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