[英]Swift return string from Array
我正在嘗試從此數組中獲取一個隨機名稱作為字符串而不是數字。
IE:數組通常返回0到9的隨機數,我寧願它返回0到9表示的字符串(如Preston或James),而不是數字本身。
下面的代碼已損壞,但我希望它可以讓您了解我正在嘗試執行的操作。
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)
謝謝您的幫助!
您不應將arc4random()
與%
運算符一起使用。 這引入了模偏差 。
在Swift 4.2和更高版本中,您應該使用randomElement()
:
let firstNames = ["Preston", "Ally", "James", "Justin", "Dave", "Bacon", "Bossy", "Edward", "Edweird"]
let randomFirstName = firstNames.randomElement()!
或者,如果數組可能為空,請不要使用強制展開運算符,而應執行以下操作:
guard let randomFirstName = firstNames.randomElement() else {
print("array was empty")
return
}
在4.2之前的Swift版本中,您應該:
通常,您應該使用arc4random_uniform
而不是arc4random
來消除模偏差。
您可能應該使用數組中項目的計數來確定可能的索引值的范圍。
從而:
guard firstNames.count > 0 else { ... }
let index = Int(arc4random_uniform(UInt32(firstNames.count)))
let randomFirstName = firstNames[index]
您應該像這樣使用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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