I'd like to convert an Int in Swift to a String with leading zeros. For example consider this code:
for myInt in 1 ... 3 {
print("\(myInt)")
}
Currently the result of it is:
1
2
3
But I want it to be:
01
02
03
Is there a clean way of doing this within the Swift standard libraries?
Assuming you want a field length of 2 with leading zeros you'd do this:
import Foundation
for myInt in 1 ... 3 {
print(String(format: "%02d", myInt))
}
output:
01
02
03
This requires import Foundation so technically it is not a part of the Swift language but a capability provided by the Foundation framework. Note that both import UIKit and import Cocoa include Foundation so it isn't necessary to import it again if you've already imported Cocoa or UIKit.
The format string can specify the format of multiple items. For instance, if you are trying to format 3 hours, 15 minutes and 7 seconds into 03:15:07 you could do it like this:
let hours = 3
let minutes = 15
let seconds = 7
print(String(format: "%02d:%02d:%02d", hours, minutes, seconds))
output:
03:15:07
With Swift 5, you may choose one of the three examples shown below in order to solve your problem.
#1. Using String's init(format:_:) initializer
Foundation provides Swift String a init(format:_:) initializer. init(format:_:) has the following declaration:
init(format: String, _ arguments: CVarArg...)
Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted.
The following Playground code shows how to create a String formatted from Int with at least two integer digits by using init(format:_:):
import Foundation
let string0 = String(format: "%02d", 0) // returns "00"
let string1 = String(format: "%02d", 1) // returns "01"
let string2 = String(format: "%02d", 10) // returns "10"
let string3 = String(format: "%02d", 100) // returns "100"
#2. Using String's init(format:arguments:) initializer
Foundation provides Swift String a init(format:arguments:) initializer. init(format:arguments:) has the following declaration:
init(format: String, arguments: [CVarArg])
Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted according to the user’s default locale.
The following Playground code shows how to create a String formatted from Int with at least two integer digits by using init(format:arguments:):
import Foundation
let string0 = String(format: "%02d", arguments: [0]) // returns "00"
let string1 = String(format: "%02d", arguments: [1]) // returns "01"
let string2 = String(format: "%02d", arguments: [10]) // returns "10"
let string3 = String(format: "%02d", arguments: [100]) // returns "100"
#3. Using NumberFormatter
Foundation provides NumberFormatter. Apple states about it:
Instances of NSNumberFormatter format the textual representation of cells that contain NSNumber objects and convert textual representations of numeric values into NSNumber objects. The representation encompasses integers, floats, and doubles; floats and doubles can be formatted to a specified decimal position.
The following Playground code shows how to create a NumberFormatter that returns String? from a Int with at least two integer digits:
import Foundation
let formatter = NumberFormatter()
formatter.minimumIntegerDigits = 2
let optionalString0 = formatter.string(from: 0) // returns Optional("00")
let optionalString1 = formatter.string(from: 1) // returns Optional("01")
let optionalString2 = formatter.string(from: 10) // returns Optional("10")
let optionalString3 = formatter.string(from: 100) // returns Optional("100")
For left padding add a string extension like this:
Swift 5.0 +
extension String {
func padLeft(totalWidth: Int, with byString: String) -> String {
let toPad = totalWidth - self.count
if toPad < 1 {
return self
}
return "".padding(toLength: toPad, withPad: byString, startingAt: 0) + self
}
}
Using this method:
for myInt in 1...3 {
print("\(myInt)".padLeft(totalWidth: 2, with: "0"))
}
Swift 3.0+
Left padding String extension similar to padding(toLength:withPad:startingAt:) in Foundation
extension String {
func leftPadding(toLength: Int, withPad: String = " ") -> String {
guard toLength > self.characters.count else { return self }
let padding = String(repeating: withPad, count: toLength - self.characters.count)
return padding + self
}
}
Usage:
let s = String(123)
s.leftPadding(toLength: 8, withPad: "0") // "00000123"
Swift 5
#imanuo answers is already great, but if you are working with an application full of number, you can consider an extension like this:
extension String {
init(withInt int: Int, leadingZeros: Int = 2) {
self.init(format: "%0\(leadingZeros)d", int)
}
func leadingZeros(_ zeros: Int) -> String {
if let int = Int(self) {
return String(withInt: int, leadingZeros: zeros)
}
print("Warning: \(self) is not an Int")
return ""
}
}
In this way you can call wherever:
String(withInt: 3)
// prints 03
String(withInt: 23, leadingZeros: 4)
// prints 0023
"42".leadingZeros(2)
// prints 42
"54".leadingZeros(3)
// prints 054
Using Swift 5’s fancy new extendible interpolation:
extension DefaultStringInterpolation {
mutating func appendInterpolation(pad value: Int, toWidth width: Int, using paddingCharacter: Character = "0") {
appendInterpolation(String(format: "%\(paddingCharacter)\(width)d", value))
}
}
let pieCount = 3
print("I ate \(pad: pieCount, toWidth: 3, using: "0") pies") // => `I ate 003 pies`
print("I ate \(pad: 1205, toWidth: 3, using: "0") pies") // => `I ate 1205 pies`
in Xcode 8.3.2, iOS 10.3
Thats is good to now
Sample1:
let dayMoveRaw = 5
let dayMove = String(format: "%02d", arguments: [dayMoveRaw])
print(dayMove) // 05
Sample2:
let dayMoveRaw = 55
let dayMove = String(format: "%02d", arguments: [dayMoveRaw])
print(dayMove) // 55
The other answers are good if you are dealing only with numbers using the format string, but this is good when you may have strings that need to be padded (although admittedly a little diffent than the question asked, seems similar in spirit). Also, be careful if the string is longer than the pad.
let str = "a str"
let padAmount = max(10, str.count)
String(repeatElement("-", count: padAmount - str.count)) + str
Output "-----a str"
The below code generates a 3 digits string with 0 padding in front:
import Foundation
var randomInt = Int.random(in: 0..<1000)
var str = String(randomInt)
var paddingZero = String(repeating: "0", count: 3 - str.count)
print(str, str.count, paddingZero + str)
Output:
5 1 005
88 2 088
647 3 647
Swift 4* and above you can try this also:
func leftPadding(valueString: String, toLength: Int, withPad: String = " ") -> String {
guard toLength > valueString.count else { return valueString }
let padding = String(repeating: withPad, count: toLength - valueString.count)
return padding + valueString
}
call the function:
leftPadding(valueString: "12", toLength: 5, withPad: "0")
Output:
"00012"
Details
Xcode 9.0.1, swift 4.0
Solutions
Data
import Foundation
let array = [0,1,2,3,4,5,6,7,8]
Solution 1
extension Int {
func getString(prefix: Int) -> String {
return "\(prefix)\(self)"
}
func getString(prefix: String) -> String {
return "\(prefix)\(self)"
}
}
for item in array {
print(item.getString(prefix: 0))
}
for item in array {
print(item.getString(prefix: "0x"))
}
Solution 2
for item in array {
print(String(repeatElement("0", count: 2)) + "\(item)")
}
Solution 3
extension String {
func repeate(count: Int, string: String? = nil) -> String {
if count > 1 {
let repeatedString = string ?? self
return repeatedString + repeate(count: count-1, string: repeatedString)
}
return self
}
}
for item in array {
print("0".repeate(count: 3) + "\(item)")
}
Unlike the other answers that use a formatter, you can also just add an "0" text in front of each number inside of the loop, like this:
for myInt in 1...3 {
println("0" + "\(myInt)")
}
But formatter is often better when you have to add suppose a designated amount of 0s for each seperate number. If you only need to add one 0, though, then it's really just your pick.
Related
I need to extract numbers from string and put them into a new array in Swift.
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
I tried to loop each characters and I have no idea to compare between Characters and Int.
Swift 3/4
let string = "0kaksd020dk2kfj2123"
if let number = Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()) {
// Do something with this number
}
You can also make an extension like:
extension Int {
static func parse(from string: String) -> Int? {
return Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
And then later use it like:
if let number = Int.parse(from: "0kaksd020dk2kfj2123") {
// Do something with this number
}
First, we split the string so we can process the single items. Then we use NSCharacterSet to select the numbers only.
import Foundation
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let strArr = str.split(separator: " ")
for item in strArr {
let part = item.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
Swift 4:
let string = "I have to buy 3 apples, 7 bananas, 10eggs"
let stringArray = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
for item in stringArray {
if let number = Int(item) {
print("number: \(number)")
}
}
Using the "regex helper function" from Swift extract regex matches:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
let regex = NSRegularExpression(pattern: regex,
options: nil, error: nil)!
let nsString = text as NSString
let results = regex.matchesInString(text,
options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
return map(results) { nsString.substringWithRange($0.range)}
}
you can achieve that easily with
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let numbersAsStrings = matchesForRegexInText("\\d+", str) // [String]
let numbersAsInts = numbersAsStrings.map { $0.toInt()! } // [Int]
println(numbersAsInts) // [3, 7, 10]
The pattern "\d+" matches one or more decimal digit.
Of course the same can be done without the use of a helper function
if you prefer that for whatever reason:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let regex = NSRegularExpression(pattern: "\\d+", options: nil, error: nil)!
let nsString = str as NSString
let results = regex.matchesInString(str, options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
let numbers = map(results) { nsString.substringWithRange($0.range).toInt()! }
println(numbers) // [3, 7, 10]
Alternative solution without regular expressions:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let digits = "0123456789"
let numbers = split(str, allowEmptySlices: false) { !contains(digits, $0) }
.map { $0.toInt()! }
println(numbers) // [3, 7, 10]
let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
NSCharacterSet
.decimalDigitCharacterSet()
.invertedSet)
.joinWithSeparator("")
That will get you a string with all the number then you can just do this:
let int = Int(intString)
Just make sure you unwrap it since let int = Int(intString) is an optional.
For me makes more sense to have it as a String extension, probably it's a matter of tastes:
extension String {
func parseToInt() -> Int? {
return Int(self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
So can be used like this:
if let number = "0kaksd020dk2kfj2123".parseToInt() {
// Do something with this number
}
Adapting from #flashadvanced's answer,
I found that the following is shorter and simpler for me.
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let component = str.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let list = component.filter({ $0 != "" }) // filter out all the empty strings in the component
print(list)
Tried in in the play ground and it works
Hope it helps :)
Swift 2.2
let strArr = str.characters.split{$0 == " "}.map(String.init)
for item in strArr {
let components = item.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let part = components.joinWithSeparator("")
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
// This will only work with single digit numbers. Works with “10eggs” (no space between number and word
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
var ints: [Int] = []
for char:Character in str {
if let int = "\(char)".toInt(){
ints.append(int)
}
}
The trick here is that you can check if a string is an integer (but you can’t check if a character is).
By looping though every character of the string, use string interpolation to create a string from the character and check if that string cas be casted as a integer.
If it can be, add it to the array.
// This will work with multi digit numbers. Does NOT work with “10 eggs” (has to have a space between number and word)
var str = "I have to buy 3 apples, 7 bananas, 10 eggs"
var ints: [Int] = []
var strArray = split(str) {$0 == " "}
for subString in strArray{
if let int = subString.toInt(){
ints.append(int)
}
}
Here we split the string at any space and create an array of every substring that is in the long string.
We again check every string to see if it is (or can be casted as) an integer.
Thanks for everyone who answered to my question.
I was looking for a block of code which uses only swift grammar, because I'm learning grammar only now..
I got an answer for my question.Maybe it is not an easier way to solve, but it uses only swift language.
var article = "I have to buy 3 apples, 7 bananas, 10 eggs"
var charArray = Array(article)
var unitValue = 0
var total = 0
for char in charArray.reverse() {
if let number = "\(char)".toInt() {
if unitValue==0 {
unitValue = 1
}
else {
unitValue *= 10
}
total += number*unitValue
}
else {
unitValue = 0
}
}
println("I bought \(total) apples.")
Swift 5:
extension String {
var allNumbers: [Int] {
let numbersInString = self.components(separatedBy: .decimalDigits.inverted).filter { !$0.isEmpty }
return numbersInString.compactMap { Int($0) }
}
}
You can get all numbers like
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
// numbers = [3, 7, 10]
numbers = str.allNumbers
So I've found issues relating to the case of converting NSRange to Range<String.Index>, but I've actually run into the opposite problem.
Quite simply, I have a String and a Range<String.Index> and need to convert the latter into an NSRange for use with an older function.
So far my only workaround has been to grab a substring instead like so:
func foo(theString: String, inRange: Range<String.Index>?) -> Bool {
let theSubString = (nil == inRange) ? theString : theString.substringWithRange(inRange!)
return olderFunction(theSubString, NSMakeRange(0, countElements(theSubString)))
}
This works of course, but it isn't very pretty, I'd much rather avoid having to grab a sub-string and just use the range itself somehow, is this possible?
If you look into the definition of String.Index you find:
struct Index : BidirectionalIndexType, Comparable, Reflectable {
/// Returns the next consecutive value after `self`.
///
/// Requires: the next value is representable.
func successor() -> String.Index
/// Returns the previous consecutive value before `self`.
///
/// Requires: the previous value is representable.
func predecessor() -> String.Index
/// Returns a mirror that reflects `self`.
func getMirror() -> MirrorType
}
So actually there is no way to convert it to Int and that for good reason. Depending on the encoding of the string the single characters occupy a different number of bytes. The only way would be to count how many successor operations are needed to reach the desired String.Index.
Edit The definition of String has changed over the various Swift versions but it's basically the same answer. To see the very current definition just CMD-click on a String definition in XCode to get to the root (works for other types as well).
The distanceTo is an extension which goes to a variety of protocols. Just look for it in the String source after the CMD-click.
let index: Int = string.startIndex.distanceTo(range.startIndex)
I don't know which version introduced it, but in Swift 4.2 you can easily convert between the two.
To convert Range<String.Index> to NSRange:
let range = s[s.startIndex..<s.endIndex]
let nsRange = NSRange(range, in: s)
To convert NSRange to Range<String.Index>:
let nsRange = NSMakeRange(0, 4)
let range = Range(nsRange, in: s)
Keep in mind that NSRange is UTF-16 based, while Range<String.Index> is Character based.
Hence you can't just use counts and positions to convert between the two!
In Swift 4, distanceTo() is deprecated. You may have to convert String to NSString to take advantage of its -[NSString rangeOfString:] method, which returns an NSRange.
Swift 4 Complete Solution:
OffsetIndexableCollection (String using Int Index)
https://github.com/frogcjn/OffsetIndexableCollection-String-Int-Indexable-
let a = "01234"
print(a[0]) // 0
print(a[0...4]) // 01234
print(a[...]) // 01234
print(a[..<2]) // 01
print(a[...2]) // 012
print(a[2...]) // 234
print(a[2...3]) // 23
print(a[2...2]) // 2
if let number = a.index(of: "1") {
print(number) // 1
print(a[number...]) // 1234
}
if let number = a.index(where: { $0 > "1" }) {
print(number) // 2
}
You can use this function and call it when ever you need convertion
extension String
{
func CnvIdxTooIntFnc(IdxPsgVal: Index) -> Int
{
return startIndex.distanceTo(IdxPsgVal)
}
}
A function in swift takes any numeric type in Swift (Int, Double, Float, UInt, etc).
the function converts the number to a string
the function signature is as follows :
func swiftNumbers <T : NumericType> (number : T) -> String {
//body
}
NumericType is a custom protocol that has been added to numeric types in Swift.
inside the body of the function, the number should be converted to a string:
I use the following
var stringFromNumber = "\(number)"
which is not so elegant, PLUS : if the absolute value of the number is strictly inferior to 0.0001 it gives this:
"\(0.000099)" //"9.9e-05"
or if the number is a big number :
"\(999999999999999999.9999)" //"1e+18"
is there a way to work around this string interpolation limitation? (without using Objective-C)
P.S :
NumberFormater doesn't work either
import Foundation
let number : NSNumber = 9_999_999_999_999_997
let formatter = NumberFormatter()
formatter.minimumFractionDigits = 20
formatter.minimumIntegerDigits = 20
formatter.minimumSignificantDigits = 40
formatter.string(from: number) // "9999999999999996.000000000000000000000000"
let stringFromNumber = String(format: "%20.20f", number) // "0.00000000000000000000"
Swift String Interpolation
1) Adding different types to a string
2) Means the string is created from a mix of constants, variables, literals or expressions.
Example:
let length:Float = 3.14
var breadth = 10
var myString = "Area of a rectangle is length*breadth"
myString = "\(myString) i.e. = \(length)*\(breadth)"
Output:
3.14
10
Area of a rectangle is length*breadth
Area of a rectangle is length*breadth i.e. = 3.14*10
Use the Swift String initializer: String(format: <#String#>, arguments: <#[CVarArgType]#>)
For example:
let stringFromNumber = String(format: "%.2f", number)
String and Characters conforms to StringInterpolationProtocol protocol which provide more power to the strings.
StringInterpolationProtocol - "Represents the contents of a string literal with interpolations while it’s being built up."
String interpolation has been around since the earliest days of Swift, but in Swift 5.0 it’s getting a massive overhaul to make it faster and more powerful.
let name = "Ashwinee Dhakde"
print("Hello, I'm \(name)")
Using the new string interpolation system in Swift 5.0 we can extend String.StringInterpolation to add our own custom interpolations, like this:
extension String.StringInterpolation {
mutating func appendInterpolation(_ value: Date) {
let formatter = DateFormatter()
formatter.dateStyle = .full
let dateString = formatter.string(from: value)
appendLiteral(dateString)
}
}
Usage: print("Today's date is \(Date()).")
We can even provide user-defined names to use String-Interpolation, let's understand with an example.
extension String.StringInterpolation {
mutating func appendInterpolation(JSON JSONData: Data) {
guard
let JSONObject = try? JSONSerialization.jsonObject(with: JSONData, options: []),
let jsonData = try? JSONSerialization.data(withJSONObject: JSONObject, options: .prettyPrinted) else {
appendInterpolation("Invalid JSON data")
return
}
appendInterpolation("\n\(String(decoding: jsonData, as: UTF8.self))")
}
}
print("The JSON is \(JSON: jsonData)")
Whenever we want to provide "JSON" in the string interpolation statement, it will print the .prettyPrinted
Isn't it cool!!
Hey guys I have string "69 - 13" How to detect "-" in the string and how to sum the numbers in the string 69+13=82 ?
There are various method to do that (componentsSeparatedByString, NSScanner, ...).
Here is one using only Swift library functions:
let str = "69 - 13"
// split string into components:
let comps = split(str, { $0 == "-" || $0 == " " }, maxSplit: Int.max, allowEmptySlices: false)
// convert strings to numbers (use zero if the conversion fails):
let nums = map(comps) { $0.toInt() ?? 0 }
// compute the sum:
let sum = reduce(nums, 0) { $0 + $1 }
println(sum)
Here is an updated implementation in Swift 4 that relies on higher order functions to perform the operation:
let string = "69+13"
let number = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
.compactMap({ Int($0) })
.reduce(0, +)
print(number) // 82
The components(separatedBy: CharacterSet.decimalDigits.inverted) removes all non-digit values and creates an array for each group of values (in this case 69 and 13)
Int($0) converts your string value into an Int
compactMap gets rid of any nil values, ensuring that only valid values are left
reduce then sums up the values that remain in your array
How could I replace nth character of a String with another one?
func replace(myString:String, index:Int, newCharac:Character) -> String {
// Write correct code here
return modifiedString
}
For example, replace("House", 2, "r") should be equal to "Horse".
Solutions that use NSString methods will fail for any strings with multi-byte Unicode characters. Here are two Swift-native ways to approach the problem:
You can use the fact that a String is a sequence of Character to convert the string to an array, modify it, and convert the array back:
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
replace("House", 2, "r")
// Horse
Alternately, you can step through the string yourself:
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var modifiedString = String()
for (i, char) in myString.characters.enumerate() {
modifiedString += String((i == index) ? newChar : char)
}
return modifiedString
}
Since these stay entirely within Swift, they're both Unicode-safe:
replace("🏠🏡🏠🏡🏠", 2, "🐴")
// 🏠🏡🐴🏡🏠
In Swift 4 it's much easier.
let newString = oldString.prefix(n) + char + oldString.dropFirst(n + 1)
This is an example:
let oldString = "Hello, playground"
let newString = oldString.prefix(4) + "0" + oldString.dropFirst(5)
where the result is
Hell0, playground
The type of newString is Substring. Both prefix and dropFirst return Substring. Substring is a slice of a string, in other words, substrings are fast because you don't need to allocate memory for the content of the string, but the same storage space as the original string is used.
I've found this solution.
var string = "Cars"
let index = string.index(string.startIndex, offsetBy: 2)
string.replaceSubrange(index...index, with: "t")
print(string)
// Cats
Please see NateCook answer for more details
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString.characters) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
For Swift 5
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
replace("House", 2, "r")
This is no longer valid and deprecated.
You can always use swift String with NSString.So you can call NSString function on swift String.
By old stringByReplacingCharactersInRange: you can do like this
var st :String = "House"
let abc = st.bridgeToObjectiveC().stringByReplacingCharactersInRange(NSMakeRange(2,1), withString:"r") //Will give Horse
For modify existing string:
extension String {
subscript(_ n: Int) -> Character {
get {
let idx = self.index(startIndex, offsetBy: n)
return self[idx]
}
set {
let idx = self.index(startIndex, offsetBy: n)
self.replaceSubrange(idx...idx, with: [newValue])
}
}
}
var s = "12345"
print(s[0])
s[0] = "9"
print(s)
I've expanded upon Nate Cooks answer and transformed it into a string extension.
extension String {
//Enables replacement of the character at a specified position within a string
func replace(_ index: Int, _ newChar: Character) -> String {
var chars = Array(characters)
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
}
usage:
let source = "House"
let result = source.replace(2,"r")
result is "Horse"
I think what #Greg was trying to achieve with his extension is this:
mutating func replace(characterAt index: Int, with newChar: Character) {
var chars = Array(characters)
if index >= 0 && index < self.characters.count {
chars[index] = newChar
let modifiedString = String(chars)
self = modifiedString
} else {
print("can't replace character, its' index out of range!")
}
}
usage:
let source = "House"
source.replace(characterAt: 2, with: "r") //gives you "Horse"
After looking at the Swift Docs, I managed to make this function:
//Main function
func replace(myString:String, index:Int, newCharac:Character) -> String {
//Looping through the characters in myString
var i = 0
for character in myString {
//Checking to see if the index of the character is the one we're looking for
if i == index {
//Found it! Now instead of adding it, add newCharac!
modifiedString += newCharac
} else {
modifiedString += character
}
i = i + 1
}
// Write correct code here
return modifiedString
}
Please note that this is untested, but it should give you the right idea.
func replace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString = myString
let range = Range<String.Index>(
start: advance(myString.startIndex, index),
end: advance(myString.startIndex, index + 1))
modifiedString.replaceRange(range, with: "\(newCharac)")
return modifiedString
}
I would prefer to pass a String than a Character though.
Here's a way to replace a single character:
var string = "This is the original string."
let offset = 27
let index = string.index(string.startIndex, offsetBy: offset)
let range = index...index
print("ORIGINAL string: " + string)
string.replaceSubrange(range, with: "!")
print("UPDATED string: " + string)
// ORIGINAL string: This is the original string.
// UPDATED string: This is the original string!
This works with multi-character strings as well:
var string = "This is the original string."
let offset = 7
let index = string.index(string.startIndex, offsetBy: offset)
let range = index...index
print("ORIGINAL string: " + string)
string.replaceSubrange(range, with: " NOT ")
print("UPDATED string: " + string)
// ORIGINAL string: This is the original string.
// UPDATED string: This is NOT the original string.
var s = "helloworld"
let index = ((s.count) / 2) // index is 4
let firstIndex = s.index(s.startIndex, offsetBy: index)
let secondIndex = s.index(s.startIndex, offsetBy: index)
s.replaceSubrange(firstIndex...secondIndex, with: "*")
print("Replaced string is: \(s)") //OUTPUT IS: hell*world
This is working fine to replace string using the index.
String class in Swift (till v5 and maybe later) is what other languages call a StringBuilder class, and for performance reasons, Swift does NOT provide setting character by index; If you don't care about performance a simple solution could be:
public static func replace(_ string: String, at index: Int, with value: String) {
let start = string.index(string.startIndex, offsetBy: index)
let end = string.index(start, offsetBy: 1)
string.replaceSubrange(start..<end, with: value)
}
Or as an extension:
extension String {
public func charAt(_ index: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: index)];
}
public mutating func setCharAt(_ index: Int, _ new: Character) {
self.setCharAt(index, String(new))
}
public mutating func setCharAt(_ index: Int, _ new: String) {
let i = self.index(self.startIndex, offsetBy: index)
self.replaceSubrange(i...i, with: new)
}
}
Note how above needs to call index(...) method to convert integer to actual-index!? It seems, Swift implements String like a linked-list, where append(...) is really fast, but even finding the index (without doing anything with it) is a linear-time operation (and gets slower based on concatenation count).
public void createEncodedSentence() {
StringBuffer buff = new StringBuffer();
int counter = 0;
char a;
for (int i = 0; i < sentence.length(); i++) {
a = sentence.charAt(i);
if (a == '.') {
buff.append('*');
}
if (a != ' ' && a != '.') {
counter++;
}
if (counter % 3 == 0) {
buff.append("");
}
buff.append(sentence.charAt(i));
}
encodedSentence = buff.toString();
}
Strings in swift don't have an accessor to read or write a single character. There's an excellent blog post by Ole Begemann describing how strings in swift work.
Note: the implementation below is wrong, read addendum
So the right way is by taking the left part of the string up to the index -1 character, append the replacing character, then append the string from index + 1 up to the end:
func myReplace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString: String
let len = countElements(myString)
if (index < len) && (index >= 0) {
modifiedString = myString.substringToIndex(index) + newCharac + myString.substringFromIndex(index + 1)
} else {
modifiedString = myString
}
return modifiedString
}
Note: in my implementation I chose to return the original string if the index is not in a valid range
Addendum Thanks to #slazyk, who found out that my implementation is wrong (see comment), I am providing a new swift only version of the function.
func replace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString: String
if (index < 0) || (index >= countElements(myString)) {
modifiedString = myString
} else {
var start = myString.startIndex
var end = advance(start, index)
modifiedString = myString[start ..< end]
modifiedString += newCharac
start = end.successor()
end = myString.endIndex
modifiedString += myString[start ... end]
}
return modifiedString
}
#codester's answer looks very good, and it's probably what I would use myself.
It would be interesting to know how performances compare though, using a fully swift solution and bridging to objective-c instead.
Here is an efficient answer :
import Foundation
func replace(myString:String, index:Int, newCharac:Character) -> String {
return myString.substringToIndex(index-1) + newCharac + myString.substringFromIndex(index)
}