基本上,我需要一个 appendContentsOf:
的版本,它不会追加重复的元素。
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
简而言之:
let unique = Array(Set(a + b))
使用Set
,在Swift中可以轻松进行称为并集的操作:
let a = [1, 2, 3]
let b = [3, 4, 5]
let set = Set(a)
let union = set.union(b)
然后您可以将这个集合转换为一个数组:
let result = Array(union)
Swift 5 更新
如果您需要组合多个数组。
func combine<T>(_ arrays: Array<T>?...) -> Set<T> {
return arrays.compactMap{$0}.compactMap{Set($0)}.reduce(Set<T>()){$0.union($1)}
}
使用例子:
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]
Swift 4.0版本
(关于it技术)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从最终数组中省略的对象数组
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: 这里需要说明的是,传递给函数的数组是最终数组中将被省略的对象数组。如果你要合并一个对象数组,其中Equatable
属性可能相同但其他属性可能不同,则这一点非常重要。
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)
}
}
我将 Sequence 和 Array 的扩展与 这个答案 结合起来,以便在仅使用单个属性合并数组时提供易于使用的语法:
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)
}
}
let unique = Array(Set(a + b)).sort()
看起来更好一些... - Dravidianb
的元素添加到a
的末尾时并不能保持a
中元素的顺序,这可能会被认为是不符合预期的行为。由于使用了Set
,所以无法维护顺序。 - George Marmaridislet unique = Array(Set(a + b)).sorted()
? - Shaybc