This question already has answers here:
Split a String into an array in Swift?
(40 answers)
Closed 6 years ago.
For example i have usual fast enumeration
for (var myChar:Character) in "Hello World!"
{//code }
This works fine and i can do whatever i want with each character of this string.
But what if i want to use string instead of character, like this
for ( var myStr : String) in "Hello World!"//this is error
{
switch myStr
{
case "Hello":
//code
case "World!":
//code
default:
break
}
}
Is it possible to implement this? Thanks
You can use componentsSeparatedByCharactersInSet to divide the string into an array of strings:
import Foundation // not necessary if you import UIKit or Cocoa
for myStr in "Hello World!".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString:" ")) {
switch myStr {
case "Hello":
println("well hello there")
case "World!":
println("the whole world")
default:
println("something else")
}
}
In place of NSCharacterSet(charactersInString:" ") can also use NSCharacterSet.whitespaceCharacterSet() or NSCharacterSet.whitespaceAndnewlineCharacterSet() depending on what kinds of characters divide your words.
If your words are separated by a single space, you can use the shorter:
for myStr in "Hello World!".componentsSeparatedByString(" ") {
...
}
You cannot just enumeratate by strings over a string. you need to tell what ranges the substring has to have.
Here it is done by words:
var str:String = "Hello World!"
str.enumerateSubstringsInRange(Range<String.Index>(start:str.startIndex , end: str.endIndex), options: NSStringEnumerationOptions.ByWords) { (substring, substringRange, enclosingRange, stop) -> () in
switch substring
{
case "Hello":
println("Hi")
case "World":
println("universe")
default:
break
}
}
but actually I am cheating. In you code you want to switch for World!, but I am using World. I do this as in word-based enumeration non-alphanumeric characters are ignored.
But we have all information to fix it
var str:String = "Hello World!"
str.enumerateSubstringsInRange(Range<String.Index>(start:str.startIndex , end: str.endIndex), options: NSStringEnumerationOptions.ByWords) { (substring, substringRange, enclosingRange, stop) -> () in
var enclosingStr = str.substringWithRange(enclosingRange)
enclosingStr = enclosingStr.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
switch enclosingStr
{
case "Hello":
println("Hi")
case "World!":
println("universe")
default:
break
}
}
While in Swift 2.0, I use
let line = "Hello, world"
for (index, value) in line.characters.enumerate() {
// do something
}
Related
I would like to get a func which will be able to reverse a string without affecting special characters, preferably using regex, ex:
Input: “Weather is cool 24/7” -> Output: “rehtaeW si looc 24/7”
Input: “abcd efgh” -> Output: “dcba hgfe”
Input: “a1bcd efg!h” -> Output: “d1cba hgf!e”
I was able to write only for all characters without exceptions, I'm a beginner, and I don't know how to use regexes
func reverseTheWord(reverse: String) -> String {
let parts = reverse.components(separatedBy: " ")
let reversed = parts.map{String($0.reversed())}
let reversedWord = reversed.joined(separator: " ")
return reversedWord
}
thanks in advance!
Here is a solution where I first check what type each word is, only letters, no letters or a mix of letters and other characters and handle each differently.
The first two are self explanatory and for the mix one I first reverse the word and remove all non letters and then reinsert the non letters at their original position
func reverseTheWords(_ string: String) -> String {
var words = string.components(separatedBy: .whitespaces)
for (index, word) in words.enumerated() {
//Only letters
if word.allSatisfy(\.isLetter) {
words[index] = String(word.reversed())
continue
}
//No letters
if !word.contains(where: \.isLetter) { continue }
//Mix
var reversed = word.reversed().filter(\.isLetter)
for (index, char) in word.enumerated() {
if !char.isLetter {
index < reversed.endIndex ? reversed.insert(char, at: index) : reversed.append(char)
}
}
words[index] = String(reversed)
}
return words.joined(separator: " ")
}
let input = "hello, world song"
I have an input string as above.
So i can easily use this partial string something like this
output.0 // hello
output.1 // world song
I tried something like this How to split a string by new lines in Swift but i could not exactly what i want.
Can someone write an extension for this please in a nice way?
I would do something like this:
extension String {
func splitAtFirst(_ separator: Character) -> (head: Substring, tail: Substring?) {
guard let indexOfSeparator = self.firstIndex(of: separator) else {
return (head: Substring(self), tail: nil)
}
let indexAfterSeparator = self.index(indexOfSeparator, offsetBy: +1, limitedBy: self.endIndex)!
return (
head: self[..<indexOfSeparator],
tail: self[indexAfterSeparator...]
)
}
}
let (head, tail) = "abc, def, ghi".splitAtFirst(",")
print(head) // abc
print(tail as Any) // Optional(" def, ghi")
This returns Substrings, which gives you an efficient way to do a lot of processing on an input string without causing a bunch of copies along the way. Of course, you should promote these substrings to full on strings after you've finished processing them.
You can use an extension as below:
extension StringProtocol {
var tupleOfSplittedString: (String,String) {
if !self.isEmpty {
let splitted = self.split(separator: ",").map { String($0)}
let firstPart = splitted[0]
let otherPart = String(splitted[1...].joined().dropFirst())
return (firstPart,otherPart)
}
return ("","")
}
}
let input = "hello, world song"
let resultOfFirstPart = input.tupleOfSplittedString.0 // hello
let resultOfOtherPart = input.tupleOfSplittedString.1 // world song
This code uses the .components method to split a string by a substring. I have tested this, and it was successful, even if the String is blank (it returns ("", "") in that case)
Line 4 of this code is a bit hard to read, but you can split it up into multiple lines of code if you would like.
import Foundation
extension String {
var tuple: (String, String) {
return self.components(separatedBy: ",").count == 2 ? ((self.components(separatedBy: ",")[0], self.components(separatedBy: ",")[1] )) : ("", "")
}
}
var input = "hello, world song"
print(input.tuple.0) //prints hello
print(input.tuple.1) //prints world song
Every example of trimming strings in Swift remove both leading and trailing whitespace, but how can only trailing whitespace be removed?
For example, if I have a string:
" example "
How can I end up with:
" example"
Every solution I've found shows trimmingCharacters(in: CharacterSet.whitespaces), but I want to retain the leading whitespace.
RegEx is a possibility, or a range can be derived to determine index of characters to remove, but I can't seem to find an elegant solution for this.
With regular expressions:
let string = " example "
let trimmed = string.replacingOccurrences(of: "\\s+$", with: "", options: .regularExpression)
print(">" + trimmed + "<")
// > example<
\s+ matches one or more whitespace characters, and $ matches
the end of the string.
In Swift 4 & Swift 5
This code will also remove trailing new lines.
It works based on a Character struct's method .isWhitespace
var trailingSpacesTrimmed: String {
var newString = self
while newString.last?.isWhitespace == true {
newString = String(newString.dropLast())
}
return newString
}
This short Swift 3 extension of string uses the .anchored and .backwards option of rangeOfCharacter and then calls itself recursively if it needs to loop. Because the compiler is expecting a CharacterSet as the parameter, you can just supply the static when calling, e.g. "1234 ".trailing(.whitespaces) will return "1234". (I've not done timings, but would expect faster than regex.)
extension String {
func trailingTrim(_ characterSet : CharacterSet) -> String {
if let range = rangeOfCharacter(from: characterSet, options: [.anchored, .backwards]) {
return self.substring(to: range.lowerBound).trailingTrim(characterSet)
}
return self
}
}
In Foundation you can get ranges of indices matching a regular expression. You can also replace subranges. Combining this, we get:
import Foundation
extension String {
func trimTrailingWhitespace() -> String {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
return self.replacingCharacters(in: trailingWs, with: "")
} else {
return self
}
}
}
You can also have a mutating version of this:
import Foundation
extension String {
mutating func trimTrailingWhitespace() {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
self.replaceSubrange(trailingWs, with: "")
}
}
}
If we match against \s* (as Martin R. did at first) we can skip the if let guard and force-unwrap the optional since there will always be a match. I think this is nicer since it's obviously safe, and remains safe if you change the regexp. I did not think about performance.
Handy String extension In Swift 4
extension String {
func trimmingTrailingSpaces() -> String {
var t = self
while t.hasSuffix(" ") {
t = "" + t.dropLast()
}
return t
}
mutating func trimmedTrailingSpaces() {
self = self.trimmingTrailingSpaces()
}
}
Swift 4
extension String {
var trimmingTrailingSpaces: String {
if let range = rangeOfCharacter(from: .whitespacesAndNewlines, options: [.anchored, .backwards]) {
return String(self[..<range.lowerBound]).trimmingTrailingSpaces
}
return self
}
}
Demosthese's answer is a useful solution to the problem, but it's not particularly efficient. This is an upgrade to their answer, extending StringProtocol instead, and utilizing Substring to remove the need for repeated copying.
extension StringProtocol {
#inline(__always)
var trailingSpacesTrimmed: Self.SubSequence {
var view = self[...]
while view.last?.isWhitespace == true {
view = view.dropLast()
}
return view
}
}
No need to create a new string when dropping from the end each time.
extension String {
func trimRight() -> String {
String(reversed().drop { $0.isWhitespace }.reversed())
}
}
This operates on the collection and only converts the result back into a string once.
It's a little bit hacky :D
let message = " example "
var trimmed = ("s" + message).trimmingCharacters(in: .whitespacesAndNewlines)
trimmed = trimmed.substring(from: trimmed.index(after: trimmed.startIndex))
Without regular expression there is not direct way to achieve that.Alternatively you can use the below function to achieve your required result :
func removeTrailingSpaces(with spaces : String) -> String{
var spaceCount = 0
for characters in spaces.characters{
if characters == " "{
print("Space Encountered")
spaceCount = spaceCount + 1
}else{
break;
}
}
var finalString = ""
let duplicateString = spaces.replacingOccurrences(of: " ", with: "")
while spaceCount != 0 {
finalString = finalString + " "
spaceCount = spaceCount - 1
}
return (finalString + duplicateString)
}
You can use this function by following way :-
let str = " Himanshu "
print(removeTrailingSpaces(with : str))
One line solution with Swift 4 & 5
As a beginner in Swift and iOS programming I really like #demosthese's solution above with the while loop as it's very easy to understand. However the example code seems longer than necessary. The following uses essentially the same logic but implements it as a single line while loop.
// Remove trailing spaces from myString
while myString.last == " " { myString = String(myString.dropLast()) }
This can also be written using the .isWhitespace property, as in #demosthese's solution, as follows:
while myString.last?.isWhitespace == true { myString = String(myString.dropLast()) }
This has the benefit (or disadvantage, depending on your point of view) that this removes all types of whitespace, not just spaces but (according to Apple docs) also including newlines, and specifically the following characters:
“\t” (U+0009 CHARACTER TABULATION)
“ “ (U+0020 SPACE)
U+2029 PARAGRAPH SEPARATOR
U+3000 IDEOGRAPHIC SPACE
Note: Even though .isWhitespace is a Boolean it can't be used directly in the while loop as it ends up being optional ? due to the chaining of the optional .last property, which returns nil if the String (or collection) is empty. The == true logic gets around this since nil != true.
I'd love to get some feedback on this, esp. in case anyone sees any issues or drawbacks with this simple single line approach.
Swift 5
extension String {
func trimTrailingWhiteSpace() -> String {
guard self.last == " " else { return self }
var tmp = self
repeat {
tmp = String(tmp.dropLast())
} while tmp.last == " "
return tmp
}
}
This question already has answers here:
What is the best way to determine if a string contains a character from a set in Swift
(11 answers)
Closed 7 years ago.
I'm trying to check whether a specific string contains letters or not.
So far I've come across NSCharacterSet.letterCharacterSet() as a set of letters, but I'm having trouble checking whether a character in that set is in the given string. When I use this code, I get an error stating:
'Character' is not convertible to 'unichar'
For the following code:
for chr in input{
if letterSet.characterIsMember(chr){
return "Woah, chill out!"
}
}
You can use NSCharacterSet in the following way :
let letters = NSCharacterSet.letters
let phrase = "Test case"
let range = phrase.rangeOfCharacter(from: characterSet)
// range will be nil if no letters is found
if let test = range {
println("letters found")
}
else {
println("letters not found")
}
Or you can do this too :
func containsOnlyLetters(input: String) -> Bool {
for chr in input {
if (!(chr >= "a" && chr <= "z") && !(chr >= "A" && chr <= "Z") ) {
return false
}
}
return true
}
In Swift 2:
func containsOnlyLetters(input: String) -> Bool {
for chr in input.characters {
if (!(chr >= "a" && chr <= "z") && !(chr >= "A" && chr <= "Z") ) {
return false
}
}
return true
}
It's up to you, choose a way. I hope this help you.
You should use the Strings built in range functions with NSCharacterSet rather than roll your own solution. This will give you a lot more flexibility too (like case insensitive search if you so desire).
let str = "Hey this is a string"
let characterSet = NSCharacterSet(charactersInString: "aeiou")
if let _ = str.rangeOfCharacterFromSet(characterSet, options: .CaseInsensitiveSearch) {
println("true")
}
else {
println("false")
}
Substitute "aeiou" with whatever letters you're looking for.
A less flexible, but fun swift note all the same, is that you can use any of the functions available for Sequences. So you can do this:
contains("abc", "c")
This of course will only work for individual characters, and is not flexible and not recommended.
The trouble with .characterIsMember is that it takes a unichar (a typealias for UInt16).
If you iterate your input using the utf16 view of the string, it will work:
let set = NSCharacterSet.letterCharacterSet()
for chr in input.utf16 {
if set.characterIsMember(chr) {
println("\(chr) is a letter")
}
}
You can also skip the loop and use the contains algorithm if you only want to check for presence/non-presence:
if contains(input.utf16, { set.characterIsMember($0) }) {
println("contains letters")
}
I have a String description that holds my sentence and want to capitalize only the first letter. I tried different things but most of them give me exceptions and errors. I'm using Xcode 6.
Here is what I tried so far:
let cap = [description.substringToIndex(advance(0,1))] as String
description = cap.uppercaseString + description.substringFromIndex(1)
It gives me:
Type 'String.Index' does not conform to protocol 'IntegerLiteralConvertible'
I tried:
func capitalizedStringWithLocale(locale:0) -> String
But I haven't figured out how to make it work.
In Swift 2, you can do
String(text.characters.first!).capitalizedString + String(text.characters.dropFirst())
Another possibility in Swift 3:
extension String {
func capitalizeFirst() -> String {
let firstIndex = self.index(startIndex, offsetBy: 1)
return self.substring(to: firstIndex).capitalized + self.substring(from: firstIndex).lowercased()
}
}
For Swift 4:
Warnings from above Swift 3 code:
'substring(to:)' is deprecated: Please use String slicing subscript
with a 'partial range upto' operator.
'substring(from:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator.
Swift 4 solution:
extension String {
var capitalizedFirst: String {
guard !isEmpty else {
return self
}
let capitalizedFirstLetter = charAt(i: 0).uppercased()
let secondIndex = index(after: startIndex)
let remainingString = self[secondIndex..<endIndex]
let capitalizedString = "\(capitalizedFirstLetter)\(remainingString)"
return capitalizedString
}
}
Swift 5.0
Answer 1:
extension String {
func capitalizingFirstLetter() -> String {
return prefix(1).capitalized + dropFirst()
}
mutating func capitalizeFirstLetter() {
self = self.capitalizingFirstLetter()
}
}
Answer 2:
extension String {
func capitalizeFirstLetter() -> String {
return self.prefix(1).capitalized + dropFirst()
}
}
Answer 3:
extension String {
var capitalizeFirstLetter:String {
return self.prefix(1).capitalized + dropFirst()
}
}
import Foundation
// A lowercase string
let description = "the quick brown fox jumps over the lazy dog."
// The start index is the first letter
let first = description.startIndex
// The rest of the string goes from the position after the first letter
// to the end.
let rest = advance(first,1)..<description.endIndex
// Glue these two ranges together, with the first uppercased, and you'll
// get the result you want. Note that I'm using description[first...first]
// to get the first letter because I want a String, not a Character, which
// is what you'd get with description[first].
let capitalised = description[first...first].uppercaseString + description[rest]
// Result: "The quick brown fox jumps over the lazy dog."
You may want to make sure there's at least one character in your sentence before you start, as otherwise you'll get a runtime error trying to advance the index beyond the end of the string.
Here is how to do it in Swift 4; just in case if it helps anybody:
extension String {
func captalizeFirstCharacter() -> String {
var result = self
let substr1 = String(self[startIndex]).uppercased()
result.replaceSubrange(...startIndex, with: substr1)
return result
}
}
It won't mutate the original String.
extension String {
var capitalizedFirstLetter:String {
let string = self
return string.replacingCharacters(in: startIndex...startIndex, with: String(self[startIndex]).capitalized)
}
}
Answer:
let newSentence = sentence.capitalizedFirstLetter
For one or each word in string, you can use String's .capitalized property.
print("foo".capitalized) //prints: Foo
print("foo foo foo".capitalized) //prints: Foo Foo Foo
Swift 4.2 version:
extension String {
var firstCharCapitalized: String {
switch count {
case 0:
return self
case 1:
return uppercased()
default:
return self[startIndex].uppercased() + self[index(after: startIndex)...]
}
}
}
Simplest soulution for Swift 4.0.
Add as a computed property extension:
extension String {
var firstCapitalized: String {
var components = self.components(separatedBy: " ")
guard let first = components.first else {
return self
}
components[0] = first.capitalized
return components.joined(separator: " ")
}
}
Usage:
"hello world".firstCapitalized