How to split an Int to its individual digits? - swift

Question

I am trying to split an Int into its individual digits, e.g. 3489 to 3 4 8 9, and then I want to put the digits in an Int array.
I have already tried putting the number into a string and then iterating over each digit, but it doesn't work:
var number = "123456"
var array = [Int]()
for digit in number {
array.append(digit)
}
Any ideas?

Answer 1

We can also extend the StringProtocol and create a computed property:
edit/update: Xcode 11.5 • Swift 5.2
extension StringProtocol {
var digits: [Int] { compactMap(\.wholeNumberValue) }
}
let string = "123456"
let digits = string.digits // [1, 2, 3, 4, 5, 6]
extension LosslessStringConvertible {
var string: String { .init(self) }
}
extension Numeric where Self: LosslessStringConvertible {
var digits: [Int] { string.digits }
}
let integer = 123
let integerDigits = integer.digits // [1, 2, 3]
let double = 12.34
let doubleDigits = double.digits // // [1, 2, 3, 4]
In Swift 5 now we can use the new Character property wholeNumberValue
let string = "123456"
let digits = string.compactMap{ $0.wholeNumberValue } // [1, 2, 3, 4, 5, 6]

Answer 2

A solution without having to convert the int to string....
Example
1234%10 = 4 <-
1234/10 = 123
123%10 = 3 <-
123/10 = 12
12%10 = 2 <-
12/10 = 1
1%10 = 1 <-
var num = 12345
var arrayInt = [Int]()
arrayInt.append(num%10)
while num >= 10 {
num = num/10
arrayInt.insert(num%10, at: 0)
}

Answer 3

Swift 5
extension Int {
func digits() -> [Int] {
var digits: [Int] = []
var num = self
digits.append(num % 10)
while num >= 10 {
num = num / 10
digits.append(num % 10)
}
return digits.reversed()
}
}

Answer 4

You can try this:
var number = "123456"
var array = number.utf8.map{Int($0)-48}
You can make use of the utf8 property of String to directly access the ASCII value of the characters in the String representation of your number.

Answer 5

this code works:
var number = "123456"
var array = number.utf8.map{Int(($0 as UInt8)) - 48}
this might be slower:
var array = number.characters.map{Int(String($0))!}
and if your number is less or equal than Int.max which is 9223372036854775807 here is the fastest variant (and if your number>Int.max you can split your long string that represents your number into 18-digit groups to make this variant work on large strings)
var array2 = [Int]()
var i = Int(number)!
while i > 0 {array2.append(i%10); i/=10}
array2 = array2.reverse()

Answer 6

You can use this
extension Int {
func numberOfDigits() -> Int {
if abs(self) < 10 {
return 1
} else {
return 1 + (self/10).numberOfDigits()
}
}
func getDigits() -> [Int] {
let num = self.numberOfDigits()
var tempNumber = self
var digitList = [Int]()
for i in (0..<num).reversed() {
let divider = Int(pow(CGFloat(10), CGFloat(i)))
let digit = tempNumber/divider
digitList.append(digit)
tempNumber -= digit*divider
}
return digitList
}
}

Answer 7

if you need convert integer to Array
let num = 12345
let array = String(num).compactMap({$0.wholeNumberValue})
print("\(array), \(type(of: array))")
//[1,2,3,4,5], Array<Int>

Answer 8

extension Int {
var digits : [Int] {
var result = [Int]()
var remaining = abs(self)
while remaining > 0 {
result.insert(remaining % 10, at: 0)
remaining /= 10
}
return result
}
}

Answer 9

You can use this:
// input -> "123456"
// output -> [1, 2, 3, 4, 5, 6]
// Get the string, convert it in an Array(),
// use a loop (convert i in Int()), to add it into a container, then return it. Done!
func getDigitsFromString(for string: String) -> [Int]{
let stringInt = Array(string)
var array = [Int]()
for i in stringInt {
if let i = Int(String(i)) {
array.append(i)
}
}
return array
}

Answer 10

There is a simpler method to achieve this. Say x = 12345 , then
var s = String(x);
var numArray = Array(s);
clean, easy-to-read code.

Answer 11

You can use:
let number = -123
let regex = /-?\d/
let values = String(number).matches(of: regex).compactMap { Int($0.output) }
print(values) // [-1, 2, 3]
This code takes into account negative values