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"]"

您还可以使用函数库Dollar在数组上执行indexOf，例如http://www.dollarswift.org/#indexof-indexof

$.indexOf([1, 2, 3, 1, 2, 3], value: 2) 
=> 1

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

如果你还在使用Swift 1.x

然后尝试,

let testArray = ["A","B","C"]

let indexOfA = find(testArray, "A") 
let indexOfB = find(testArray, "B")
let indexOfC = find(testArray, "C")

斯威夫特2.1

var array = ["0","1","2","3"]

if let index = array.indexOf("1") {
   array.removeAtIndex(index)
}

print(array) // ["0","2","3"]

斯威夫特3

var array = ["0","1","2","3"]

if let index = array.index(of: "1") {
    array.remove(at: index)
}
array.remove(at: 1)