[英]Swift return string from Array
我正在尝试从此数组中获取一个随机名称作为字符串而不是数字。
IE:数组通常返回0到9的随机数,我宁愿它返回0到9表示的字符串(如Preston或James),而不是数字本身。
下面的代码已损坏,但我希望它可以让您了解我正在尝试执行的操作。
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)
谢谢您的帮助!
您不应将arc4random()
与%
运算符一起使用。 这引入了模偏差 。
在Swift 4.2和更高版本中,您应该使用randomElement()
:
let firstNames = ["Preston", "Ally", "James", "Justin", "Dave", "Bacon", "Bossy", "Edward", "Edweird"]
let randomFirstName = firstNames.randomElement()!
或者,如果数组可能为空,请不要使用强制展开运算符,而应执行以下操作:
guard let randomFirstName = firstNames.randomElement() else {
print("array was empty")
return
}
在4.2之前的Swift版本中,您应该:
通常,您应该使用arc4random_uniform
而不是arc4random
来消除模偏差。
您可能应该使用数组中项目的计数来确定可能的索引值的范围。
从而:
guard firstNames.count > 0 else { ... }
let index = Int(arc4random_uniform(UInt32(firstNames.count)))
let randomFirstName = firstNames[index]
您应该像这样使用arc4random
:
let firstRandom = Int(arc4random() % 10)
let secondRandom = Int(arc4random() % 6)
var standardIdent = "First Name:\(firstName[firstRandom]) Last Name:\(lastName[secondRandom]) \n Age:\(rand())"
更好的是:使用arc4random_uniform
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())"
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