有没有什么简单的方法可以快速合并两个数组以及删除重复项？

Question

基本上我需要一个不附加 duplicate 元素的appendContentsOf: 版本。

示例

var a = [1, 2, 3]
let b = [3, 4, 5]

a.mergeElements(b)
//gives a = [1, 2, 3, 4, 5] //order does not matter

Answer 1

可以创建一个数组扩展来执行此操作。

extension Array where Element : Equatable{

    public mutating func mergeElements<C : CollectionType where C.Generator.Element == Element>(newElements: C){
       let filteredList = newElements.filter({!self.contains($0)})
       self.appendContentsOf(filteredList)
   }
}

当然，这仅对 Equatable 元素有用。

Answer 2

简单地说：

let unique = Array(Set(a + b))

Answer 3

这通常称为union，在 Swift 中可以使用Set：

let a = [1, 2, 3]
let b = [3, 4, 5]

let set = Set(a)
let union = set.union(b)

然后你可以将集合转换为数组：

let result = Array(union)

Answer 4

Swift 3.0 版本的已接受答案。

extension Array where Element : Equatable{

  public mutating func mergeElements<C : Collection>(newElements: C) where C.Generator.Element == Element{
    let filteredList = newElements.filter({!self.contains($0)})
    self.append(contentsOf: filteredList)
  }

}

注意：这里值得一提的是，传递给函数的数组是最终数组中将省略的对象数组。如果您要合并一组对象，其中 Equatable 属性可能相同但其他对象可能不同，这点很重要。

Answer 5

Swift 4.0 版本

extension Array where Element : Equatable {

  public mutating func mergeElements<C : Collection>(newElements: C) where C.Iterator.Element == Element{
    let filteredList = newElements.filter({!self.contains($0)})
    self.append(contentsOf: filteredList)
  }

}

如前所述：传递给函数的数组是将最终数组中省略的对象数组

Answer 6

斯威夫特 5 更新了

如果您需要合并多个数组。

func combine<T>(_ arrays: Array<T>?...) -> Set<T> {
    return arrays.compactMap{$0}.compactMap{Set($0)}.reduce(Set<T>()){$0.union($1)}
}

用法示例：

1.

    let stringArray1 = ["blue", "red", "green"]
    let stringArray2 = ["white", "blue", "black"]
    
    let combinedStringSet = combine(stringArray1, stringArray2)

    // Result: {"green", "blue", "red", "black", "white"}

    let numericArray1 = [1, 3, 5, 7]
    let numericArray2 = [2, 4, 6, 7, 8]
    let numericArray3 = [2, 9, 6, 10, 8]
    let numericArray4: Array<Int>? = nil
    
    let combinedNumericArray = Array(combine(numericArray1, numericArray2, numericArray3, numericArray4)).sorted()

    // Result: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Answer 7

我将我对序列和数组的扩展与this answer 结合起来，以便在通过单个属性将数组与自定义对象合并时提供简单的语法：

extension Dictionary {
    init<S>(_ values: S, uniquelyKeyedBy keyPath: KeyPath<S.Element, Key>) where S : Sequence, S.Element == Value {
        let keys = values.map { $0[keyPath: keyPath] }

        self.init(uniqueKeysWithValues: zip(keys, values))
    }
}

// Unordered example
extension Sequence {
    func merge<T: Sequence, U: Hashable>(mergeWith: T, uniquelyKeyedBy: KeyPath<T.Element, U>) -> [Element] where T.Element == Element {
        let dictOld = Dictionary(self, uniquelyKeyedBy: uniquelyKeyedBy)
        let dictNew = Dictionary(mergeWith, uniquelyKeyedBy: uniquelyKeyedBy)

        return dictNew.merging(dictOld, uniquingKeysWith: { old, new in old }).map { $0.value }
    }
}

// Ordered example
extension Array {
    mutating func mergeWithOrdering<U: Hashable>(mergeWith: Array, uniquelyKeyedBy: KeyPath<Array.Element, U>) {
        let dictNew = Dictionary(mergeWith, uniquelyKeyedBy: uniquelyKeyedBy)

        for (key, value) in dictNew {
            guard let index = firstIndex(where: { $0[keyPath: uniquelyKeyedBy] == key }) else {
                append(value)
                continue
            }

            self[index] = value
        }
    }
}

测试：

@testable import // Your project name
import XCTest

struct SomeStruct: Hashable {
    let id: Int
    let name: String
}

class MergeTest: XCTestCase {
    let someStruct1 = SomeStruct(id: 1, name: "1")
    let someStruct2 = SomeStruct(id: 2, name: "2")
    let someStruct3 = SomeStruct(id: 2, name: "3")
    let someStruct4 = SomeStruct(id: 4, name: "4")

    var arrayA: [SomeStruct]!
    var arrayB: [SomeStruct]!

    override func setUp() {
        arrayA = [someStruct1, someStruct2]
        arrayB = [someStruct3, someStruct4]
    }

    func testMerging() {
        arrayA = arrayA.merge(mergeWith: arrayB, uniquelyKeyedBy: \.id)

        XCTAssert(arrayA.count == 3)
        XCTAssert(arrayA.contains(someStruct1))
        XCTAssert(arrayA.contains(someStruct3))
        XCTAssert(arrayA.contains(someStruct4))
    }

    func testMergingWithOrdering() {
        arrayA.mergeWithOrdering(mergeWith: arrayB, uniquelyKeyedBy: \.id)

        XCTAssert(arrayA.count == 3)
        XCTAssert(arrayA[0] == someStruct1)
        XCTAssert(arrayA[1] == someStruct3)
        XCTAssert(arrayA[2] == someStruct4)
    }
}