Replace characters in swift 3 string

Question

I need a function to replace numbers 0..9 with characters A..J consecutively. How can I write it in Swift 3? Swift 3 string manipulation is so weird :| I tried accessing characters and then adding and offset (shifting) characters, but have no success.

Answer 1

Here is a possible solution to your problem which is also Unicode safe:

let string = "0123456789 abc xyz 9876543210"
// upper case A...J use 65; lower case a...j use 97
let shift = 65
let convertedString = String(string.characters.lazy.map{ char in
    Int(String(char)).map{ Character(UnicodeScalar($0 + shift)!) } ?? char
})

Explanation

Idea/Problem

Map each character of the string to another (or same) character: 0...9 -> A...J, everything else stays

Implementation

Use a lazy character collection for performance reasons for large strings. A non lazy one would create an intermediate Array.

The mapping from one to another character goes through the following process:

1. Try to convert the character to an Int (Here Int?)

   1.1 If it fails (is nil) the nil coalescing operator (??) comes in to play and the input character is returned

2. The Int value ($0) now gets converted back to a Character through its (shifted) ASCII representation

3. Convert the lazily mapped collection to a String. Swifts declaration of the initialiser:

   public init<S : Sequence>(_ characters: S)
       where S.Iterator.Element == Character { ... }