- swift 提供了array，set， dictionary 等三個(gè)收藏類型。
- 創(chuàng)建一個(gè)空數(shù)組 array:就是數(shù)組
var someInts = [Int]()
print("數(shù)組里有\(zhòng)(someInts.count)個(gè)元素")
//會(huì)打印數(shù)組里有0個(gè)元素
someInts.append(7)
//現(xiàn)在someInts包含了一個(gè)值
someInts = []
//現(xiàn)在它是一個(gè)空數(shù)組了，他的類型也是依然int
創(chuàng)建有默認(rèn)值得數(shù)組
swift 有一個(gè)數(shù)組初始化器，他能創(chuàng)建特定大小的數(shù)組
var threeDoubles = Array(repeatElement(7.7, count: 3))
//兩個(gè)特定的數(shù)組還可以相加
var anotherThreeDoubles = Array(repeatElement(3.3, count: 3))
var sixDoubles = threeDoubles + anotherThreeDoubles
字符串?dāng)?shù)組
var shoppingList: [String] = ["Eggs","Milik"]
var list = ["book" ,"12"]
//下面代碼可以簡(jiǎn)寫數(shù)組
var shoppingList1 = ["apple","orange","banana"]
訪問和修改數(shù)組
//count 能查找數(shù)組中的項(xiàng)目數(shù)
print("shoppinglist 數(shù)組有\(zhòng)(shoppingList.count)項(xiàng)目")
//數(shù)組是不是等于零的檢查方式
if shoppingList.isEmpty{
print("no one")
}else{
print("數(shù)組不是空的")
}
//如果我想添加一個(gè)新成員的話會(huì)調(diào)用append(_:)注意他會(huì)添加到最后一個(gè)位置
shoppingList.append("phone")
//我還可以 += 來添加新成員
shoppingList += ["beer"]
索引來修改數(shù)組的值
shoppingList[0] = "hony"
//如果想特定的位置加成員,我想位置1 和 2的成員改掉
shoppingList[1...2] = ["butter","beef"]
//我要插進(jìn)去一個(gè)成員的話呢用這個(gè)方法
shoppingList.insert("coffe", at: 3)
shoppingList.insert("tea", at: 3)
//相反remove（at：）方法來刪掉特定位置的成員
shoppingList.remove(at: 3)
shoppingList.removeLast()
shoppingList.count
//for item in array 是遍歷整個(gè)數(shù)組的
for nl_item in shoppingList{
print(nl_item)
}
set 集合
//set 存儲(chǔ)沒有特定循序的,相同類型的不同的值
//創(chuàng)建和初始化一個(gè)空集
var letters = Set<Character>()
print("\(letters.count)")
//會(huì)打印零
letters.insert("m")
letters = []
//現(xiàn)在集又空了 ，屬性依然是Set<Character>
創(chuàng)建一個(gè)字符串的集合
//var favoriteGenres: Set<String> = ["rock","hiphop","classical"]
//可以簡(jiǎn)寫
var favoriteGenres: Set = ["rock","hiphop","classical"]
let oddDigits: Set = [1,3,5,7,9]
let evenDigits: Set = [0,2,4,6,8]
let singleDigitNumber: Set = [2,3,5,7]
oddDigits.union(evenDigits).sorted()
//最后的sorted（）方法將排序集合元祖如果把它刪除不會(huì)報(bào)錯(cuò)就是沒有循序
//intersection 中文意思：交叉；十字路口；交集；交叉點(diǎn) 它就是交集！
oddDigits.intersection(evenDigits).sorted()
// subtracting:差集
oddDigits.subtracting(evenDigits).sorted()
字典
- 創(chuàng)建一 個(gè)空字典
//其實(shí)他也與數(shù)組一樣
var namesOfinteger = [Int: String]()
//上面一行代碼可以這麼理解：我要頂一個(gè)字典，他key的類型是int，value的類型是string
namesOfinteger[27] = "is my age"
//下面這一行代碼是空字典
namesOfinteger = [:]
- 創(chuàng)建一個(gè)字符串字典
//[key1：value1, key2: value2, key3: value3]
//var airports: [String: String] = ["0998" : "喀什","0991":"烏魯木齊","0995":"吐魯番"]
//上面代碼中數(shù)組是地區(qū)號(hào)
//簡(jiǎn)寫的話可以這么些
var airports = ["0998" : "喀什","0991":"烏魯木齊","0995":"吐魯番"]
訪問和修改字典
- 數(shù)組一樣用count 來查找其中的item數(shù)
print("the air ports dictionary contains \(airports.count) item")
//同樣適用isEmptyE來檢查是否空
if airports.isEmpty{
print("the airport dictionary is empty.")
}else{
print("the airport dictionary is not empty .")
}
//你拿字典的key 修改它的值
airports["0998"] = "喀什人民歡迎你"
if let oldValue = airports.updateValue("烏魯木齊人民歡迎你", forKey: "0991"){
print("the old value for 烏魯木齊 was\(oldValue)")
}
//if let airportName = airports["0991"]{
// print("the name of the airport is \(airportName)")
//}
- 你可以遍歷字典的每一個(gè)key和value
for (airportCode,airportName) in airports{
print("\(airportCode):\(airportName)")
}
for airportCode in airports.keys{
print("airportCode: \(airportCode)")
}
for airportName in airports.values{
print("airportName:\(airportName)")
}
初始化
let airportCode1 = [String](airports.keys)
let airportName1 = [String](airports.values)
print(airportCode1)
print(airportName1)
