Question

我试图从这个数组中获取一个随机名称作为字符串而不是数字。

I.E：数组通常会返回一个0到9的随机数，我希望它返回0到9表示的字符串，如Preston或James，而不是数字本身。

以下代码已被破解，但我希望它可以让您看到我尝试做的事情。

 var firstName : [String] = ["Preston", "Ally", "James", "Justin", "Dave", "Bacon", "Bossy",     "Edward", "Edweird" ]

var standardIdent = "First Name:\(firstName[random(0...9)]) Last Name:\(lastName[random(0...5)]) \n Age:\(rand())"
println(standardIdent)

感谢您的帮助！

Answer 1

您不应将arc4random()与%运算符一起使用。这引入了modulo bias。

在Swift 4.2及更高版本中，您应该使用randomElement()：

let firstNames = ["Preston", "Ally", "James", "Justin", "Dave", "Bacon", "Bossy", "Edward", "Edweird"]
let randomFirstName = firstNames.randomElement()!

或者，如果阵列可能是空的，请不要使用强制解包操作符，而是执行：

guard let randomFirstName = firstNames.randomElement() else {
    print("array was empty")
    return
}

在4.2之前的Swift版本中，你应该：

您通常应使用arc4random_uniform而不是arc4random来消除模偏差。
您应该使用数组中项目的计数来确定可能的索引值的范围。

因此：

guard firstNames.count > 0 else { ... }
let index = Int(arc4random_uniform(UInt32(firstNames.count)))
let randomFirstName = firstNames[index]

Answer 2

您应该使用arc4random之类的：

let firstRandom = Int(arc4random() % 10)
let secondRandom = Int(arc4random() % 6)
var standardIdent = "First Name:\(firstName[firstRandom]) Last Name:\(lastName[secondRandom]) \n Age:\(rand())"

Answer 3

更好：使用arc4random_uniform来避免模偏差：

var firstName : [String] = ["Preston", "Ally", "James", "Justin", "Dave", "Bacon", "Bossy", "Edward", "Edweird" ]

var lastName : [String] = ["Miller", "Jones", "Jackson", "Smith"]

let firstRandom = Int(arc4random_uniform(UInt32(firstName.count)))
let secondRandom = Int(arc4random_uniform(UInt32(lastName.count)))
var standardIdent = "First Name:\(firstName[firstRandom]) Last Name:\(lastName[secondRandom]) \n Age:\(rand())"

Swift从Array返回字符串

3 个答案: