在OC中数组,字典,集合有自己的表示方法,分别是Array,Dictionary,Set 与 String 都属于数值类型变量,他们都属于结构体。 使用简介灵活多变,个人感觉可读性变差了很多,用起来由点别扭,还需要慢慢适应。 基本上功能都是照办OC的,主要包括创建,增,删,改,查,遍历。下面从数组开始说起。

    1. Array ,数组,线性机构,按顺序存储结构,有对应的下表标

       创建方式贼多,大家根据自己的喜好随便选一种吧,       

var firstArray:Array<Int> = Array<Int>() 
        var secondArray:Array = Array<Int>()
        var thirdArray = Array<Int>()
        var fourthArray:Array<Int> = Array()
        var fivethArray:Array<Int> = [Int]()
        var sixthArray:[Int] = [Int]()
        var seventhArray:[Int] = []
        var eighthArray = [1,2,3,4,5,5,6]
        var ninethArray:[Int]   //最后一种只生成没有初试化

    2. 增加数据  append()    +      =        

firstArray.append(250)      OC:    addObject:   
        firstArray[0] = "500"  
        firstArray += ["180","360","720","a","g"]  //添加另外一个数组
        firstArray.count
        firstArray.isEmpty

    3.  插入元素  inset()   

firstArray.insert("iphone6 plus",at Index:1) 
        其它元素需要后移动,比较消耗性能

    4. 删除元素  removeAtIndex(1)   它的返回值为移除的数组元素类型   

var res: String = firstArray.removeAtIndex(1)
       res = firstArray.removeFirst()
       res = firstArray.removeLast()
       var range = Range(start:0,  end:2)
       firstArray.removeRange(Range)

   5. 修改元素   

通过下标直接修改
       shoppingList[1] = "price"
       shoppingList.removeAll()

    6. 从数组中获取部分元素同时声称一个数组

   

var subList = shoppingList[0..<3]   
      /**遍历*/
     for elements in subList {
        print(elements)
     }
 
    for var i = 0; i < subList.count;i++ {
        print("第\(i)个元素是\(subList[i])")
    }
    for (index,value) in subList.enumerate() {
        print("第\(index) \(value)") 
    }
    subList
   subList[0]

2. 字典部分,创建方式

 

var dictionary1:Dictionary<String,Int> = Dictionary<String,Int>()
    var dictionary2 = Dictionary <String,Int>()
    var dictionary3:Dictionary  = Dictionary<String,Int>()
    var dictionary4:Dictionary<String,Int> = Dictionary()
    var dictionary5:Dictionary<String,Int> = [String:Int]()
    var dictionary6:Dictionary<String,Int> = ["age":23,"age2":18]
    var dicitonary7 = ["age":18,"age2":33]
    //不能直接使用 nil 作为变量,但是可以使用可选值座位值
    var dictionary8:[String:Int?] = ["age":23,"age":nil]
    //未初始化,只申明
    var dictionary9:[String:Int]

3. 基本操作方法: 

/**操作字典的方法*/
var airports:[String:String] = ["PEK":"Beijing airport","CAN":"GuangzhouBaiYun airport","SHA":"ShangHaiHongQiao airport"]
airports.count
airports["SZA"] = "Baoan airport"
//如果eky不存在就是添加,如果存在就是修改
airports.count

4 . updataValue( value:xxx  ,forkey:xxx) ,注意Swift key必须写成String类型

airports.updateValue("ShangHaiPuDong", forKey: "PVG")
airports.count
airports.isEmpty
airports["TRA"] = "DaNei airport"
 
/**删除机场*/
airports.count
airports["SZA"] = nil
airports.count

5 .删除字典元素,注意不管是删除,还是添加过程中,都可以返回一个可选值Value 

removeValueForKey  removeAll
if let airport = airports.removeValueForKey("TRA"){
   print("airport: \(airport)delete successfully")
}else {
   print("meiyou duiying jichang")
}

 6. 获取字典中所有的keys 和所有的Values,使用数组类型进行强转

let  airportCodes = [String] (airports.keys)
    let  airportValues:[String] = [String] (airports.values)
 
//集合部分 Set,它具有无序性和唯一性,个人感觉基本没怎么用,除了网络并发请求外

 1. Standard style  

var letters:Set<Character> = Set<Character>()

     其余写法大家自己琢磨去吧,之要不报错基本都行,写法太多了,和字典数组都差不多。

     其实只要保证左右两边结合起来能确定 它自身的类型(Set,Dictionary,Array),元素类型(Int,String,Character),然后右边必须有一个() 或者[] 就能满足要求了

 2. 增,删,交,并,差,异或

   

var musics:Set<String> = ["Rock","Classical","Jazz"]
    music.isEmpty
    music.count
    music.insert("Jazz")
    if let removeMusic = musics.remove("Jazz")  {
      print ("\(removeMusic) delete successfully")
} else {
  print("element is not exist")
} 
    musics.contains("rock")
   for music in musics {print(music)}
   //排序
   for music  in music.sort() { print(music) }
    let oddDigits:Set = [1,3,4,5,6]
    let evemDigits:Set = [0,1,3,4,5]
    //union 并集
     var newNumbers = oddDigits.union(evenDigits)
     print(newNumbers)
    //intersect 交集
     newNumbers = oddDigits.intersect(evenDigits)
     //subtract 差集
     newNumbers = oddDigits.subtract(evenDigits)
     //异或集合,把不相同的部分提取出来
     newNumbers = oddDigits.exclusiveOr(evenDigits)
      记住好四个单词, uinion,intersect,subract,exclusiveOr 就好了