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Swift 3.0 学习笔记之Collection (Array

Swift 3.0 学习笔记之Collection (Array

作者: WavesFisherman | 来源:发表于2016-12-07 10:51 被阅读0次

    Array 数组  相当于OC中的可变数组

    var someInts = [Int]() //初始化                                                                          someInts.append(3) // 添加一个元素                                                                            someInts= [] // 制空数组 

    var threeDoubles = Array(repeating:0.0,count:3) // 初始化数组                                                var anotherThreeDoubles = Array(repeating:2.5,count:3)                                                           var sixDoubles = threeDoubles + anotherThreeDoubles // 合并两个数组 ( 相当于 第一个数组 依次添加第二个数组里面的元素)

    var shoppingList: [String] = ["Eggs","Milk"] //用字面量形式 初始化数组                                  var shoppingList = ["Eggs","Milk"]       shoppingList[4...6] = ["Bananas","Apples"]//改变一定范围内的元素    shoppingList.insert("Maple Syrup",at:0)//插入元素                                       let mapleSyrup = shoppingList.remove(at:0) //移除元素

    for (index,value) in shoppingList.enumerated() {//获取数组元素以及对应的索引

           print("Item\(index+1):\(value)")

    }

    Sets 集合

    var letters = Set<Character>() //初始化集合  letters.insert("a")//插入元素 letters = [] // 制空集合  var favoriteGenres:Set<String> = ["Rock","Classical","Hip hop"] // 字面量初始化              var favoriteGenres:Set= ["Rock","Classical","Hip hop"] // 自行推断是字符串类型的集合          favoriteGenres.isEmpty// 判空 favoriteGenres.insert("Jazz") // 插入元素 favoriteGenres.remove("Rock") // 删除元素 favoriteGenres.contains("Funk")//包含

    for genre in favoriteGenres.sorted() { //按照索引排序 和数组一样

          print("\(genre)")

    }

    a.intersection(b)// 求a和b的交集  a.symmetricDifference(b)// a和b的非交集 a.union(b) // a和b的并集 a.subtracting(b) // a中除了b的元素剩下的

    let houseAnimals:Set= ["🐶","🐱"]

    let farmAnimals:Set= ["🐮","🐔","🐑","🐶","🐱"]

    let cityAnimals:Set= ["🐦","🐭"]

    houseAnimals.isSubset(of:farmAnimals) //是否是子集    // true

    farmAnimals.isSuperset(of:houseAnimals) //是否包含    // true

    farmAnimals.isDisjoint(with:cityAnimals) // 是否有交集     // true

    Dictionary 字典

    var namesOfIntegers = [Int:String]() //字典初始化 // namesOfIntegers is an empty [Int: String] dictionary  [ key : value ]    namesOfIntegers[16] ="sixteen" // namesOfIntegers now contains 1 key-value pair    namesOfIntegers= [:] // namesOfIntegers is once again an empty dictionary of type [Int: String]  var airports: [String:String] = ["YYZ":"Toronto Pearson","DUB":"Dublin"] //字典初始化 var airports= ["YYZ":"Toronto Pearson","DUB":"Dublin"] //和上句一样 自行推断类型  

    if airports.isEmpty { // 判空

        print("The airports dictionary is empty.")

    }else{

        print("The airports dictionary is not empty.")

    }    // Prints "The airports dictionary is not empty."

    airports["LHR"] ="London" // 访问值

    if let oldValue = airports.updateValue("Dublin Airport",forKey:"DUB") { // 更换对应key 的 value

          print("The old value for DUB was\(oldValue).")

    }    // Prints "The old value for DUB was Dublin."

    if let removedValue = airports.removeValue(forKey:"DUB") { // 移除对应key的value

           print("The removed airport's name is\(removedValue).")

    }else{

           print("The airports dictionary does not contain a value for DUB.")

    }    // Prints "The removed airport's name is Dublin Airport."

    for (airportCode,airportName) in  airports{ // 遍历字典

            print("\(airportCode):\(airportName)")

    }     // YYZ: Toronto Pearson       // LHR: London Heathrow

    for airportCode in airports.keys { // 遍历字典的所有key

        print("Airport code:\(airportCode)")

    }    // Airport code: YYZ        // Airport code: LHR

    for airportName in airports.values{ // 遍历字典所有的value

         print("Airport name:\(airportName)")

    }     // Airport name: Toronto Pearson        // Airport name: London Heathrow

    let airportCodes = [String](airports.keys) //  获取字典的所有的key 返回一个数组

    // airportCodes is ["YYZ", "LHR"]

    let airportNames = [String](airports.values) // 获取字典所有的value 返回一个数组

    // airportNames is ["Toronto Pearson", "London Heathrow"]

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