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。

我的解决方案，它似乎可以在O(n)时间，因为哈希映射访问是O(1)，过滤器是O(n)。它还使用by闭包来选择按顺序区分元素的属性。

extension Sequence {

    func distinct<T: Hashable>(by: (Element) -> T) -> [Element] {
        var seen: [T: Bool] = [:]
        return self.filter { seen.updateValue(true, forKey: by($0)) == nil }
    }
}

斯威夫特5.7

使用有序集

您可以将具有重复元素的数组传递给以下通用函数，该函数处理有序Set并返回没有重复元素的新数组。

import Foundation

internal func withoutDuplicates<T>(_ array: [T]) -> [T] {
    
    let orderedSet: NSMutableOrderedSet = []
    var modifiedArray = [T]()
    
    orderedSet.addObjects(from: array)
    
    for i in 0...(orderedSet.count - 1) {
        modifiedArray.append(orderedSet[i] as! T)
    }
    return modifiedArray
}

////////////////////////////////////////////////////////////

let arrayOfStrings: [String] = ["A","A","A","B","B","C","C"]
let arrayOfIntegers: [UInt8] = [1, 1, 1, 2, 2, 2, 3, 3]
let arrayOfBooleans: [Bool] = [true, false, false, true]

let ordered_01 = withoutDuplicates(arrayOfStrings)
let ordered_02 = withoutDuplicates(arrayOfIntegers)
let ordered_03 = withoutDuplicates(arrayOfBooleans)

结果:

// ordered_01  –––>  ["A","B","C"]
// ordered_02  –––>  [1, 2, 3]
// ordered_03  –––>  [true, false]

使用无序集

如果新数组中元素的顺序对您来说并不重要，则在处理时使用无序集。无序集中的元素类型必须符合哈希协议。

import UIKit

fileprivate func noDuplicates<T: Hashable>(_ array: [T]) -> [T] {
    
    var unorderedSet = Set<T>()
    var modifiedArray: [T] = []
    
    for i in 0...(array.count - 1) {
        unorderedSet.insert(array[i])
    }
    for i in unorderedSet.indices {
        modifiedArray.append(unorderedSet[i])
    }
    return modifiedArray
}

////////////////////////////////////////////////////////////

let arrayOfInts: [Int] = [10, 5, 7, 200, -500, 10, 7, 5]
let arrayOfStrs: [String] = ["A","A","A","B","B","C","C"]
let arrayOfBools: [Bool] = [true, false, false, true]

let unordered_01 = noDuplicates(arrayOfInts)
let unordered_02 = noDuplicates(arrayOfStrs)
let unordered_03 = noDuplicates(arrayOfBools)

结果:

// unordered_01  –––>  [200, 7, 10, -500, 5]
// unordered_02  –––>  ["B", "C", "A"]
// unordered_03  –––>  [false, true]

包含相等性检查，而插入检查哈希，最安全的检查方式如下:

extension Array where Element: Hashable {

    /// Big O(N) version. Updated since @Adrian's comment. 
    var uniques: Array {
        // Go front to back, add element to buffer if it isn't a repeat.
         var buffer: [Element] = []
         var dictionary: [Element: Int] = [:]
         for element in self where dictionary[element] == nil {
             buffer.append(element)
             dictionary[element] = 1
         }
         return buffer
    }
}

在Swift 3.0中，我发现了最简单和最快的解决方案，可以在保持顺序的同时消除重复的元素:

extension Array where Element:Hashable {
    var unique: [Element] {
        var set = Set<Element>() //the unique list kept in a Set for fast retrieval
        var arrayOrdered = [Element]() //keeping the unique list of elements but ordered
        for value in self {
            if !set.contains(value) {
                set.insert(value)
                arrayOrdered.append(value)
            }
        }

        return arrayOrdered
    }
}

斯威夫特4. x:

extension Sequence where Iterator.Element: Hashable {
  func unique() -> [Iterator.Element] {
    return Array(Set<Iterator.Element>(self))
  }

  func uniqueOrdered() -> [Iterator.Element] {
    return reduce([Iterator.Element]()) { $0.contains($1) ? $0 : $0 + [$1] }
  }
}

用法:

["Ljubljana", "London", "Los Angeles", "Ljubljana"].unique()

or

["Ljubljana", "London", "Los Angeles", "Ljubljana"].uniqueOrdered()