使用map函数

let array = ["11", "43", "26", "11", "45", "40"]
let intArray = array.map { Int($0)!} // [11, 43, 26, 11, 45, 40]

在UIViewController之类的类中使用

let array = ["11", "43", "26", "11", "45", "40"]
var intArray = Array<Int>!

override func viewDidLoad() {
  super.viewDidLoad()
  intArray = array.map { Int($0)!} // [11, 43, 26, 11, 45, 40]
}

如果数组包含不同的类型，您可以使用flatMap（Swift 2）或compactMap（Swift 4.1+）来仅考虑可转换为Int的项

let array = ["11", "43", "26", "Foo", "11", "45", "40"]
let intArray = array.compactMap { Int($0) } // [11, 43, 26, 11, 45, 40]

雨燕4、5

使用compactMap并强制转换为Int，解决方案不带“！”。

let array = ["1","foo","0","bar","100"]
let arrayInt = array.compactMap { Int($0) }

print(arrayInt)
// [1, 0, 100]

雨燕4、5

如果你想把字符串数字转换成int类型的数组（在一个我曾经经历过的特殊情况下），最快的方法是：

let pinString = "123456"
let pin = pinString.map { Int(String($0))! }

而对于你的问题是：

let pinArrayString = ["1","2","3","4","5","6"]
let pinArrayInt = pinArrayString.map { Int($0)! }

我建议用一种不同的方法

let stringarr = ["1","foo","0","bar","100"]
let res = stringarr.map{ Int($0) }.enumerate().flatMap { (i,j) -> (Int,String,Int)? in
    guard let value = j else {
        return nil
    }
    return (i, stringarr[i],value)
}
// now i have an access to (index in orig [String], String, Int) without any optionals and / or default values
print(res)
// [(0, "1", 1), (2, "0", 0), (4, "100", 100)]