Capitalize strings at given indexes with Swift - swift

I want to be able to capitalizes characters from a string from an array of indexes.
let string = "format"
let indexes = [0, 5]
Expected: "FormaT"
Can this be accomplished in Swift?

You can try and get an array of the characters, and uppercase the ones in your indexes array like this:
let string = "format"
let indexes = [0, 5]
var array = string.map{ String($0) }
indexes.forEach{ array[$0] = array[$0].uppercased() }
let fixedString = array.joined()
print(fixedString)

The following String extension should work for you.
extension String {
func capitalized(at characters: [Int]) -> String {
var formattedString = self
for character in characters {
formattedString = formattedString.prefix(character) + String(self[index(startIndex, offsetBy: character)]).capitalized + formattedString.dropFirst(character + 1)
}
return formattedString
}
}
And then using it:
print("format".capitalized(at: [0,2,5]))
Result: FoRmaT

You just need to map through the String and at each character, either return the character itself or its uppercased version in case the index of the character is contained in the indices you want to modify.
extension String {
func uppercased(at indices: [Int]) -> String {
return enumerated().map { indices.contains($0.offset) ? $0.element.uppercased() : String($0.element) }.joined()
}
}
let string = "format"
let indices = [0, 5]
print(string.uppercased(at: indices)) // "FormaT"

You can iterate your indices, get the index at the desired offset and replace the subrange by its uppercase:
extension StringProtocol where Self: RangeReplaceableCollection {
mutating func uppercase(at indices: [Int]) {
for offset in indices {
let index = self.index(startIndex, offsetBy: offset)
replaceSubrange(index...index, with: self[index...index].uppercased())
}
}
func uppercased(at indices: [Int]) -> Self {
var string = self
string.uppercase(at: indices)
return string
}
}
var string = "format"
let indices = [0, 5]
string.uppercase(at: indices) // "FormaT"
string
print(string.uppercased(at: [1,2])) // "FORmaT\n"

Related

Building Unicode scalar String in Swift

How can I make "\u{3A9}" String with coding? here what I tried, but did not worked!
let omegaHexadecimal: String = "3A9"
let omega = "\u{" + omegaHexadecimal + "}"
Or:
let omegaHexadecimal: String = "3A9"
let omega = "\u{\(omegaHexadecimal)}"
Update:
extension StringProtocol where Self: RangeReplaceableCollection {
private var decodingUnicodeCharacters: String { applyingTransform(.init("Hex-Any"), reverse: false) ?? "" }
func stringToUniCodeHexConvertor(upTo lenght: Int = 4, using character: Character = "0") -> String {
return ("\\u" + repeatElement(character, count: Swift.max(0,lenght-count)) + self).decodingUnicodeCharacters
}
}
let omegaHexadecimal: String = "3A9"
let omega = omegaHexadecimal.stringToUniCodeHexConvertor()
print(omega) // "Ω"
You can pad your string up to 4 hexa digits (2 bytes UInt16), add \u prefix \uXXXX and use a string transform to convert your unicode hexa value to the corresponding character:
extension StringProtocol where Self: RangeReplaceableCollection {
func paddingToLeft(upTo lenght: Int = 4, using character: Character = "0") -> Self {
repeatElement(character, count: Swift.max(0,lenght-count)) + self
}
var decodingUnicodeCharacters: String { applyingTransform(.init("Hex-Any"), reverse: false) ?? "" }
}
let omegaHexadecimal: String = "3A9"
let omega = "\\u" + omegaHexadecimal.paddingToLeft() // "\\u03A9"
omega.decodingUnicodeCharacters // "Ω"
Why so complicated? Work with numbers directly:
let omega = UnicodeScalar(0x3A9)!
print(String(omega)) // Ω

substring(with:)' is deprecated: Please use String slicing subscript [duplicate]

I have the following simple code written in Swift 3:
let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)
From Xcode 9 beta 5, I get the following warning:
'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator.
How can this slicing subscript with partial range from be used in Swift 4?
You should leave one side empty, hence the name "partial range".
let newStr = str[..<index]
The same stands for partial range from operators, just leave the other side empty:
let newStr = str[index...]
Keep in mind that these range operators return a Substring. If you want to convert it to a string, use String's initialization function:
let newStr = String(str[..<index])
You can read more about the new substrings here.
Convert Substring (Swift 3) to String Slicing (Swift 4)
Examples In Swift 3, 4:
let newStr = str.substring(to: index) // Swift 3
let newStr = String(str[..<index]) // Swift 4
let newStr = str.substring(from: index) // Swift 3
let newStr = String(str[index...]) // Swift 4
let range = firstIndex..<secondIndex // If you have a range
let newStr = = str.substring(with: range) // Swift 3
let newStr = String(str[range]) // Swift 4
Swift 5, 4
Usage
let text = "Hello world"
text[0] // H
text[...3] // "Hell"
text[6..<text.count] // world
text[NSRange(location: 6, length: 3)] // wor
Code
import Foundation
public extension String {
subscript(value: Int) -> Character {
self[index(at: value)]
}
}
public extension String {
subscript(value: NSRange) -> Substring {
self[value.lowerBound..<value.upperBound]
}
}
public extension String {
subscript(value: CountableClosedRange<Int>) -> Substring {
self[index(at: value.lowerBound)...index(at: value.upperBound)]
}
subscript(value: CountableRange<Int>) -> Substring {
self[index(at: value.lowerBound)..<index(at: value.upperBound)]
}
subscript(value: PartialRangeUpTo<Int>) -> Substring {
self[..<index(at: value.upperBound)]
}
subscript(value: PartialRangeThrough<Int>) -> Substring {
self[...index(at: value.upperBound)]
}
subscript(value: PartialRangeFrom<Int>) -> Substring {
self[index(at: value.lowerBound)...]
}
}
private extension String {
func index(at offset: Int) -> String.Index {
index(startIndex, offsetBy: offset)
}
}
Shorter in Swift 4/5:
let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"
Swift5
(Java's substring method):
extension String {
func subString(from: Int, to: Int) -> String {
let startIndex = self.index(self.startIndex, offsetBy: from)
let endIndex = self.index(self.startIndex, offsetBy: to)
return String(self[startIndex..<endIndex])
}
}
Usage:
var str = "Hello, Nick Michaels"
print(str.subString(from:7,to:20))
// print Nick Michaels
The conversion of your code to Swift 4 can also be done this way:
let str = "Hello, playground"
let index = str.index(of: ",")!
let substr = str.prefix(upTo: index)
You can use the code below to have a new string:
let newString = String(str.prefix(upTo: index))
substring(from: index) Converted to [index...]
Check the sample
let text = "1234567890"
let index = text.index(text.startIndex, offsetBy: 3)
text.substring(from: index) // "4567890" [Swift 3]
String(text[index...]) // "4567890" [Swift 4]
Some useful extensions:
extension String {
func substring(from: Int, to: Int) -> String {
let start = index(startIndex, offsetBy: from)
let end = index(start, offsetBy: to - from)
return String(self[start ..< end])
}
func substring(range: NSRange) -> String {
return substring(from: range.lowerBound, to: range.upperBound)
}
}
Example of uppercasedFirstCharacter convenience property in Swift3 and Swift4.
Property uppercasedFirstCharacterNew demonstrates how to use String slicing subscript in Swift4.
extension String {
public var uppercasedFirstCharacterOld: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = substring(to: splitIndex).uppercased()
let sentence = substring(from: splitIndex)
return firstCharacter + sentence
} else {
return self
}
}
public var uppercasedFirstCharacterNew: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = self[..<splitIndex].uppercased()
let sentence = self[splitIndex...]
return firstCharacter + sentence
} else {
return self
}
}
}
let lorem = "lorem".uppercasedFirstCharacterOld
print(lorem) // Prints "Lorem"
let ipsum = "ipsum".uppercasedFirstCharacterNew
print(ipsum) // Prints "Ipsum"
You can create your custom subString method using extension to class String as below:
extension String {
func subString(startIndex: Int, endIndex: Int) -> String {
let end = (endIndex - self.count) + 1
let indexStartOfText = self.index(self.startIndex, offsetBy: startIndex)
let indexEndOfText = self.index(self.endIndex, offsetBy: end)
let substring = self[indexStartOfText..<indexEndOfText]
return String(substring)
}
}
Creating SubString (prefix and suffix) from String using Swift 4:
let str : String = "ilike"
for i in 0...str.count {
let index = str.index(str.startIndex, offsetBy: i) // String.Index
let prefix = str[..<index] // String.SubSequence
let suffix = str[index...] // String.SubSequence
print("prefix \(prefix), suffix : \(suffix)")
}
Output
prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix :
If you want to generate a substring between 2 indices , use :
let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex
I have written a string extension for replacement of 'String: subString:'
extension String {
func sliceByCharacter(from: Character, to: Character) -> String? {
let fromIndex = self.index(self.index(of: from)!, offsetBy: 1)
let toIndex = self.index(self.index(of: to)!, offsetBy: -1)
return String(self[fromIndex...toIndex])
}
func sliceByString(from:String, to:String) -> String? {
//From - startIndex
var range = self.range(of: from)
let subString = String(self[range!.upperBound...])
//To - endIndex
range = subString.range(of: to)
return String(subString[..<range!.lowerBound])
}
}
Usage : "Date(1511508780012+0530)".sliceByString(from: "(", to: "+")
Example Result : "1511508780012"
PS: Optionals are forced to unwrap. Please add Type safety check wherever necessary.
If you are trying to just get a substring up to a specific character, you don't need to find the index first, you can just use the prefix(while:) method
let str = "Hello, playground"
let subString = str.prefix { $0 != "," } // "Hello" as a String.SubSequence
When programming I often have strings with just plain A-Za-z and 0-9. No need for difficult Index actions. This extension is based on the plain old left / mid / right functions.
extension String {
// LEFT
// Returns the specified number of chars from the left of the string
// let str = "Hello"
// print(str.left(3)) // Hel
func left(_ to: Int) -> String {
return "\(self[..<self.index(startIndex, offsetBy: to)])"
}
// RIGHT
// Returns the specified number of chars from the right of the string
// let str = "Hello"
// print(str.left(3)) // llo
func right(_ from: Int) -> String {
return "\(self[self.index(startIndex, offsetBy: self.length-from)...])"
}
// MID
// Returns the specified number of chars from the startpoint of the string
// let str = "Hello"
// print(str.left(2,amount: 2)) // ll
func mid(_ from: Int, amount: Int) -> String {
let x = "\(self[self.index(startIndex, offsetBy: from)...])"
return x.left(amount)
}
}
Hope this will help little more :-
var string = "123456789"
If you want a substring after some particular index.
var indexStart = string.index(after: string.startIndex )// you can use any index in place of startIndex
var strIndexStart = String (string[indexStart...])//23456789
If you want a substring after removing some string at the end.
var indexEnd = string.index(before: string.endIndex)
var strIndexEnd = String (string[..<indexEnd])//12345678
you can also create indexes with the following code :-
var indexWithOffset = string.index(string.startIndex, offsetBy: 4)
with this method you can get specific range of string.you need to pass start index and after that total number of characters you want.
extension String{
func substring(fromIndex : Int,count : Int) -> String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(self.startIndex, offsetBy: fromIndex + count)
let range = startIndex..<endIndex
return String(self[range])
}
}
This is my solution, no warning, no errors, but perfect
let redStr: String = String(trimmStr[String.Index.init(encodedOffset: 0)..<String.Index.init(encodedOffset: 2)])
let greenStr: String = String(trimmStr[String.Index.init(encodedOffset: 3)..<String.Index.init(encodedOffset: 4)])
let blueStr: String = String(trimmStr[String.Index.init(encodedOffset: 5)..<String.Index.init(encodedOffset: 6)])
var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))
You can try in this way and will get proper results.
the simples way that I use is :
String(Array(str)[2...4])
Swift 4, 5, 5+
Substring from Last
let str = "Hello World"
let removeFirstSix = String(str.dropFirst(6))
print(removeFirstSix) //World
Substring from First
let removeLastSix = String(str.dropLast(6))
print(removeLastSix) //Hello
Hope it would be helpful.
extension String {
func getSubString(_ char: Character) -> String {
var subString = ""
for eachChar in self {
if eachChar == char {
return subString
} else {
subString += String(eachChar)
}
}
return subString
}
}
let str: String = "Hello, playground"
print(str.getSubString(","))

How can I use String substring in Swift 4? 'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator

I have the following simple code written in Swift 3:
let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)
From Xcode 9 beta 5, I get the following warning:
'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator.
How can this slicing subscript with partial range from be used in Swift 4?
You should leave one side empty, hence the name "partial range".
let newStr = str[..<index]
The same stands for partial range from operators, just leave the other side empty:
let newStr = str[index...]
Keep in mind that these range operators return a Substring. If you want to convert it to a string, use String's initialization function:
let newStr = String(str[..<index])
You can read more about the new substrings here.
Convert Substring (Swift 3) to String Slicing (Swift 4)
Examples In Swift 3, 4:
let newStr = str.substring(to: index) // Swift 3
let newStr = String(str[..<index]) // Swift 4
let newStr = str.substring(from: index) // Swift 3
let newStr = String(str[index...]) // Swift 4
let range = firstIndex..<secondIndex // If you have a range
let newStr = = str.substring(with: range) // Swift 3
let newStr = String(str[range]) // Swift 4
Swift 5, 4
Usage
let text = "Hello world"
text[0] // H
text[...3] // "Hell"
text[6..<text.count] // world
text[NSRange(location: 6, length: 3)] // wor
Code
import Foundation
public extension String {
subscript(value: Int) -> Character {
self[index(at: value)]
}
}
public extension String {
subscript(value: NSRange) -> Substring {
self[value.lowerBound..<value.upperBound]
}
}
public extension String {
subscript(value: CountableClosedRange<Int>) -> Substring {
self[index(at: value.lowerBound)...index(at: value.upperBound)]
}
subscript(value: CountableRange<Int>) -> Substring {
self[index(at: value.lowerBound)..<index(at: value.upperBound)]
}
subscript(value: PartialRangeUpTo<Int>) -> Substring {
self[..<index(at: value.upperBound)]
}
subscript(value: PartialRangeThrough<Int>) -> Substring {
self[...index(at: value.upperBound)]
}
subscript(value: PartialRangeFrom<Int>) -> Substring {
self[index(at: value.lowerBound)...]
}
}
private extension String {
func index(at offset: Int) -> String.Index {
index(startIndex, offsetBy: offset)
}
}
Shorter in Swift 4/5:
let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"
Swift5
(Java's substring method):
extension String {
func subString(from: Int, to: Int) -> String {
let startIndex = self.index(self.startIndex, offsetBy: from)
let endIndex = self.index(self.startIndex, offsetBy: to)
return String(self[startIndex..<endIndex])
}
}
Usage:
var str = "Hello, Nick Michaels"
print(str.subString(from:7,to:20))
// print Nick Michaels
The conversion of your code to Swift 4 can also be done this way:
let str = "Hello, playground"
let index = str.index(of: ",")!
let substr = str.prefix(upTo: index)
You can use the code below to have a new string:
let newString = String(str.prefix(upTo: index))
substring(from: index) Converted to [index...]
Check the sample
let text = "1234567890"
let index = text.index(text.startIndex, offsetBy: 3)
text.substring(from: index) // "4567890" [Swift 3]
String(text[index...]) // "4567890" [Swift 4]
Some useful extensions:
extension String {
func substring(from: Int, to: Int) -> String {
let start = index(startIndex, offsetBy: from)
let end = index(start, offsetBy: to - from)
return String(self[start ..< end])
}
func substring(range: NSRange) -> String {
return substring(from: range.lowerBound, to: range.upperBound)
}
}
Example of uppercasedFirstCharacter convenience property in Swift3 and Swift4.
Property uppercasedFirstCharacterNew demonstrates how to use String slicing subscript in Swift4.
extension String {
public var uppercasedFirstCharacterOld: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = substring(to: splitIndex).uppercased()
let sentence = substring(from: splitIndex)
return firstCharacter + sentence
} else {
return self
}
}
public var uppercasedFirstCharacterNew: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = self[..<splitIndex].uppercased()
let sentence = self[splitIndex...]
return firstCharacter + sentence
} else {
return self
}
}
}
let lorem = "lorem".uppercasedFirstCharacterOld
print(lorem) // Prints "Lorem"
let ipsum = "ipsum".uppercasedFirstCharacterNew
print(ipsum) // Prints "Ipsum"
You can create your custom subString method using extension to class String as below:
extension String {
func subString(startIndex: Int, endIndex: Int) -> String {
let end = (endIndex - self.count) + 1
let indexStartOfText = self.index(self.startIndex, offsetBy: startIndex)
let indexEndOfText = self.index(self.endIndex, offsetBy: end)
let substring = self[indexStartOfText..<indexEndOfText]
return String(substring)
}
}
Creating SubString (prefix and suffix) from String using Swift 4:
let str : String = "ilike"
for i in 0...str.count {
let index = str.index(str.startIndex, offsetBy: i) // String.Index
let prefix = str[..<index] // String.SubSequence
let suffix = str[index...] // String.SubSequence
print("prefix \(prefix), suffix : \(suffix)")
}
Output
prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix :
If you want to generate a substring between 2 indices , use :
let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex
I have written a string extension for replacement of 'String: subString:'
extension String {
func sliceByCharacter(from: Character, to: Character) -> String? {
let fromIndex = self.index(self.index(of: from)!, offsetBy: 1)
let toIndex = self.index(self.index(of: to)!, offsetBy: -1)
return String(self[fromIndex...toIndex])
}
func sliceByString(from:String, to:String) -> String? {
//From - startIndex
var range = self.range(of: from)
let subString = String(self[range!.upperBound...])
//To - endIndex
range = subString.range(of: to)
return String(subString[..<range!.lowerBound])
}
}
Usage : "Date(1511508780012+0530)".sliceByString(from: "(", to: "+")
Example Result : "1511508780012"
PS: Optionals are forced to unwrap. Please add Type safety check wherever necessary.
If you are trying to just get a substring up to a specific character, you don't need to find the index first, you can just use the prefix(while:) method
let str = "Hello, playground"
let subString = str.prefix { $0 != "," } // "Hello" as a String.SubSequence
When programming I often have strings with just plain A-Za-z and 0-9. No need for difficult Index actions. This extension is based on the plain old left / mid / right functions.
extension String {
// LEFT
// Returns the specified number of chars from the left of the string
// let str = "Hello"
// print(str.left(3)) // Hel
func left(_ to: Int) -> String {
return "\(self[..<self.index(startIndex, offsetBy: to)])"
}
// RIGHT
// Returns the specified number of chars from the right of the string
// let str = "Hello"
// print(str.left(3)) // llo
func right(_ from: Int) -> String {
return "\(self[self.index(startIndex, offsetBy: self.length-from)...])"
}
// MID
// Returns the specified number of chars from the startpoint of the string
// let str = "Hello"
// print(str.left(2,amount: 2)) // ll
func mid(_ from: Int, amount: Int) -> String {
let x = "\(self[self.index(startIndex, offsetBy: from)...])"
return x.left(amount)
}
}
Hope this will help little more :-
var string = "123456789"
If you want a substring after some particular index.
var indexStart = string.index(after: string.startIndex )// you can use any index in place of startIndex
var strIndexStart = String (string[indexStart...])//23456789
If you want a substring after removing some string at the end.
var indexEnd = string.index(before: string.endIndex)
var strIndexEnd = String (string[..<indexEnd])//12345678
you can also create indexes with the following code :-
var indexWithOffset = string.index(string.startIndex, offsetBy: 4)
with this method you can get specific range of string.you need to pass start index and after that total number of characters you want.
extension String{
func substring(fromIndex : Int,count : Int) -> String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(self.startIndex, offsetBy: fromIndex + count)
let range = startIndex..<endIndex
return String(self[range])
}
}
This is my solution, no warning, no errors, but perfect
let redStr: String = String(trimmStr[String.Index.init(encodedOffset: 0)..<String.Index.init(encodedOffset: 2)])
let greenStr: String = String(trimmStr[String.Index.init(encodedOffset: 3)..<String.Index.init(encodedOffset: 4)])
let blueStr: String = String(trimmStr[String.Index.init(encodedOffset: 5)..<String.Index.init(encodedOffset: 6)])
var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))
You can try in this way and will get proper results.
the simples way that I use is :
String(Array(str)[2...4])
Swift 4, 5, 5+
Substring from Last
let str = "Hello World"
let removeFirstSix = String(str.dropFirst(6))
print(removeFirstSix) //World
Substring from First
let removeLastSix = String(str.dropLast(6))
print(removeLastSix) //Hello
Hope it would be helpful.
extension String {
func getSubString(_ char: Character) -> String {
var subString = ""
for eachChar in self {
if eachChar == char {
return subString
} else {
subString += String(eachChar)
}
}
return subString
}
}
let str: String = "Hello, playground"
print(str.getSubString(","))

In Swift, how to modify a character in string with subscript?

Like in C, we can simply do
str[i] = str[j]
But how to write the similar logic in swift?
Here is my code, but got error:
Cannot assign through subscript: subscript is get-only
let indexI = targetString.index(targetString.startIndex, offsetBy: i)
let indexJ = targetString.index(targetString.startIndex, offsetBy: j)
targetString[indexI] = targetString[indexJ]
I know it may work by using this method, but it's too inconvenient
replaceSubrange(, with: )
In C, a string (char *) can be treated as an array of characters. In Swift, you can convert the String to an [Character], do the modifications you want, and then convert the [Character] back to String.
For example:
let str = "hello"
var strchars = Array(str)
strchars[0] = strchars[4]
let str2 = String(strchars)
print(str2) // "oello"
This might seem like a lot of work for a single modification, but if you are moving many characters this way, you only have to convert once each direction.
Reverse a String
Here's an example of a simple algorithm to reverse a string. By converting to an array of characters first, this algorithm is similar to the way you might do it in C:
let str = "abcdefg"
var strchars = Array(str)
var start = 0
var end = strchars.count - 1
while start < end {
let temp = strchars[start]
strchars[start] = strchars[end]
strchars[end] = temp
start += 1
end -= 1
}
let str2 = String(strchars)
print(str2) // "gfedcba"
Dealing with String with Swift is major pain in the a**. Unlike most languages I know that treat string as an array of characters, Swift treats strings as collection of extended grapheme clusters and the APIs to access them is really clumsy. Changes are coming in Swift 4 but that manifesto lost me about 10 paragraphs in.
Back to your question... you can replace the character like this:
var targetString = "Hello world"
let i = 0
let j = 1
let indexI = targetString.index(targetString.startIndex, offsetBy: i)
let indexJ = targetString.index(targetString.startIndex, offsetBy: j)
targetString.replaceSubrange(indexI...indexI, with: targetString[indexJ...indexJ])
print(targetString) // eello world
I was quite shocked as well by the fact that swift makes string indexing so damn complicated. For that reason, I have built some string extensions that enable you to retrieve and change parts of strings based on indices, closed ranges, and open ranges, PartialRangeFrom, PartialRangeThrough, and PartialRangeUpTo. You can download the repository I created here
You can also pass in negative numbers in order to access characters from the end backwards.
public extension String {
/**
Enables passing in negative indices to access characters
starting from the end and going backwards.
if num is negative, then it is added to the
length of the string to retrieve the true index.
*/
func negativeIndex(_ num: Int) -> Int {
return num < 0 ? num + self.count : num
}
func strOpenRange(index i: Int) -> Range<String.Index> {
let j = negativeIndex(i)
return strOpenRange(j..<(j + 1), checkNegative: false)
}
func strOpenRange(
_ range: Range<Int>, checkNegative: Bool = true
) -> Range<String.Index> {
var lower = range.lowerBound
var upper = range.upperBound
if checkNegative {
lower = negativeIndex(lower)
upper = negativeIndex(upper)
}
let idx1 = index(self.startIndex, offsetBy: lower)
let idx2 = index(self.startIndex, offsetBy: upper)
return idx1..<idx2
}
func strClosedRange(
_ range: CountableClosedRange<Int>, checkNegative: Bool = true
) -> ClosedRange<String.Index> {
var lower = range.lowerBound
var upper = range.upperBound
if checkNegative {
lower = negativeIndex(lower)
upper = negativeIndex(upper)
}
let start = self.index(self.startIndex, offsetBy: lower)
let end = self.index(start, offsetBy: upper - lower)
return start...end
}
// MARK: - Subscripts
/**
Gets and sets a character at a given index.
Negative indices are added to the length so that
characters can be accessed from the end backwards
Usage: `string[n]`
*/
subscript(_ i: Int) -> String {
get {
return String(self[strOpenRange(index: i)])
}
set {
let range = strOpenRange(index: i)
replaceSubrange(range, with: newValue)
}
}
/**
Gets and sets characters in an open range.
Supports negative indexing.
Usage: `string[n..<n]`
*/
subscript(_ r: Range<Int>) -> String {
get {
return String(self[strOpenRange(r)])
}
set {
replaceSubrange(strOpenRange(r), with: newValue)
}
}
/**
Gets and sets characters in a closed range.
Supports negative indexing
Usage: `string[n...n]`
*/
subscript(_ r: CountableClosedRange<Int>) -> String {
get {
return String(self[strClosedRange(r)])
}
set {
replaceSubrange(strClosedRange(r), with: newValue)
}
}
/// `string[n...]`. See PartialRangeFrom
subscript(r: PartialRangeFrom<Int>) -> String {
get {
return String(self[strOpenRange(r.lowerBound..<self.count)])
}
set {
replaceSubrange(strOpenRange(r.lowerBound..<self.count), with: newValue)
}
}
/// `string[...n]`. See PartialRangeThrough
subscript(r: PartialRangeThrough<Int>) -> String {
get {
let upper = negativeIndex(r.upperBound)
return String(self[strClosedRange(0...upper, checkNegative: false)])
}
set {
let upper = negativeIndex(r.upperBound)
replaceSubrange(
strClosedRange(0...upper, checkNegative: false), with: newValue
)
}
}
/// `string[...<n]`. See PartialRangeUpTo
subscript(r: PartialRangeUpTo<Int>) -> String {
get {
let upper = negativeIndex(r.upperBound)
return String(self[strOpenRange(0..<upper, checkNegative: false)])
}
set {
let upper = negativeIndex(r.upperBound)
replaceSubrange(
strOpenRange(0..<upper, checkNegative: false), with: newValue
)
}
}
}
Usage:
let text = "012345"
print(text[2]) // "2"
print(text[-1] // "5"
print(text[1...3]) // "123"
print(text[2..<3]) // "2"
print(text[3...]) // "345"
print(text[...3]) // "0123"
print(text[..<3]) // "012"
print(text[(-3)...] // "345"
print(text[...(-2)] // "01234"
All of the above works with assignment as well. All subscripts have getters and setters.
a new extension added,
since String conforms to BidirectionalCollection Protocol
extension String{
subscript(at i: Int) -> String? {
get {
if i < count{
let idx = index(startIndex, offsetBy: i)
return String(self[idx])
}
else{
return nil
}
}
set {
if i < count{
let idx = index(startIndex, offsetBy: i)
remove(at: idx)
if let new = newValue, let first = new.first{
insert(first, at: idx)
}
}
}
}
}
call like this:
var str = "fighter"
str[at: 2] = "6"

Swift find all occurrences of a substring

I have an extension here of the String class in Swift that returns the index of the first letter of a given substring.
Can anybody please help me make it so it will return an array of all occurrences instead of just the first one?
Thank you.
extension String {
func indexOf(string : String) -> Int {
var index = -1
if let range = self.range(of : string) {
if !range.isEmpty {
index = distance(from : self.startIndex, to : range.lowerBound)
}
}
return index
}
}
For example instead of a return value of 50 I would like something like [50, 74, 91, 103]
You just keep advancing the search range until you can't find any more instances of the substring:
extension String {
func indicesOf(string: String) -> [Int] {
var indices = [Int]()
var searchStartIndex = self.startIndex
while searchStartIndex < self.endIndex,
let range = self.range(of: string, range: searchStartIndex..<self.endIndex),
!range.isEmpty
{
let index = distance(from: self.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
}
let keyword = "a"
let html = "aaaa"
let indicies = html.indicesOf(string: keyword)
print(indicies) // [0, 1, 2, 3]
I know we aren't playing code golf here, but for anyone interested in a functional style one-line implementation that doesn't use vars or loops, this is another possible solution:
extension String {
func indices(of string: String) -> [Int] {
return indices.reduce([]) { $1.encodedOffset > ($0.last ?? -1) && self[$1...].hasPrefix(string) ? $0 + [$1.encodedOffset] : $0 }
}
}
Here are 2 functions. One returns [Range<String.Index>], the other returns [Range<Int>]. If you don't need the former, you can make it private. I've designed it to mimic the range(of:options:range:locale:) method, so it supports all the same features.
import Foundation
extension String {
public func allRanges(
of aString: String,
options: String.CompareOptions = [],
range: Range<String.Index>? = nil,
locale: Locale? = nil
) -> [Range<String.Index>] {
// the slice within which to search
let slice = (range == nil) ? self[...] : self[range!]
var previousEnd = s.startIndex
var ranges = [Range<String.Index>]()
while let r = slice.range(
of: aString, options: options,
range: previousEnd ..< s.endIndex,
locale: locale
) {
if previousEnd != self.endIndex { // don't increment past the end
previousEnd = self.index(after: r.lowerBound)
}
ranges.append(r)
}
return ranges
}
public func allRanges(
of aString: String,
options: String.CompareOptions = [],
range: Range<String.Index>? = nil,
locale: Locale? = nil
) -> [Range<Int>] {
return allRanges(of: aString, options: options, range: range, locale: locale)
.map(indexRangeToIntRange)
}
private func indexRangeToIntRange(_ range: Range<String.Index>) -> Range<Int> {
return indexToInt(range.lowerBound) ..< indexToInt(range.upperBound)
}
private func indexToInt(_ index: String.Index) -> Int {
return self.distance(from: self.startIndex, to: index)
}
}
let s = "abc abc abc abc abc"
print(s.allRanges(of: "abc") as [Range<String.Index>])
print()
print(s.allRanges(of: "abc") as [Range<Int>])
There's not really a built-in function to do this, but we can implement a modified Knuth-Morris-Pratt algorithm to get all the indices of the string we want to match. It should also be very performant as we don't need to repeatedly call range on the string.
extension String {
func indicesOf(string: String) -> [Int] {
// Converting to an array of utf8 characters makes indicing and comparing a lot easier
let search = self.utf8.map { $0 }
let word = string.utf8.map { $0 }
var indices = [Int]()
// m - the beginning of the current match in the search string
// i - the position of the current character in the string we're trying to match
var m = 0, i = 0
while m + i < search.count {
if word[i] == search[m+i] {
if i == word.count - 1 {
indices.append(m)
m += i + 1
i = 0
} else {
i += 1
}
} else {
m += 1
i = 0
}
}
return indices
}
}
Please check the following answer for finding multiple items in multiple locations
func indicesOf(string: String) -> [Int] {
var indices = [Int]()
var searchStartIndex = self.startIndex
while searchStartIndex < self.endIndex,
let range = self.range(of: string, range: searchStartIndex..<self.endIndex),
!range.isEmpty
{
let index = distance(from: self.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
func attributedStringWithColor(_ strings: [String], color: UIColor, characterSpacing: UInt? = nil) -> NSAttributedString {
let attributedString = NSMutableAttributedString(string: self)
for string in strings {
let indexes = self.indicesOf(string: string)
for index in indexes {
let range = NSRange(location: index, length: string.count)
attributedString.addAttribute(NSAttributedString.Key.foregroundColor, value: color, range: range)
}
}
guard let characterSpacing = characterSpacing else {return attributedString}
attributedString.addAttribute(NSAttributedString.Key.kern, value: characterSpacing, range: NSRange(location: 0, length: attributedString.length))
return attributedString
}
can be used as follows :
let message = "Item 1 + Item 2 + Item 3"
message.attributedStringWithColor(["Item", "+"], color: UIColor.red)
and gets the result
This could be done with recursive method. I used a numeric string to test it. It returns an optional array of Int, meaning it will be nil if no substring can be found.
extension String {
func indexes(of string: String, offset: Int = 0) -> [Int]? {
if let range = self.range(of : string) {
if !range.isEmpty {
let index = distance(from : self.startIndex, to : range.lowerBound) + offset
var result = [index]
let substr = self.substring(from: range.upperBound)
if let substrIndexes = substr.indexes(of: string, offset: index + distance(from: range.lowerBound, to: range.upperBound)) {
result.append(contentsOf: substrIndexes)
}
return result
}
}
return nil
}
}
let numericString = "01234567890123456789012345678901234567890123456789012345678901234567890123456789"
numericString.indexes(of: "3456")
I have tweaked the accepted answer so that case sensitivity can be configured
extension String {
func allIndexes(of subString: String, caseSensitive: Bool = true) -> [Int] {
let subString = caseSensitive ? subString : subString.lowercased()
let mainString = caseSensitive ? self : self.lowercased()
var indices = [Int]()
var searchStartIndex = mainString.startIndex
while searchStartIndex < mainString.endIndex,
let range = mainString.range(of: subString, range: searchStartIndex..<mainString.endIndex),
!range.isEmpty
{
let index = distance(from: mainString.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
}