So I got stuck on a coding challenge that I almost knew the answer too. And I think I have to use the subString call in Swift 4 to get it 100%. I want to reverse every OTHER word in a string, but ignore or keep the punctuation in its original place( index ).
var sample = "lets start. And not worry about proper sentences."
func reverseString(inputString: String) -> String {
let oldSentence = sample.components(separatedBy: " ")
var newSentence = ""
for index in 0...oldSentence.count - 1 {
let word = oldSentence[index]
if newSentence != "" {
newSentence += " "
}
if index % 2 == 1 {
let reverseWord = String(word.reversed())
newSentence += reverseWord
} else {
newSentence += word
}
}
return newSentence
}
reverseString(inputString: sample)
And this would be the expected output.
"lets trats. And ton worry tuoba proper secnetnes."
Notice the punctuation is not reversed.
You shouldn't use components(separatedBy: ) to split a string in words. See this article for the reason. Use enumerateSubstrings and pass in the appropriate option:
func reverseString(inputString: String) -> String {
var index = 1
var newSentence = inputString
inputString.enumerateSubstrings(in: inputString.startIndex..., options: .byWords) { substr, range, _, stop in
guard let substr = substr else { return }
if index % 2 == 0 {
newSentence = newSentence.replacingCharacters(in: range, with: String(substr.reversed()))
}
index += 1
}
return newSentence
}
print(reverseString(inputString: "lets start. And not worry about proper sentences."))
// lets trats. And ton worry tuoba proper secnetnes.
print(reverseString(inputString: "I think, therefore I'm"))
// I kniht, therefore m'I
I am using the following code to get a String substring from an NSRange:
func substring(with nsrange: NSRange) -> String? {
guard let range = Range.init(nsrange)
else { return nil }
let start = UTF16Index(range.lowerBound)
let end = UTF16Index(range.upperBound)
return String(utf16[start..<end])
}
(via: https://mjtsai.com/blog/2016/12/19/nsregularexpression-and-swift/)
When I compile with Swift 4 (Xcode 9b4), I get the following errors for the two lines that declare start and end:
'init' is unavailable
'init' was obsoleted in Swift 4.0
I am confused, since I am not using an init.
How can I fix this?
Use Range(_, in:) to convert an NSRange to a Range in Swift 4.
extension String {
func substring(with nsrange: NSRange) -> Substring? {
guard let range = Range(nsrange, in: self) else { return nil }
return self[range]
}
}
With Swift 4 we can get substrings this way.
Substring from index
let originStr = "Test"
let offset = 1
let str = String(originStr.suffix(from: String.Index.init(encodedOffset: offset)))
Substring to index
let originStr = "Test"
let offset = 1
String(self.prefix(index))
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"))
That is what i am trying to do:
var i = 0
var string = "abcdef"
for value in string
{
value.[Put value of variable i here] = "a"
i++
}
How can i insert the value of i in the code?
Easiest is probably just convert it to an NSMutableString:
let string = "abcdef".mutableCopy() as NSMutableString
println( "\(string)")
for var i = 0; i < string.length; ++i {
string.replaceCharactersInRange(NSMakeRange(i, 1), withString: "a")
}
println( "\(string)")
Yes, it's a bit ugly but it works.
A much cleaner way is to use Swifts map function:
var string = "abcdef"
let result = map(string) { (c) -> Character in
"a"
}
println("\(result)") // aaaaaa
You should just be able to use the following but this doesn't compile:
map(string) { "a" }
In you comments you mention you want to split up the string on a space, you can just use this for that:
let stringWithSpace = "abcdef 012345"
let splitString = stringWithSpace.componentsSeparatedByString(" ")
println("\(splitString[0])") // abcdef
println("\(splitString[1])") // 012345
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)
}