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Swift 4.0 数组、集合Set、字典相关常用用法总结

Swift 4.0 数组、集合Set、字典相关常用用法总结

作者: 少年想多了 | 来源:发表于2017-11-07 19:14 被阅读23次

    数组用法

    //初始化数组
    let emptyArray = [String]()
    //Array初始化4个一样的字符串,repeating 是Array.Element 类型
    let f4Doubles = Array(repeating: "hello", count: 4)
    //输出[1, 2, 3, 4, 5, 6, 7] 使用范围初始化
    let numbers = Array(1...7)
    //一般使用到的初始化
    let array1:[String] = ["3","6","9","4"]
    let array2:[Any] = ["2",3,"e",4]
    var appendArray = array1 + array2//出错
    //只有类型一样才可以+
    let numbers1 = Array(1...3)
    let numbers2 = Array(5...8)
    let ss = numbers1 + numbers2 
    //一般用法
    var testArray = [2,5,3,1,9,6,4,8,7]   
    //给数组倒序 注意此时下标的值没有变化
    testArray.reverse()
    for i in testArray {
        print("每个元素值为:\(i)")
    }
    for (index,value) in testArray.enumerated() {
        print("下标 = \(index), 值 = \(value)")
    }    
    testArray.append(10)//追加元素
    print("testArray = \(testArray)")
    if testArray.isEmpty {
        print("判断是否为空")
    }
    let count = testArray.count
    print("数组元素个数 = \(count)")
    testArray += [21] //使用+=必须里面的类型一致
    print(testArray)
    //修改元素
    var newArray = ["hello","you","jk","test"]
    newArray[0] = "修改了第0个元素"
    //根据下标范围修改
    newArray[2...4] = ["范围2","范围4"]
    newArray[2..<4] = ["范围2","范围4"]
    //在对应的位置插入元素
    newArray.insert("我是插入的元素", at: 2)
    //在某个位置插入新的数组
    newArray.insert(contentsOf: ["111","222"], at: 3)
    let cast = ["Vivien", "Marlon", "Kim", "Karl"]
    let list = cast.joined(separator: ", ")//list这时候返回的是一个字符串
    // Prints "Vivien, Marlon, Kim, Karl"
    let nestedNumbers = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
    let joined = nestedNumbers.joined(separator: [-1, -2]) 
    //这时候joined是JoinedSequence序列类型
    print(Array(joined))
    //[1, 2, 3, -1, -2, 4, 5, 6, -1, -2, 7, 8, 9]
    //也可以通过范围来joined
    let ranges = [0..<3, 8..<10, 15..<17]
    for index in ranges.joined() {
        print(index, terminator: " ")
    }
    // Prints: "0 1 2 8 9 15 16"
    let removeItem = newArray.remove(at: 2)
    print("被删除的元素为:\(removeItem)")
    //删除第一个元素
    let removeFirst = newArray.removeFirst()
    //删除第二个元素
    let removeLast = newArray.removeLast()
    //删除前3个元素
    newArray.removeFirst(3)
    //删除后3个元素
    newArray.removeLast(3)
    //删除1...4的元素
    newArray.removeSubrange(1...4)
    //删除全部 keepingCapacity是否保存内存空间
    _ = newArray.removeAll(keepingCapacity: true)
    print(newArray.capacity)
    
    

    集合Set用法

    Set中的类型必须是Hashable的
    let letters = Set<String>()
    let count = letters.count
    print(count)
    //Set 无序且相同元素只出现一次
    let set:Set<String> = ["你好","hello","你好","name","hhh","cad","name"]
    print("\(set)")
    //重复的不会输出["name", "hhh", "你好", "hello", "cad"]
    set.insert("我是插入的")
    if set.isEmpty {
        print("集合为空")
    }
    //因为一个集合不是一个有序集合,所以“first”元素可能不是第一个添加到集合中的元素。 该集合不能为空。
    let removeFirst = set.removeFirst()
    //同理数组的其他几个也一样,在集合里面
    // Set 的remove 返回的是一个可选值
    if let removed = ingredients.remove("hhh") { 
        print("The hhh is now delete") 
    }
    if set.contains("你好") {
        print("参数是否在集合中")
    }
    //遍历
    for gen in set {
        print(gen)
    }
    //"Swift’s Set type does not have a defined ordering. To iterate over the values of a set in a specific order, use the sorted() method, which returns the set’s elements as an array sorted using the < operator"
    //"Swift的Set类型没有定义的顺序。 要按特定顺序迭代集合的值,请使用sorted()方法,该方法将集合的元素作为使用<运算符排序的数组。"
    for gen in set.sorted() {
        print(gen)
    }
    //集合本身的forEach
    set.forEach { (value) in
        print(value)
    }
    //创建一个只有两个集合通用(交集)的值的新集合
    intersection(_:) 
    //在集合中创建一个新集合,但不能同时创建集合
    symmetricDifference(_:)
    //用两个集合中的所有值创建一个新集合
    union(_:)
    //创建一个新的集合,其值不在指定的集合中
    subtracting(_:)
    //
    let set2:Set<String> = ["你好","name","hhh"]
    let set3:Set = ["qp","ss","lo"]
    let isTrue = set2.isSubset(of: setEqual)
    if isTrue {
        print("一个集合的所有值包含在指定的集合中")
    }
        
    let isTrue2 = setEqual.isSuperset(of: set2)
    if isTrue2 {
        print("一个集合包含指定集合中的所有值")
    }
        
    //isStrictSubset(of :)或isStrictSuperset(of :)方法来确定一个集合是一个子集还是超集,但不等于一个指定的集合
    let isTure3 = set2.isStrictSubset(of: setEqual)
    if isTure3 {
        print("set2 是 setEqual的子集")
    }
    let isTure4 = setEqual.isStrictSuperset(of: set2)
    if isTure4 {
        print("setEqual 是 set2的超集")
    }
        
    let isTure5 = set3.isDisjoint(with: set)
    if isTure5 {
        print("确定两个集合没有共同的值")
    }
    
    

    字典用法

    var testDic = [Int: String]()
    testDic[12] = "hello"
    testDic[9] = "world"
    testDic = [:]
    if testDic.isEmpty {
        print("字典为空")
    }
    print(testDic)
        
    var testDic2:[String: String] = ["token":"221212","type":"21"]
    print(testDic2)
    let count = testDic2.count
    print("字典的数量\(count)")
    //赋值
    testDic2["method"] = "test"
    //修改
    testDic2["type"] = "100"
    print(testDic2)
    //updateValue 方法
    if let oldValue = testDic2.updateValue("newValue", forKey: "token") {
        print("获取到的旧值:\(oldValue)")
    }
    print(testDic2)
    //删除值
    testDic2["type"] = nil
    print(testDic2)
    if let removeValue = testDic2.removeValue(forKey: "token") {
        print("被移除的值\(removeValue)")
    }
    print(testDic2)
    //删除所有
    testDic2.removeAll()
        
    let testDic3:[String: String] = ["token":"221212","type1":"hsaj","type2":"hej","type3":"21","type4":"109"]
    
    for (keyv,valuek) in testDic3 {
        print("key = \(keyv) value = \(valuek)")
    }
    //返回所有的key
    let keys = testDic3.keys
    for key in keys {
        print(key)
    }
    //返回所有的values
    let values = testDic3.values
    for value in values {
        print(value)
    }
    //通过keys构造数组
    let keysArray = Array(keys)
    print(keysArray)
    

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