Finding index of character in Swift String
StringSwiftString Problem Overview
It's time to admit defeat...
In Objective-C, I could use something like:
NSString* str = @"abcdefghi";
[str rangeOfString:@"c"].location; // 2
In Swift, I see something similar:
var str = "abcdefghi"
str.rangeOfString("c").startIndex
...but that just gives me a String.Index, which I can use to subscript back into the original string, but not extract a location from.
FWIW, that String.Index has a private ivar called _position that has the correct value in it. I just don't see how it's exposed.
I know I could easily add this to String myself. I'm more curious about what I'm missing in this new API.
String Solutions
Solution 1 - String
You are not the only one who couldn't find the solution.
String doesn't implement RandomAccessIndexType. Probably because they enable characters with different byte lengths. That's why we have to use string.characters.count (count or countElements in Swift 1.x) to get the number of characters. That also applies to positions. The _position is probably an index into the raw array of bytes and they don't want to expose that. The String.Index is meant to protect us from accessing bytes in the middle of characters.
That means that any index you get must be created from String.startIndex or String.endIndex (String.Index implements BidirectionalIndexType). Any other indices can be created using successor or predecessor methods.
Now to help us with indices, there is a set of methods (functions in Swift 1.x):
Swift 4.x
let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.index(text.startIndex, offsetBy: 2)
let lastChar2 = text[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)
Swift 3.0
let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.characters.index(text.characters.startIndex, offsetBy: 2)
let lastChar2 = text.characters[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)
Swift 2.x
let text = "abc"
let index2 = text.startIndex.advancedBy(2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let lastChar2 = text.characters[index2] //will do the same as above
let range: Range<String.Index> = text.rangeOfString("b")!
let index: Int = text.startIndex.distanceTo(range.startIndex) //will call successor/predecessor several times until the indices match
Swift 1.x
let text = "abc"
let index2 = advance(text.startIndex, 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let range = text.rangeOfString("b")
let index: Int = distance(text.startIndex, range.startIndex) //will call succ/pred several times
Working with String.Index is cumbersome but using a wrapper to index by integers (see https://stackoverflow.com/a/25152652/669586) is dangerous because it hides the inefficiency of real indexing.
Note that Swift indexing implementation has the problem that indices/ranges created for one string cannot be reliably used for a different string, for example:
Swift 2.x
let text: String = "abc"
let text2: String = "πΎππ"
let range = text.rangeOfString("b")!
//can randomly return a bad substring or throw an exception
let substring: String = text2[range]
//the correct solution
let intIndex: Int = text.startIndex.distanceTo(range.startIndex)
let startIndex2 = text2.startIndex.advancedBy(intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]
Swift 1.x
let text: String = "abc"
let text2: String = "πΎππ"
let range = text.rangeOfString("b")
//can randomly return nil or a bad substring
let substring: String = text2[range]
//the correct solution
let intIndex: Int = distance(text.startIndex, range.startIndex)
let startIndex2 = advance(text2.startIndex, intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]
Solution 2 - String
Swift 3.0 makes this a bit more verbose:
let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.index(of: needle) {
let pos = string.characters.distance(from: string.startIndex, to: idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
Extension:
extension String {
public func index(of char: Character) -> Int? {
if let idx = characters.index(of: char) {
return characters.distance(from: startIndex, to: idx)
}
return nil
}
}
In Swift 2.0 this has become easier:
let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.indexOf(needle) {
let pos = string.startIndex.distanceTo(idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
Extension:
extension String {
public func indexOfCharacter(char: Character) -> Int? {
if let idx = self.characters.indexOf(char) {
return self.startIndex.distanceTo(idx)
}
return nil
}
}
Swift 1.x implementation:
For a pure Swift solution one can use:
let string = "Hello.World"
let needle: Character = "."
if let idx = find(string, needle) {
let pos = distance(string.startIndex, idx)
println("Found \(needle) at position \(pos)")
}
else {
println("Not found")
}
As an extension to String:
extension String {
public func indexOfCharacter(char: Character) -> Int? {
if let idx = find(self, char) {
return distance(self.startIndex, idx)
}
return nil
}
}
Solution 3 - String
extension String {
// MARK: - sub String
func substringToIndex(index:Int) -> String {
return self.substringToIndex(advance(self.startIndex, index))
}
func substringFromIndex(index:Int) -> String {
return self.substringFromIndex(advance(self.startIndex, index))
}
func substringWithRange(range:Range<Int>) -> String {
let start = advance(self.startIndex, range.startIndex)
let end = advance(self.startIndex, range.endIndex)
return self.substringWithRange(start..<end)
}
subscript(index:Int) -> Character{
return self[advance(self.startIndex, index)]
}
subscript(range:Range<Int>) -> String {
let start = advance(self.startIndex, range.startIndex)
let end = advance(self.startIndex, range.endIndex)
return self[start..<end]
}
// MARK: - replace
func replaceCharactersInRange(range:Range<Int>, withString: String!) -> String {
var result:NSMutableString = NSMutableString(string: self)
result.replaceCharactersInRange(NSRange(range), withString: withString)
return result
}
}
Solution 4 - String
Swift 5.0
public extension String {
func indexInt(of char: Character) -> Int? {
return firstIndex(of: char)?.utf16Offset(in: self)
}
}
Swift 4.0
public extension String {
func indexInt(of char: Character) -> Int? {
return index(of: char)?.encodedOffset
}
}
Solution 5 - String
I have found this solution for swift2:
var str = "abcdefghi"
let indexForCharacterInString = str.characters.indexOf("c") //returns 2
Solution 6 - String
I'm not sure how to extract the position from String.Index, but if you're willing to fall back on some Objective-C frameworks, you can bridge to objective-c and do it the same way you used to.
"abcdefghi".bridgeToObjectiveC().rangeOfString("c").location
It seems like some NSString methods haven't yet been (or maybe won't be) ported to String. Contains also comes to mind.
Solution 7 - String
Here is a clean String extention that answers the question:
Swift 3:
extension String {
var length:Int {
return self.characters.count
}
func indexOf(target: String) -> Int? {
let range = (self as NSString).range(of: target)
guard range.toRange() != nil else {
return nil
}
return range.location
}
func lastIndexOf(target: String) -> Int? {
let range = (self as NSString).range(of: target, options: NSString.CompareOptions.backwards)
guard range.toRange() != nil else {
return nil
}
return self.length - range.location - 1
}
func contains(s: String) -> Bool {
return (self.range(of: s) != nil) ? true : false
}
}
Swift 2.2:
extension String {
var length:Int {
return self.characters.count
}
func indexOf(target: String) -> Int? {
let range = (self as NSString).rangeOfString(target)
guard range.toRange() != nil else {
return nil
}
return range.location
}
func lastIndexOf(target: String) -> Int? {
let range = (self as NSString).rangeOfString(target, options: NSStringCompareOptions.BackwardsSearch)
guard range.toRange() != nil else {
return nil
}
return self.length - range.location - 1
}
func contains(s: String) -> Bool {
return (self.rangeOfString(s) != nil) ? true : false
}
}
Solution 8 - String
You can also find indexes of a character in a single string like this,
extension String {
func indexes(of character: String) -> [Int] {
precondition(character.count == 1, "Must be single character")
return self.enumerated().reduce([]) { partial, element in
if String(element.element) == character {
return partial + [element.offset]
}
return partial
}
}
}
Which gives the result in [String.Distance] ie. [Int], like
"apple".indexes(of: "p") // [1, 2]
"element".indexes(of: "e") // [0, 2, 4]
"swift".indexes(of: "j") // []
Solution 9 - String
If you want to use familiar NSString, you can declare it explicitly:
var someString: NSString = "abcdefghi"
var someRange: NSRange = someString.rangeOfString("c")
I'm not sure yet how to do this in Swift.
Solution 10 - String
Swift 5
Find index of substring
let str = "abcdecd"
if let range: Range<String.Index> = str.range(of: "cd") {
let index: Int = str.distance(from: str.startIndex, to: range.lowerBound)
print("index: ", index) //index: 2
}
else {
print("substring not found")
}
Find index of Character
let str = "abcdecd"
if let firstIndex = str.firstIndex(of: "c") {
let index: Int = str.distance(from: str.startIndex, to: firstIndex)
print("index: ", index) //index: 2
}
else {
print("symbol not found")
}
Solution 11 - String
If you want to know the position of a character in a string as an int value use this:
let loc = newString.range(of: ".").location
Solution 12 - String
This worked for me,
var loc = "abcdefghi".rangeOfString("c").location
NSLog("%d", loc);
this worked too,
var myRange: NSRange = "abcdefghi".rangeOfString("c")
var loc = myRange.location
NSLog("%d", loc);
Solution 13 - String
I know this is old and an answer has been accepted, but you can find the index of the string in a couple lines of code using:
var str : String = "abcdefghi"
let characterToFind: Character = "c"
let characterIndex = find(str, characterToFind) //returns 2
Some other great information about Swift strings here [Strings in Swift][1]
[1]: http://oleb.net/blog/2014/07/swift-strings/ "Strings in Swift"
Solution 14 - String
Variable type String in Swift contains different functions compared to NSString in Objective-C . And as Sulthan mentioned,
> Swift String doesn't implement RandomAccessIndex
What you can do is downcast your variable of type String to NSString (this is valid in Swift). This will give you access to the functions in NSString.
var str = "abcdefghi" as NSString
str.rangeOfString("c").locationx // returns 2
Solution 15 - String
If you think about it, you actually don't really need the exact Int version of the location. The Range or even the String.Index is enough to get the substring out again if needed:
let myString = "hello"
let rangeOfE = myString.rangeOfString("e")
if let rangeOfE = rangeOfE {
myString.substringWithRange(rangeOfE) // e
myString[rangeOfE] // e
// if you do want to create your own range
// you can keep the index as a String.Index type
let index = rangeOfE.startIndex
myString.substringWithRange(Range<String.Index>(start: index, end: advance(index, 1))) // e
// if you really really need the
// Int version of the index:
let numericIndex = distance(index, advance(index, 1)) // 1 (type Int)
}
Solution 16 - String
The Simplest Way is:
In Swift 3:
var textViewString:String = "HelloWorld2016"
guard let index = textViewString.characters.index(of: "W") else { return }
let mentionPosition = textViewString.distance(from: index, to: textViewString.endIndex)
print(mentionPosition)
Solution 17 - String
String is a bridge type for NSString, so add
import Cocoa
to your swift file and use all the "old" methods.
Solution 18 - String
In terms of thinking this might be called an INVERSION. You discover the world is round instead of flat. "You don't really need to know the INDEX of the character to do things with it." And as a C programmer I found that hard to take too! Your line "let index = letters.characters.indexOf("c")!" is enough by itself. For example to remove the c you could use...(playground paste in)
var letters = "abcdefg"
//let index = letters.rangeOfString("c")!.startIndex //is the same as
let index = letters.characters.indexOf("c")!
range = letters.characters.indexOf("c")!...letters.characters.indexOf("c")!
letters.removeRange(range)
letters
However, if you want an index you need to return an actual INDEX not an Int as an Int value would require additional steps for any practical use. These extensions return an index, a count of a specific character, and a range which this playground plug-in-able code will demonstrate.
extension String
{
public func firstIndexOfCharacter(aCharacter: Character) -> String.CharacterView.Index? {
for index in self.characters.indices {
if self[index] == aCharacter {
return index
}
}
return nil
}
public func returnCountOfThisCharacterInString(aCharacter: Character) -> Int? {
var count = 0
for letters in self.characters{
if aCharacter == letters{
count++
}
}
return count
}
public func rangeToCharacterFromStart(aCharacter: Character) -> Range<Index>? {
for index in self.characters.indices {
if self[index] == aCharacter {
let range = self.startIndex...index
return range
}
}
return nil
}
}
var MyLittleString = "MyVery:important String"
var theIndex = MyLittleString.firstIndexOfCharacter(":")
var countOfColons = MyLittleString.returnCountOfThisCharacterInString(":")
var theCharacterAtIndex:Character = MyLittleString[theIndex!]
var theRange = MyLittleString.rangeToCharacterFromStart(":")
MyLittleString.removeRange(theRange!)
Solution 19 - String
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
}
Solution 20 - String
extension String {
//Fucntion to get the index of a particular string
func index(of target: String) -> Int? {
if let range = self.range(of: target) {
return characters.distance(from: startIndex, to: range.lowerBound)
} else {
return nil
}
}
//Fucntion to get the last index of occurence of a given string
func lastIndex(of target: String) -> Int? {
if let range = self.range(of: target, options: .backwards) {
return characters.distance(from: startIndex, to: range.lowerBound)
} else {
return nil
}
}
}
Solution 21 - String
You can find the index number of a character in a string with this:
var str = "abcdefghi"
if let index = str.firstIndex(of: "c") {
let distance = str.distance(from: str.startIndex, to: index)
// distance is 2
}
Solution 22 - String
If you are looking for easy way to get index of Character or String checkout this library http://www.dollarswift.org/#indexof-char-character-int
You can get the indexOf from a string using another string as well or regex pattern
Solution 23 - String
To get index of a substring in a string with Swift 2:
let text = "abc"
if let range = text.rangeOfString("b") {
var index: Int = text.startIndex.distanceTo(range.startIndex)
...
}
Solution 24 - String
In swift 2.0
var stringMe="Something In this.World"
var needle="."
if let idx = stringMe.characters.indexOf(needle) {
let pos=stringMe.substringFromIndex(idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
Solution 25 - String
In Swift 2.0, the following function returns a substring before a given character.
func substring(before sub: String) -> String {
if let range = self.rangeOfString(sub),
let index: Int = self.startIndex.distanceTo(range.startIndex) {
return sub_range(0, index)
}
return ""
}
Solution 26 - String
let mystring:String = "indeep";
let findCharacter:Character = "d";
if (mystring.characters.contains(findCharacter))
{
let position = mystring.characters.indexOf(findCharacter);
NSLog("Position of c is \(mystring.startIndex.distanceTo(position!))")
}
else
{
NSLog("Position of c is not found");
}
Solution 27 - String
I play with following
extension String {
func allCharactes() -> [Character] {
var result: [Character] = []
for c in self.characters {
result.append(c)
}
return
}
}
until I understand the provided one's now it's just Character array
and with
let c = Array(str.characters)
Solution 28 - String
If you only need the index of a character the most simple, quick solution (as already pointed out by Pascal) is:
let index = string.characters.index(of: ".")
let intIndex = string.distance(from: string.startIndex, to: index)
Solution 29 - String
On the subject of turning a String.Index into an Int, this extension works for me:
public extension Int {
/// Creates an `Int` from a given index in a given string
///
/// - Parameters:
/// - index: The index to convert to an `Int`
/// - string: The string from which `index` came
init(_ index: String.Index, in string: String) {
self.init(string.distance(from: string.startIndex, to: index))
}
}
Example usage relevant to this question:
var testString = "abcdefg"
Int(testString.range(of: "c")!.lowerBound, in: testString) // 2
testString = "π¨π¦πΊπΈπ©πͺπ©βπ©βπ§βπ¦\u{1112}\u{1161}\u{11AB}"
Int(testString.range(of: "π¨π¦πΊπΈπ©πͺ")!.lowerBound, in: testString) // 0
Int(testString.range(of: "π©βπ©βπ§βπ¦")!.lowerBound, in: testString) // 1
Int(testString.range(of: "αα
‘α«")!.lowerBound, in: testString) // 5
Important:
As you can tell, it groups extended grapheme clusters and joined characters differently than String.Index. Of course, this is why we have String.Index. You should keep in mind that this method considers clusters to be singular characters, which is closer to correct. If your goal is to split a string by Unicode codepoint, this is not the solution for you.
Solution 30 - String
As my perspective, The better way with knowing the logic itself is below
let testStr: String = "I love my family if you Love us to tell us I'm with you"
var newStr = ""
let char:Character = "i"
for value in testStr {
if value == char {
newStr = newStr + String(value)
}
}
print(newStr.count)
Solution 31 - String
extension String{
func contains(find: String)->Bool{
return self.range(of: find) != nil
}
}
func check(n:String, h:String)->Int{
let n1 = n.lowercased()
let h1 = h.lowercased()//lowercase to make string case insensitive
var pos = 0 //postion of substring
if h1.contains(n1){
// checking if sub string exists
if let idx = h1.firstIndex(of:n1.first!){
let pos1 = h1.distance(from: h1.startIndex, to: idx)
pos = pos1
}
return pos
}
else{
return -1
}
}
print(check(n:"@", h:"hithisispushker,he is 99 a good Boy"))//put substring in n: and string in h
Solution 32 - String
Swift 3
extension String {
func substring(from:String) -> String
{
let searchingString = from
let rangeOfSearchingString = self.range(of: searchingString)!
let indexOfSearchingString: Int = self.distance(from: self.startIndex, to: rangeOfSearchingString.upperBound )
let trimmedString = self.substring(start: indexOfSearchingString , end: self.count)
return trimmedString
}
}
Solution 33 - String
// Using Swift 4, the code below works.
// The problem is that String.index is a struct. Use dot notation to grab the integer part of it that you want: ".encodedOffset"
let strx = "0123456789ABCDEF"
let si = strx.index(of: "A")
let i = si?.encodedOffset // i will be an Int. You need "?" because it might be nil, no such character found.
if i != nil { // You MUST deal with the optional, unwrap it only if not nil.
print("i = ",i)
print("i = ",i!) // "!" str1ps off "optional" specification (unwraps i).
// or
let ii = i!
print("ii = ",ii)
}
// Good luck.