How to find out if letter is Alphanumeric or Digit in Swift

Question

I want to count the number of letters, digits and special characters in the following string:

let phrase = "The final score was 32-31!"

I tried:

for tempChar in phrase {
    if (tempChar >= "a" && tempChar <= "z") {
       letterCounter++
    }
// etc.

but I'm getting errors. I tried all sorts of other variations on this - still getting error - such as:

could not find an overload for '<=' that accepts the supplied arguments

Answer 1

For Swift 5 see rustylepord's answer.

Update for Swift 3:

let letters = CharacterSet.letters
let digits = CharacterSet.decimalDigits

var letterCount = 0
var digitCount = 0

for uni in phrase.unicodeScalars {
    if letters.contains(uni) {
        letterCount += 1
    } else if digits.contains(uni) {
        digitCount += 1
    }
}

---

(Previous answer for older Swift versions)

A possible Swift solution:

var letterCounter = 0
var digitCount = 0
let phrase = "The final score was 32-31!"
for tempChar in phrase.unicodeScalars {
    if tempChar.isAlpha() {
        letterCounter++
    } else if tempChar.isDigit() {
        digitCount++
    }
}

---

Update: The above solution works only with characters in the ASCII character set, i.e. it does not recognize Ä, é or ø as letters. The following alternative solution uses NSCharacterSet from the Foundation framework, which can test characters based on their Unicode character classes:

let letters = NSCharacterSet.letterCharacterSet()
let digits = NSCharacterSet.decimalDigitCharacterSet()

var letterCount = 0
var digitCount = 0

for uni in phrase.unicodeScalars {
    if letters.longCharacterIsMember(uni.value) {
        letterCount++
    } else if digits.longCharacterIsMember(uni.value) {
        digitCount++
    }
}

---

Update 2: As of Xcode 6 beta 4, the first solution does not work anymore, because the isAlpha() and related (ASCII-only) methods have been removed from Swift. The second solution still works.

Answer 2

Use the values of unicodeScalars

let phrase = "The final score was 32-31!"
var letterCounter = 0, digitCounter = 0
for scalar in phrase.unicodeScalars {
    let value = scalar.value
    if (value >= 65 && value <= 90) || (value >= 97 && value <= 122) {++letterCounter}
    if (value >= 48 && value <= 57) {++digitCounter}
}
println(letterCounter)
println(digitCounter)

Answer 3

For Swift 5 you can do the following for simple strings, but be vigilant about handling characters like "1️⃣" , "④" these would be treated as numbers as well.

let phrase = "The final score was 32-31!"

var numberOfDigits = 0;
var numberOfLetters = 0;
var numberOfSymbols = 0;

phrase.forEach {

    if ($0.isNumber) {
        numberOfDigits += 1;
    }
    else if ($0.isLetter)  {
        numberOfLetters += 1
    }
    else if ($0.isSymbol || $0.isPunctuation || $0.isCurrencySymbol || $0.isMathSymbol) {
        numberOfSymbols += 1;
    }
}

print(#"\#(numberOfDigits)  || \#(numberOfLetters) || \#(numberOfSymbols)"#);

Answer 4

In case you only need one information (letter or number or sign) you can do it in one line:

let phrase = "The final score was 32-31!"

let count = phrase.filter{ $0.isLetter }.count
print(count) // "16\n"

But doing phrase.filter several times is inefficient because it loops through the whole string.

Answer 5

I've created a short extension for letter and digits count for a String

extension String {
  var letterCount : Int {
    return self.unicodeScalars.filter({ CharacterSet.letters.contains($0) }).count
  }

  var digitCount : Int {
   return self.unicodeScalars.filter({ CharacterSet.decimalDigits.contains($0) }).count
  }
}

or a function to get a count for any CharacterSet you put in

extension String {    
  func characterCount(for set: CharacterSet) -> Int {
    return self.unicodeScalars.filter({ set.contains($0) }).count
  }
}

usage:

let phrase = "the final score is 23-13!"
let letterCount = phrase.characterCount(for: .letters)