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()

这是swift 4.2中最简单的方法，如下所示的代码

let keyarray:NSMutableArray = NSMutableArray()

for  object in dataArr
{
    if !keysArray.contains(object){
        keysArray.add(object)
    }
}

print(keysArray)

Xcode 10.1 - Swift 4.2简单而强大的解决方案

func removeDuplicates(_ nums: inout [Int]) -> Int {
    nums = Set(nums).sorted()
    return nums.count
}

例子

var arr = [1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9]
removeDuplicates(&arr)

print(arr) // [1,2,3,4,5,6,7,8,9]

在Swift 5中

 var array: [String] =  ["Aman", "Sumit", "Aman", "Sumit", "Mohan", "Mohan", "Amit"]

 let uniq = Array(Set(array))
 print(uniq)

输出将是

 ["Sumit", "Mohan", "Amit", "Aman"]

最简单的方法是使用NSOrderedSet，它存储唯一的元素并保持元素的顺序。如:

func removeDuplicates(from items: [Int]) -> [Int] {
    let uniqueItems = NSOrderedSet(array: items)
    return (uniqueItems.array as? [Int]) ?? []
}

let arr = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
removeDuplicates(from: arr)

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。