我可能有一个像下面这样的数组:

[1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]

或者,实际上,任何类似类型的数据部分的序列。我要做的是确保每个相同的元素只有一个。例如,上面的数组将变成:

[1, 4, 2, 6, 24, 15, 60]

请注意,删除了2、6和15的重复项,以确保每个相同的元素中只有一个。Swift是否提供了一种容易做到这一点的方法,还是我必须自己做?


当前回答

受https://www.swiftbysundell.com/posts/the-power-of-key-paths-in-swift的启发,我们可以声明一个更强大的工具,它能够过滤任何keyPath上的唯一性。感谢Alexander关于复杂性的各种答案的评论,下面的解决方案应该是最优的。

Non-mutating解决方案

我们扩展了一个函数,它可以过滤任意keyPath上的唯一性:

extension RangeReplaceableCollection {
    /// Returns a collection containing, in order, the first instances of
    /// elements of the sequence that compare equally for the keyPath.
    func unique<T: Hashable>(for keyPath: KeyPath<Element, T>) -> Self {
        var unique = Set<T>()
        return filter { unique.insert($0[keyPath: keyPath]).inserted }
    }
}

注意:在你的对象不符合RangeReplaceableCollection,但符合Sequence的情况下,你可以有这个额外的扩展,但返回类型将始终是一个数组:

extension Sequence {
    /// Returns an array containing, in order, the first instances of
    /// elements of the sequence that compare equally for the keyPath.
    func unique<T: Hashable>(for keyPath: KeyPath<Element, T>) -> [Element] {
        var unique = Set<T>()
        return filter { unique.insert($0[keyPath: keyPath]).inserted }
    }
}

使用

如果我们想要元素本身的唯一性,如问题中所示,我们使用keyPath \.self:

let a = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
let b = a.unique(for: \.self)
/* b is [1, 4, 2, 6, 24, 15, 60] */

如果我们想要其他东西的唯一性(比如对象集合的id),那么我们使用我们选择的keyPath:

let a = [CGPoint(x: 1, y: 1), CGPoint(x: 2, y: 1), CGPoint(x: 1, y: 2)]
let b = a.unique(for: \.y)
/* b is [{x 1 y 1}, {x 1 y 2}] */

变异的解决方案

我们扩展了一个变异函数,它能够过滤任意keyPath上的唯一性:

extension RangeReplaceableCollection {
    /// Keeps only, in order, the first instances of
    /// elements of the collection that compare equally for the keyPath.
    mutating func uniqueInPlace<T: Hashable>(for keyPath: KeyPath<Element, T>) {
        var unique = Set<T>()
        removeAll { !unique.insert($0[keyPath: keyPath]).inserted }
    }
}

使用

如果我们想要元素本身的唯一性,如问题中所示,我们使用keyPath \.self:

var a = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
a.uniqueInPlace(for: \.self)
/* a is [1, 4, 2, 6, 24, 15, 60] */

如果我们想要其他东西的唯一性(比如对象集合的id),那么我们使用我们选择的keyPath:

var a = [CGPoint(x: 1, y: 1), CGPoint(x: 2, y: 1), CGPoint(x: 1, y: 2)]
a.uniqueInPlace(for: \.y)
/* a is [{x 1 y 1}, {x 1 y 2}] */

其他回答

我认为这是更好的理解逻辑的方法

var arrayOfInts = [2, 2, 4, 4]
var mainArray = [Int]()

for value in arrayOfInts {

if mainArray.contains(value) != true  {
    
    mainArray.append(value)
    print("mainArray:\(mainArray)")
}}

正如WWDC 2021所指出的那样,Swift有社区开发的算法、集合和数字包。算法包有一个唯一的()算法。

这些还不是Swift标准库的一部分。你目前可以从苹果的Github页面下载它们和/或通过Swift软件包管理器安装它们。

WWDC视频:

https://developer.apple.com/videos/play/wwdc2021/10256/

Github页面:

https://github.com/apple/swift-algorithms

unique()和unique (on:)文档:

https://github.com/apple/swift-algorithms/blob/main/Guides/Unique.md

你可以自己卷,比如这样:

func unique<S : Sequence, T : Hashable>(source: S) -> [T] where S.Iterator.Element == T {
    var buffer = [T]()
    var added = Set<T>()
    for elem in source {
        if !added.contains(elem) {
            buffer.append(elem)
            added.insert(elem)
        }
    }
    return buffer
}

let vals = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
let uniqueVals = uniq(vals) // [1, 4, 2, 6, 24, 15, 60]

作为Array的扩展:

extension Array where Element: Hashable {
    func uniqued() -> Array {
        var buffer = Array()
        var added = Set<Element>()
        for elem in self {
            if !added.contains(elem) {
                buffer.append(elem)
                added.insert(elem)
            }
        }
        return buffer
    }
}

或者更优雅一点(Swift 4/5):

extension Sequence where Element: Hashable {
    func uniqued() -> [Element] {
        var set = Set<Element>()
        return filter { set.insert($0).inserted }
    }
}

将被使用:

[1,2,4,2,1].uniqued()  // => [1,2,4]

如果你把两个扩展都放在你的代码中,更快的Hashable版本将在可能的情况下使用,Equatable版本将用作备用版本。

public extension Sequence where Element: Hashable {
  /// The elements of the sequence, with duplicates removed.
  /// - Note: Has equivalent elements to `Set(self)`.
  @available(
  swift, deprecated: 5.4,
  message: "Doesn't compile without the constant in Swift 5.3."
  )
  var firstUniqueElements: [Element] {
    let getSelf: (Element) -> Element = \.self
    return firstUniqueElements(getSelf)
  }
}

public extension Sequence where Element: Equatable {
  /// The elements of the sequence, with duplicates removed.
  /// - Note: Has equivalent elements to `Set(self)`.
  @available(
  swift, deprecated: 5.4,
  message: "Doesn't compile without the constant in Swift 5.3."
  )
  var firstUniqueElements: [Element] {
    let getSelf: (Element) -> Element = \.self
    return firstUniqueElements(getSelf)
  }
}

public extension Sequence {
  /// The elements of the sequences, with "duplicates" removed
  /// based on a closure.
  func firstUniqueElements<Hashable: Swift.Hashable>(
    _ getHashable: (Element) -> Hashable
  ) -> [Element] {
    var set: Set<Hashable> = []
    return filter { set.insert(getHashable($0)).inserted }
  }

  /// The elements of the sequence, with "duplicates" removed,
  /// based on a closure.
  func firstUniqueElements<Equatable: Swift.Equatable>(
    _ getEquatable: (Element) -> Equatable
  ) -> [Element] {
    reduce(into: []) { uniqueElements, element in
      if zip(
        uniqueElements.lazy.map(getEquatable),
        AnyIterator { [equatable = getEquatable(element)] in equatable }
      ).allSatisfy(!=) {
        uniqueElements.append(element)
      }
    }
  }
}

如果顺序不重要,那么你总是可以使用这个Set初始化式。

这里有一个解决方案

不使用传统NS类型 对于O(n)来说是相当快的 简洁 保持元素顺序

extension Array where Element: Hashable {

    var uniqueValues: [Element] {
        var allowed = Set(self)
        return compactMap { allowed.remove($0) }
    }
}