I am using the below extension in swift 3, but I know it is incorrect because I still get errors. The point of the extension is to get a range of a string using the syntax string[0...2] That will return the string from 0. This is the error I am getting
extension String {
subscript (i: Int) -> Character {
return self[self.characters.index(self.startIndex, offsetBy: i)]
}
subscript (i: Int) -> String {
return String(self[i] as Character)
}
subscript (r: Range<Int>) -> String {
let start = characters.index(startIndex, offsetBy: r.lowerBound)
let end = characters.index(start, offsetBy: r.upperBound - r.lowerBound)
return self[(start ..< end)]
}
}
In Swift 3, the closed range operator ... generates CountableClosedRange for Ints. So you need to define one more overload for subscript.
subscript (r: CountableClosedRange<Int>) -> String {
let start = characters.index(startIndex, offsetBy: r.lowerBound)
let end = characters.index(start, offsetBy: r.upperBound - r.lowerBound)
return self[(start ... end)]
}
Updated in Swift 4:
extension String{
subscript (r: CountableClosedRange<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(start, offsetBy: r.upperBound - r.lowerBound)
return String(self[start...end])
}
}
This question already has an answer here:
how to find the index of a character in a string from specific position
(1 answer)
Closed 6 years ago.
How can I find the first position of a character in a substring. Not in the string overall, but the first after a specified character position.
Example:
var str = "This is a test string"
//find the position of first "i" after "is"
let position = str.firstPositionOfIAfterPosition(5) // returns 18
I know I can find the overall first position with code below. How can I extend this to start looking only after a specified character position?
let position = str.rangeOfString("i").startIndex
var s = "This is a test string"
var targetRange = s.characters.indices
targetRange.startIndex = targetRange.startIndex.advancedBy(6) // skip past
let r = s.rangeOfString("i", options: [], range: targetRange, locale: nil)
// 18..<19
var str = "This is a test string"
func getIndexAfterString(string: String) -> Int {
let firstIndex = str.rangeOfString(string)?.startIndex.advancedBy(string.characters.count)
let i: Int = str.startIndex.distanceTo(firstIndex!)
let secondIndex = str.substringFromIndex(firstIndex!).rangeOfString("i")?.startIndex
let j: Int = str.startIndex.distanceTo(secondIndex!)
return i + j
}
let index: Int = getIndexAfterString(" is ") //18
Similar to matt's answer, but as String extension and with error handling
extension String {
func firstPositionOf(string: String, afterPosition index: Int) -> String.Index?
{
if self.isEmpty || self.characters.count - 1 < index { return nil }
let subRange = Range<String.Index>(self.startIndex.advancedBy(index + 1)..<self.endIndex)
guard let foundRange = self.rangeOfString(string, options: [], range: subRange) else { return nil }
return foundRange.startIndex
}
}
let str = "This is a test string"
let position = str.firstPositionOf("i", afterPosition:5) // -> 18
I don't understand what to do with the issue reported by the compiler. I tried to create a Range, but it says Index is not known:
//let range = matches.first!.range.location
let range = Range(
start:matches.first!.range.location,
end: matches.first!.range.location+matches.first!.range.length
)
id = text[range]
var t = text
t.removeRange(range)
return t
Compiler says: Cannot invoke 'removeRange' with an argument list of type '(Range)' on t.removeRange(range).
I'm pretty sure it's evident, but I lost a great deal of time on such a small issue… any help highly appreciated!
As your error says that:
Cannot invoke 'removeRange' with an argument list of type '(Range)'
Means there is a problem with your range instance type and removeRange function will only accept an argument with type Range<String.Index> and its syntax is :
/// Remove the indicated `subRange` of characters
///
/// Invalidates all indices with respect to `self`.
///
/// Complexity: O(\ `count(self)`\ ).
mutating func removeRange(subRange: Range<String.Index>)
And here is working example with removeRange:
var welcome = "hello there"
let range = advance(welcome.endIndex, -6)..<welcome.endIndex
welcome.removeRange(range)
println(welcome) //hello
Hope this will help.
Swift 2.2 example of removing first 4 characters:
let range = text.startIndex..<text.startIndex.advancedBy(4)
text.removeRange(range)
That first line feels verbose. I hope newer Swift versions improve upon it.
Here is the working equivalent snippet:
static func unitTest() {
let text = "a👿bbbbb🇩🇪c"
let tag = Tag(id: "🇩🇪")
tag.regex = "👿b+"
print ("Unit test tag.foundIn(\(text)) ? = \(tag.foundIn(text))")
}
func foundIn(text: String) -> (id:String, remainingText:String)? {
// if a regex is provided, use it to capture, and keep the capture as a tag ID
if let regex = regex {
let r = Regex(regex) // text =~ regex
let matches = r.matches(text)
if matches.count >= 1 {
let first = matches.first!.range
let start = advance(text.startIndex, first.location)
let end = advance(start, first.length-1)
let range = Range(start: start, end: end)
id = text[range]
var t = text
t.removeRange(range)
return (id, t)
}
return nil
}
else if let range = text.rangeOfString(id) {
var t = text
t.removeRange(range)
return (id, t)
}
else {
return nil
}
}
The unit test returns :
Unit test tag.foundIn(a👿bbbbb🇩🇪c) ? = Optional(("👿bbbbb", "a🇩🇪c"))
I need an implementation of lastIndexOf that is as fast as possible.
I am finding that the String advance function is extremely slow.
I tried using the c function strrchr, and tried copying the string to NSData and using pointers but I can't get the syntax right.
My string will always have 1 byte characters and the string i'm searching for "|" is always 1 byte also.
Any implementation using advance will be too slow but here is the fastest example I could find:
func indexOf(target: String, startIndex: Int) -> Int
{
var startRange = advance(self.startIndex, startIndex)
var range = self.rangeOfString(target, options: NSStringCompareOptions.LiteralSearch, range: Range<String.Index>(start: startRange, end: self.endIndex))
if let range = range {
return distance(self.startIndex, range.startIndex)
} else {
return -1
}
}
func lastIndexOf(target: String) -> Int
{
var index = -1
var stepIndex = self.indexOf(target)
while stepIndex > -1
{
index = stepIndex
if stepIndex + target.length < self.length
{
stepIndex = indexOf(target, startIndex: stepIndex + target.length)
}
else
{
stepIndex = -1
}
}
return index
}
This is an example of the string I need to parse.
var str:String = "4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|Room1|2|72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4sageReceived:4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|Room1|2|72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|Room1|2|72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4TCPListener.onReceived: 4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|6|0|3259744|6352141|46|14|1|0|7|7|3259744|6352141|4|1|0|8|8|3259744|6352141|4|0|22|9|0|3259744|6352141|2|3|Room1|2|72|86330534|1|0|10|9|3259744|6352141|4|1|0|11|10|3259744|6352141|4|1|0|12|11|3259744|6352141|4|1|0|13|12|3259744|6352141|4|0|4|14|0|3259744|6352141|46|24|0|5|15|0|3259744|6352141|46|654|0|66|0|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.2212834|0|4|16|0|3259744|6352141|46|4preParse
4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|Mc02|efland,nc|36.027992|-79.221283"
Here is a Swift 2.0 Answer
func lastIndexOf(s: String) -> Int? {
if let r: Range<Index> = self.rangeOfString(s, options: .BackwardsSearch) {
return self.startIndex.distanceTo(r.startIndex)
}
return Optional<Int>()
}
Tests
func testStringLastIndexOf() {
let lastIndex = "0|2|45|7|9".lastIndexOf("|")
XCTAssertEqual(lastIndex, 8)
}
func testStringLastIndexOfNotFound() {
let lastIndex = "0123456789".lastIndexOf("|")
XCTAssertEqual(lastIndex, nil);
}
You can use strrchr in Swift
import Darwin
let str = "4|0|66|5|0|3259744|6352141|1|3259744"
func stringLastIndexOf(src:String, target:UnicodeScalar) -> Int? {
let c = Int32(bitPattern: target.value)
return src.withCString { s -> Int? in
let pos = strrchr(s, c)
return pos != nil ? pos - s : nil
}
}
stringLastIndexOf(str, "|") // -> {Some 28}
stringLastIndexOf(str, ",") // -> nil
You can use Objective C files in a Swift project; in these you can use plain C code and make a function which uses strrchr. Then you can call this from Swift.
If you do this in order to get all substring delimited by "|", you might test this approach:
import Foundation
let s = "4|0|66|5|0|3259744|6352141|1|3259744|WSMxt208L54yZ5irtHC3|..."
let a = s.componentsSeparatedByString("|")
The built in functions are sometimes very fast and you may be getting the required performance even by using String.
If you really need to get only the position of the last "|", you could work with utf16 representation, where advancing over the characters should be faster.
I think this should work:
let utf16String = s.utf16
var i = s.utf16Count - 1
while i >= 0 {
if utf16String[i] == 124 {
break
}
i--
}
println(i)
If the characters are guaranteed as single byte, the data is huge and performance is critical then it may be worth converting to an array of bytes (UInt8) and perform the operations directly on them. You can then convert the part that you need back to a String.
Also note that Optimised builds may be much faster than Debug builds so you should do any performance testing with the optimiser on. It may also be worth checking that the optimised versions are too slow at the moment.
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)
}