How to convert a String (numeric) in a Int array in Swift

Question

I'd like to know how can I convert a String in an Int array in Swift. In Java I've always done it like this:

String myString = "123456789";
int[] myArray = new int[myString.lenght()];
for(int i=0;i<myArray.lenght;i++){
   myArray[i] = Integer.parseInt(myString.charAt(i));
}

Thanks everyone for helping!

I'm removing the java tag, since this question isn't actually about Java. — yshavit
– yshavit, Commented Feb 19, 2015 at 16:27

rintaro · Accepted Answer · 2015-02-19 17:29:07Z

19

let str = "123456789"
let intArray = map(str) { String($0).toInt() ?? 0 }

map() iterates Characters in str
String($0) converts Character to String
.toInt() converts String to Int. If failed(??), use 0.

If you prefer for loop, try:

let str = "123456789"
var intArray: [Int] = []

for chr in str {
    intArray.append(String(chr).toInt() ?? 0)
}

OR, if you want to iterate indices of the String:

let str = "123456789"
var intArray: [Int] = []

for i in indices(str) {
    intArray.append(String(str[i]).toInt() ?? 0)
}

edited Feb 19, 2015 at 17:29

answered Feb 19, 2015 at 16:48

rintaro

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1 Comment

Luca Angeletti Over a year ago

toInt() is no longer available in String in Swift 2.x.

Leo Dabus · Accepted Answer · 2023-03-30 01:10:35Z

19

You can use flatMap to convert the characters into a string and coerce the character strings into an integer:

Swift 2 or 3

let string = "123456789"
let digits = string.characters.flatMap{Int(String($0))}
print(digits)   // [1, 2, 3, 4, 5, 6, 7, 8, 9]"

Swift 4

let string = "123456789"
let digits = string.flatMap{Int(String($0))}
print(digits)   // [1, 2, 3, 4, 5, 6, 7, 8, 9]"

Swift 4.1

let digits = string.compactMap{Int(String($0))}

Swift 5 or later

We can use the new Character Property wholeNumberValue https://developer.apple.com/documentation/swift/character/3127025-wholenumbervalue

let digits = string.compactMap{$0.wholeNumberValue}

Swift 5.2 or later Key Path Expressions as Functions

let digits = string.compactMap(\.wholeNumberValue)

edited Mar 30, 2023 at 1:10

answered Feb 19, 2015 at 16:32

Leo Dabus

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3 Comments

leanne Over a year ago

Swift 4.1 deprecates .flatMap for this use case. Use .compactMap instead: let digits = string.compactMap{Int(String($0))}

JKvr Over a year ago

let digits = string.compactMap(\.wholeNumberValue) is also a nice way to write it (Swift 5)

Leo Dabus Over a year ago

@JKvr yes it is called keyPath stackoverflow.com/a/30415964/2303865 Swift 5.2 or later. It is not available for Swift 5.0

Arjun Yadav · Accepted Answer · 2017-05-16 06:41:57Z

2

Swift 3

Int array to String

let arjun = [1,32,45,5]
    print(self.get_numbers(array: arjun))

 func get_numbers(array:[Int]) -> String {
        let stringArray = array.flatMap { String(describing: $0) }
        return stringArray.joined(separator: ",")

String to Int Array

let arjun = "1,32,45,5"
    print(self.get_numbers(stringtext: arjun))

    func get_numbers(stringtext:String) -> [Int] {
    let StringRecordedArr = stringtext.components(separatedBy: ",")
    return StringRecordedArr.map { Int($0)!}   
}

answered May 16, 2017 at 6:41

Arjun Yadav

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Comments

Aaron Rasmussen · Accepted Answer · 2015-02-19 19:00:07Z

1

@rintaro's answer is correct, but I just wanted to add that you can use reduce to weed out any characters that can't be converted to an Int, and even display a warning message if that happens:

let str = "123456789"
let intArray = reduce(str, [Int]()) { (var array: [Int], char: Character) -> [Int] in
    if let i = String(char).toInt() {
        array.append(i)
    } else {
        println("Warning: could not convert character \(char) to an integer")
    }
    return array
}

The advantages are:

if intArray contains zeros you will know that there was a 0 in str, and not some other character that turned into a zero
you will get told if there is a non-Int character that is possibly screwing things up.

answered Feb 19, 2015 at 19:00

Aaron Rasmussen

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Comments

Nate Cook · Accepted Answer · 2015-02-19 17:00:00Z

0

var myString = "123456789"
var myArray:[Int] = []

for index in 0..<countElements(myString) {
    var myChar = myString[advance(myString.startIndex, index)]
    myArray.append(String(myChar).toInt()!)
}

println(myArray)   // [1, 2, 3, 4, 5, 6, 7, 8, 9]"

To get the iterator pointing to a char from the string you can use advance

The method to convert string to int in Swift is toInt()

edited Feb 19, 2015 at 17:00

Nate Cook

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answered Feb 19, 2015 at 16:19

Jérôme

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1 Comment

Nate Cook Over a year ago

You have a loop in your loop—calling advance over and over like that is a bad idea, since you're making the program step through the string from the beginning each time.

iCode · Accepted Answer · 2017-01-10 11:16:50Z

0

Swift 3 update:

@appzYourLife : That's correct toInt() method is no longer available for String in Swift 3. As an alternative what you can do is :

intArray.append(Int(String(chr)) ?? 0)

Enclosing it within Int() converts it to Int.

answered Jan 10, 2017 at 11:16

iCode

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

Swift 3: Functional Approach

Split the String into separate String instances using: components(separatedBy separator: String) -> [String]

Reference: Returns an array containing substrings from the String that have been divided by a given separator.

Use the flatMap Array method to bypass the nil coalescing while converting to Int

Reference: Returns an array containing the non-nil results of calling the given transformation with each element of this sequence.

Implementation

let string = "123456789"
let intArray = string.components(separatedBy: "").flatMap { Int($0) }

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered May 19, 2017 at 6:31

bajocode

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3 Comments

leanne Over a year ago

Swift 4.1 deprecates .flatMap for this use case. Instead, use .compactMap: let intArray = string.components(separatedBy: "").compactMap { Int($0) }

Leo Dabus Over a year ago

@leanne this is wrong it would result in an array with a single element of the same value of the input [123456789]. The first method components(separatedBy: "") would result in a single string element ["123456789"]

leanne Over a year ago

I simply replaced the deprecated .flatMap with the new .compactMap, @LeoDabus. Both results are the same. As printed in Xcode's debug console: [123456789] and [123456789]. Note that I am not judging whether the result is appropriate for the original question, but simply remarking for anyone who comes here that .flatMap has been deprecated in favor of .compactMap in cases where one is attempting to achieve a result that removes nil values.

FatMonkey · Accepted Answer · 2020-07-25 12:29:01Z

0

let array = "0123456789".compactMap{ Int(String($0)) }
print(array)

answered Jul 25, 2020 at 12:29

FatMonkey

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1 Comment

Mark Rotteveel Over a year ago

Please don't post only code as answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually more helpful and of better quality, and are more likely to attract upvotes.

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