Sort alphanumeric array, consecutive numbers should reside at last

Question

I want to sort an alphanumeric, but numbers should come at end instead of beginning. for instance

let array = [\"1\", \"a\", \"b\", \"z\", \"3\", \"!\"]
let sor         


        

Answer 1

Here's a solution that works for your case :

The expected result is :

• letters first ;
• then numbers ;
• then others ;

Here's a way to do it.

// Works with both arrays
var array = ["1", "b", "a", "z", "3", "!"]
//var array = ["1", "b", "A", "Z", "3", "!"]

func isLetter(char: String) -> Bool {
    return ("a"..."z").contains(char) || ("A"..."Z").contains(char)
}

func isNumber(char: String) -> Bool {
    return Int(char) != nil
}

let letters = array.filter(isLetter).sort{$0.lowercaseString < $1.lowercaseString}
let numbers = array.filter(isNumber).sort{Int($0) < Int($1)}
let others = Array(Set(array).subtract(letters).subtract(numbers)).sort{$0 < $1}

let sortedArray = letters + numbers + others

The first array would be

["a", "b", "z", "1", "3", "!"]

The second would be

["A", "b", "Z", "1", "3", "!"]

It does what you want. Include unit tests and wrap that inside a method and you're good to go. Buuuut, it's not "clean" at all.