范围0..<count - 1 。

一个解决scheme是使用startIndex和endIndex而不是0并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是零）。

请注意，根据新的“Swift APIdevise指南” ， shuffle()是突变的shuffle方法的“正确”名称， shuffle()是返回数组的非突变副本。

 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（甚至更一般）Swift 3版本已被添加到如何在Swift中洗牌数组？ 同时。

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对于Swift 4（Xcode 9），必须通过调用集合的swapAt()方法来调用swap()函数。 此外， Indextypes的限制不再需要：

 extension MutableCollection { /// 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 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(_:_:)的更多信息，请参阅SE-0173添加MutableCollection.swapAt(_:_:) 。

Answer 2

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，这样每次提供相同的种子时，就可以得到相同的伪随机序列值，以防需要重新创build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) 

Answer 3

我会build议简单地洗牌数组，而不是试图扩展到一般的集合：

 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]) } } } 

Answer 4

你可以使用GameplayKit框架的NSArray Extension来实现这个function：

 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]