这里我对对象做了O(n)解。不是少行解决方案，而是……

struct DistinctWrapper <T>: Hashable {
    var underlyingObject: T
    var distinctAttribute: String
    var hashValue: Int {
        return distinctAttribute.hashValue
    }
}
func distinct<S : SequenceType, T where S.Generator.Element == T>(source: S,
                                                                distinctAttribute: (T) -> String,
                                                                resolution: (T, T) -> T) -> [T] {
    let wrappers: [DistinctWrapper<T>] = source.map({
        return DistinctWrapper(underlyingObject: $0, distinctAttribute: distinctAttribute($0))
    })
    var added = Set<DistinctWrapper<T>>()
    for wrapper in wrappers {
        if let indexOfExisting = added.indexOf(wrapper) {
            let old = added[indexOfExisting]
            let winner = resolution(old.underlyingObject, wrapper.underlyingObject)
            added.insert(DistinctWrapper(underlyingObject: winner, distinctAttribute: distinctAttribute(winner)))
        } else {
            added.insert(wrapper)
        }
    }
    return Array(added).map( { return $0.underlyingObject } )
}
func == <T>(lhs: DistinctWrapper<T>, rhs: DistinctWrapper<T>) -> Bool {
    return lhs.hashValue == rhs.hashValue
}

// tests
// case : perhaps we want to get distinct addressbook list which may contain duplicated contacts like Irma and Irma Burgess with same phone numbers
// solution : definitely we want to exclude Irma and keep Irma Burgess
class Person {
    var name: String
    var phoneNumber: String
    init(_ name: String, _ phoneNumber: String) {
        self.name = name
        self.phoneNumber = phoneNumber
    }
}

let persons: [Person] = [Person("Irma Burgess", "11-22-33"), Person("Lester Davidson", "44-66-22"), Person("Irma", "11-22-33")]
let distinctPersons = distinct(persons,
    distinctAttribute: { (person: Person) -> String in
        return person.phoneNumber
    },
    resolution:
    { (p1, p2) -> Person in
        return p1.name.characters.count > p2.name.characters.count ? p1 : p2
    }
)
// distinctPersons contains ("Irma Burgess", "11-22-33") and ("Lester Davidson", "44-66-22")

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。

Daniel Krom的Swift 2答案的更简洁的语法版本，使用了一个尾随闭包和简写参数名，这似乎是基于Airspeed Velocity的原始答案:

func uniq<S: SequenceType, E: Hashable where E == S.Generator.Element>(source: S) -> [E] {
  var seen = [E: Bool]()
  return source.filter { seen.updateValue(true, forKey: $0) == nil }
}

实现一个可以与uniq(_:)一起使用的自定义类型的示例(必须符合Hashable，因此符合Equatable，因为Hashable扩展了Equatable):

func ==(lhs: SomeCustomType, rhs: SomeCustomType) -> Bool {
  return lhs.id == rhs.id // && lhs.someOtherEquatableProperty == rhs.someOtherEquatableProperty
}

struct SomeCustomType {

  let id: Int

  // ...

}

extension SomeCustomType: Hashable {

  var hashValue: Int {
    return id
  }

}

在上面的代码中…

在==重载中使用的id可以是任何Equatable类型(或返回Equatable类型的方法，例如someMethodThatReturnsAnEquatableType())。注释掉的代码演示了扩展相等性检查，其中someOtherEquatableProperty是Equatable类型的另一个属性(但也可以是返回Equatable类型的方法)。

在hashValue计算属性中使用的id(必须符合Hashable)可以是任何Hashable(因此是Equatable)属性(或返回Hashable类型的方法)。

使用uniq(_:)的示例:

var someCustomTypes = [SomeCustomType(id: 1), SomeCustomType(id: 2), SomeCustomType(id: 3), SomeCustomType(id: 1)]

print(someCustomTypes.count) // 4

someCustomTypes = uniq(someCustomTypes)

print(someCustomTypes.count) // 3

这里有很多答案，但我错过了这个简单的扩展，适合Swift 2及以上:

extension Array where Element:Equatable {
    func removeDuplicates() -> [Element] {
        var result = [Element]()

        for value in self {
            if result.contains(value) == false {
                result.append(value)
            }
        }

        return result
    }
}

这非常简单。可以这样调用:

let arrayOfInts = [2, 2, 4, 4]
print(arrayOfInts.removeDuplicates()) // Prints: [2, 4]

基于属性的过滤

要根据属性筛选数组，你可以使用这个方法:

extension Array {

    func filterDuplicates(@noescape includeElement: (lhs:Element, rhs:Element) -> Bool) -> [Element]{
        var results = [Element]()

        forEach { (element) in
            let existingElements = results.filter {
                return includeElement(lhs: element, rhs: $0)
            }
            if existingElements.count == 0 {
                results.append(element)
            }
        }

        return results
    }
}

你可以这样调用它:

let filteredElements = myElements.filterDuplicates { $0.PropertyOne == $1.PropertyOne && $0.PropertyTwo == $1.PropertyTwo }

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]

让我提出一个类似于斯科特·加德纳的答案，但使用了更简洁的reduce语法。 此解决方案从自定义对象数组中删除重复项(保持初始顺序)

// Custom Struct. Can be also class. 
// Need to be `equitable` in order to use `contains` method below
struct CustomStruct : Equatable {
      let name: String
      let lastName : String
    }

// conform to Equatable protocol. feel free to change the logic of "equality"
func ==(lhs: CustomStruct, rhs: CustomStruct) -> Bool {
  return (lhs.name == rhs.name && lhs.lastName == rhs.lastName)
}

let categories = [CustomStruct(name: "name1", lastName: "lastName1"),
                  CustomStruct(name: "name2", lastName: "lastName1"),
                  CustomStruct(name: "name1", lastName: "lastName1")]
print(categories.count) // prints 3

// remove duplicates (and keep initial order of elements)
let uniq1 : [CustomStruct] = categories.reduce([]) { $0.contains($1) ? $0 : $0 + [$1] }
print(uniq1.count) // prints 2 - third element has removed

如果你想知道这个约简魔法是如何工作的，这里是完全相同的，只是使用了更扩展的约简语法

let uniq2 : [CustomStruct] = categories.reduce([]) { (result, category) in
  var newResult = result
  if (newResult.contains(category)) {}
  else {
    newResult.append(category)
  }
  return newResult
}
uniq2.count // prints 2 - third element has removed

你可以简单地复制粘贴这段代码到Swift Playground中。