Question

基本上我需要一个appendContentsOf:的版本，它不附加重复的元素。

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

简单地说：

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

Answer 2

这通常称为union，使用Set可以在Swift中使用：

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 3

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)
  }

}

如上所述：传递给函数的数组是将从最终数组

中省略的对象数组

Answer 4

已接受答案的

Swift 3.0 版本。

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：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。重要的是，如果合并一个对象数组，其中Equatable属性可能相同但其他属性可能不同。

Answer 5

可以创建一个Array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，这仅适用于Equatable元素。

Answer 6

Swift 4

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

Answer 7

我将Sequence和Array的扩展名与this answer结合使用，以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)
    }
}

是否有任何简单的方法可以在swift中合并两个数组并删除重复项？

7 个答案: