斯威夫特5

func firstIndex(of element: Element) -> Int?

var alphabets = ["A", "B", "E", "D"]

例二

let index = alphabets.firstIndex(where: {$0 == "A"})

Example2

if let i = alphabets.firstIndex(of: "E") {
    alphabets[i] = "C" // i is the index
}
print(alphabets)
// Prints "["A", "B", "C", "D"]"

你可以用闭包来过滤数组:

var myList = [1, 2, 3, 4]
var filtered = myList.filter { $0 == 3 }  // <= returns [3]

你可以对数组进行计数:

filtered.count // <= returns 1

所以你可以通过组合这些来确定数组是否包含你的元素:

myList.filter { $0 == 3 }.count > 0  // <= returns true if the array includes 3

如果你想找这个职位，我不认为有什么花哨的方法，但你可以这样做:

var found: Int?  // <= will hold the index if it was found, or else will be nil
for i in (0..x.count) {
    if x[i] == 3 {
        found = i
    }
}

EDIT

当我们这样做的时候，为了一个有趣的练习，让我们扩展Array来有一个find方法:

extension Array {
    func find(includedElement: T -> Bool) -> Int? {
        for (idx, element) in enumerate(self) {
            if includedElement(element) {
                return idx
            }
        }
        return nil
    }
}

现在我们可以这样做:

myList.find { $0 == 3 }
// returns the index position of 3 or nil if not found

由于swift在某些方面比面向对象的函数性更强(数组是结构体，而不是对象)，使用函数"find"对数组进行操作，返回一个可选值，所以准备好处理nil值:

let arr:Array = ["a","b","c"]
find(arr, "c")!              // 2
find(arr, "d")               // nil

使用firstIndex和lastIndex -这取决于你是在寻找项目的第一个索引还是最后一个索引:

let arr = ["a","b","c","a"]

let indexOfA = arr.firstIndex(of: "a") // 0
let indexOfB = arr.lastIndex(of: "a") // 3

只需使用firstIndex方法。

array.firstIndex(where: { $0 == searchedItem })

这些方法对我都适用

这是我对Swift 4的解决方案:

let monday = Day(name: "M")
let tuesday = Day(name: "T")
let friday = Day(name: "F")

let days = [monday, tuesday, friday]

let index = days.index(where: { 
            //important to test with === to be sure it's the same object reference
            $0 === tuesday
        })

斯威夫特4

假设你想从名为cardButtons的数组中存储一个数字到cardNumber中，你可以这样做:

let cardNumber = cardButtons.index(of: sender)

Sender是按钮的名称