I'm a little confused about the best practices for Swift 4 string manipulation.
How do you handle the following:
let str = "test"
let start = str.index(str.startIndex, offsetBy: 7)
Thread 1: Fatal error: cannot increment beyond endIndex
Imagine that you do not know the length of the variable 'str' above. And since 'start' is not an optional value, what is the best practice to prevent that crash?
If you use the variation with limitedBy parameter, that will return an optional value:
if let start = str.index(str.startIndex, offsetBy: 7, limitedBy: str.endIndex) {
...
}
That will gracefully detect whether the offset moves the index past the endIndex. Obviously, handle this optional however best in your scenario (if let, guard let, nil coalescing operator, etc.).
Your code doesn't do any range checking:
let str = "test"
let start = str.index(str.startIndex, offsetBy: 7)
Write a function that tests the length of the string first. In fact, you could create an extension on String that lets you use integer subscripts, and returns a Character?:
extension String {
//Allow string[Int] subscripting. WARNING: Slow O(n) performance
subscript(index: Int) -> Character? {
guard index < self.count else { return nil }
return self[self.index(self.startIndex, offsetBy: index)]
}
}
This code:
var str = "test"
print("str[7] = \"\(str[7])\"")
Would display:
str[7] = "nil"
##EDIT:
Be aware, as Alexander pointed out in a comment below, that the subscript extension above has up to O(n) performance (it takes longer and longer as the index value goes up, up to the length of the string.)
If you need to loop through all the characters in a string code like this:
for i in str.count { doSomething(string: str[i]) }
would have O(n^2) (Or n-squared) performance, which is really, really bad. in that case, you should instead first convert the string to an array of characters:
let chars = Array(str.characters)
for i in chars.count { doSomething(string: chars[i]) }
or
for aChar in chars { //do something with aChar }
With that code you pay the O(n) time cost of converting the string to an array of characters once, and then you can do operations on the array of characters with maximum speed. The downside of that approach is that it would more than double the memory requirements.
Related
I am solving a programming problem in Swift and I found a solution online which I don't totally understand, the problem is: Write a function that reverses characters in (possibly nested) parentheses in the input string. the solution is
var inputString = "foo(bar)baz(ga)kjh"
var s = inputString
while let openIdx = s.lastIndex(of: "(") {
let closeIdx = s[openIdx...].firstIndex(of:")")!
s.replaceSubrange(openIdx...closeIdx, with: s[s.index(after: openIdx)..<closeIdx].reversed())
}
print (s) // output: foorabbazagkjh (the letters inside the braces are reversed)
I d like to have details about: lastIndex(of: does in this case
and what let closeIdx = s[openIdx...].firstIndex(of:")")! does as well
The best place to experiment with these kinds of questions would Playground. Also, check out the documentation.
Now let go through each of the statement:
let openIdx = s.lastIndex(of: "(") // it will find the last index of "(", the return type here is Array.Index?
so if I print the value after with index including till end of string, it would be
print(s[openIdx!...]) // `!` exclamation is used for forced casting
// (ga)kjh
Now for your second question;
let closeIdx = s[openIdx...].firstIndex(of:")")!
Let break it down s[openIdx...] is equal to (ga)kjh in first iteration and so it will return the index of ) after a.
The suggestion would be always break the statement and learn what each expression is doing.
Consider this function to build a string of random characters:
func makeToken(length: Int) -> String {
let chars: String = "abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ"
var result: String = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let idxEnd = idx + 1
let range: Range = idx..<idxEnd
let char = chars.substring(with: range)
result += char
}
return result
}
This throws an error on the substring method:
Cannot convert value of type 'Range<Int>' to expected argument
type 'Range<String.Index>' (aka 'Range<String.CharacterView.Index>')
I'm confused why I can't simply provide a Range with 2 integers, and why it's making me go the roundabout way of making a Range<String.Index>.
So I have to change the Range creation to this very over-complicated way:
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let start = chars.index(chars.startIndex, offsetBy: idx)
let end = chars.index(chars.startIndex, offsetBy: idx + 1)
let range: Range = start..<end
Why isn't it good enough for Swift for me to simply create a range with 2 integers and the half-open range operator? (..<)
Quite the contrast to "swift", in javascript I can simply do chars.substr(idx, 1)
I suggest converting your String to [Character] so that you can index it easily with Int:
func makeToken(length: Int) -> String {
let chars = Array("abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ".characters)
var result = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.count)))
result += String(chars[idx])
}
return result
}
Swift takes great care to provide a fully Unicode-compliant, type-safe, String abstraction.
Indexing a given Character, in an arbitrary Unicode string, is far from a trivial task. Each Character is a sequence of one or more Unicode scalars that (when combined) produce a single human-readable character. In particular, hiding all this complexity behind a simple Int based indexing scheme might result in the wrong performance mental model for programmers.
Having said that, you can always convert your string to a Array<Character> once for easy (and fast!) indexing. For instance:
let chars: String = "abcdefghijklmnop"
var charsArray = Array(chars.characters)
...
let resultingString = String(charsArray)
I have a String "000". I want to change this to "0.00".
I took a look at the insert function.
var str = "000"
str.insert(".", at: str.endIndex)
How do I get the index of 2 before the end index?
I tried:
str.insert(".", at: str.endIndex - 1)
but this didn't work at all.
You could also use Strings character property. Its basically an array made up of all the characters (duh) in the String.
So you would:
var str = "000"
let index = str.characters.index(str.characters.startIndex, offsetBy: 1) //here you define a place (index) to insert at
str.characters.insert(".", at: index) //and here you insert
Unfortunately you have to create an index first, as .insert does not allow you to specify the position using an Int.
Since Swift 2, String does no longer conform to SequenceType. In other words, you can not iterate through a string with a for...in loop.
The simple and easy way is to convert String to Array to get the benefit of the index just like that:
let input = Array(str)
I remember when I tried to index into String without using any conversion. I was really frustrated that I couldn’t come up with or reach a desired result, and was about to give up.
But I ended up creating my own workaround solution, and here is the full code of the extension:
extension String {
subscript (_ index: Int) -> String {
get {
String(self[self.index(startIndex, offsetBy: index)])
}
set {
if index >= count {
insert(Character(newValue), at: self.index(self.startIndex, offsetBy: count))
} else {
insert(Character(newValue), at: self.index(self.startIndex, offsetBy: index))
}
}
}
}
Now that you can read and write a single character from string using its index just like you originally wanted to:
var str = "car"
car[3] = "d"
print(str)
It’s simple and useful way to use it and get through Swift’s String access model.
Now that you’ll feel it’s smooth sailing next time when you can loop through the string just as it is, not casting it into Array.
Try it out, and see if it can help!
I've been updating some of my old code and answers with Swift 3 but when I got to Swift Strings and Indexing it has been a pain to understand things.
Specifically I was trying the following:
let str = "Hello, playground"
let prefixRange = str.startIndex..<str.startIndex.advancedBy(5) // error
where the second line was giving me the following error
'advancedBy' is unavailable: To advance an index by n steps call 'index(_:offsetBy:)' on the CharacterView instance that produced the index.
I see that String has the following methods.
str.index(after: String.Index)
str.index(before: String.Index)
str.index(String.Index, offsetBy: String.IndexDistance)
str.index(String.Index, offsetBy: String.IndexDistance, limitedBy: String.Index)
These were really confusing me at first so I started playing around with them until I understood them. I am adding an answer below to show how they are used.
All of the following examples use
var str = "Hello, playground"
startIndex and endIndex
startIndex is the index of the first character
endIndex is the index after the last character.
Example
// character
str[str.startIndex] // H
str[str.endIndex] // error: after last character
// range
let range = str.startIndex..<str.endIndex
str[range] // "Hello, playground"
With Swift 4's one-sided ranges, the range can be simplified to one of the following forms.
let range = str.startIndex...
let range = ..<str.endIndex
I will use the full form in the follow examples for the sake of clarity, but for the sake of readability, you will probably want to use the one-sided ranges in your code.
after
As in: index(after: String.Index)
after refers to the index of the character directly after the given index.
Examples
// character
let index = str.index(after: str.startIndex)
str[index] // "e"
// range
let range = str.index(after: str.startIndex)..<str.endIndex
str[range] // "ello, playground"
before
As in: index(before: String.Index)
before refers to the index of the character directly before the given index.
Examples
// character
let index = str.index(before: str.endIndex)
str[index] // d
// range
let range = str.startIndex..<str.index(before: str.endIndex)
str[range] // Hello, playgroun
offsetBy
As in: index(String.Index, offsetBy: String.IndexDistance)
The offsetBy value can be positive or negative and starts from the given index. Although it is of the type String.IndexDistance, you can give it an Int.
Examples
// character
let index = str.index(str.startIndex, offsetBy: 7)
str[index] // p
// range
let start = str.index(str.startIndex, offsetBy: 7)
let end = str.index(str.endIndex, offsetBy: -6)
let range = start..<end
str[range] // play
limitedBy
As in: index(String.Index, offsetBy: String.IndexDistance, limitedBy: String.Index)
The limitedBy is useful for making sure that the offset does not cause the index to go out of bounds. It is a bounding index. Since it is possible for the offset to exceed the limit, this method returns an Optional. It returns nil if the index is out of bounds.
Example
// character
if let index = str.index(str.startIndex, offsetBy: 7, limitedBy: str.endIndex) {
str[index] // p
}
If the offset had been 77 instead of 7, then the if statement would have been skipped.
Why is String.Index needed?
It would be much easier to use an Int index for Strings. The reason that you have to create a new String.Index for every String is that Characters in Swift are not all the same length under the hood. A single Swift Character might be composed of one, two, or even more Unicode code points. Thus each unique String must calculate the indexes of its Characters.
It is possible to hide this complexity behind an Int index extension, but I am reluctant to do so. It is good to be reminded of what is actually happening.
I appreciate this question and all the info with it. I have something in mind that's kind of a question and an answer when it comes to String.Index.
I'm trying to see if there is an O(1) way to access a Substring (or Character) inside a String because string.index(startIndex, offsetBy: 1) is O(n) speed if you look at the definition of index function. Of course we can do something like:
let characterArray = Array(string)
then access any position in the characterArray however SPACE complexity of this is n = length of string, O(n) so it's kind of a waste of space.
I was looking at Swift.String documentation in Xcode and there is a frozen public struct called Index. We can initialize is as:
let index = String.Index(encodedOffset: 0)
Then simply access or print any index in our String object as such:
print(string[index])
Note: be careful not to go out of bounds`
This works and that's great but what is the run-time and space complexity of doing it this way? Is it any better?
func change(string: inout String) {
var character: Character = .normal
enum Character {
case space
case newLine
case normal
}
for i in stride(from: string.count - 1, through: 0, by: -1) {
// first get index
let index: String.Index?
if i != 0 {
index = string.index(after: string.index(string.startIndex, offsetBy: i - 1))
} else {
index = string.startIndex
}
if string[index!] == "\n" {
if character != .normal {
if character == .newLine {
string.remove(at: index!)
} else if character == .space {
let number = string.index(after: string.index(string.startIndex, offsetBy: i))
if string[number] == " " {
string.remove(at: number)
}
character = .newLine
}
} else {
character = .newLine
}
} else if string[index!] == " " {
if character != .normal {
string.remove(at: index!)
} else {
character = .space
}
} else {
character = .normal
}
}
// startIndex
guard string.count > 0 else { return }
if string[string.startIndex] == "\n" || string[string.startIndex] == " " {
string.remove(at: string.startIndex)
}
// endIndex - here is a little more complicated!
guard string.count > 0 else { return }
let index = string.index(before: string.endIndex)
if string[index] == "\n" || string[index] == " " {
string.remove(at: index)
}
}
Create a UITextView inside of a tableViewController. I used function: textViewDidChange and then checked for return-key-input.
then if it detected return-key-input, delete the input of return key and dismiss keyboard.
func textViewDidChange(_ textView: UITextView) {
tableView.beginUpdates()
if textView.text.contains("\n"){
textView.text.remove(at: textView.text.index(before: textView.text.endIndex))
textView.resignFirstResponder()
}
tableView.endUpdates()
}
The following was possible with Swift 2.2:
let m = "alpha"
for i in m.startIndex..<m.endIndex {
print(m[i])
}
a
l
p
h
a
With 3.0, we get the following error:
Type 'Range' (aka 'Range') does not conform to protocol 'Sequence'
I am trying to do a very simple operation with strings in swift -- simply traverse through the first half of the string (or a more generic problem: traverse through a range of a string).
I can do the following:
let s = "string"
var midIndex = s.index(s.startIndex, offsetBy: s.characters.count/2)
let r = Range(s.startIndex..<midIndex)
print(s[r])
But here I'm not really traversing the string. So the question is: how do I traverse through a range of a given string. Like:
for i in Range(s.startIndex..<s.midIndex) {
print(s[i])
}
You can traverse a string by using indices property of the characters property like this:
let letters = "string"
let middle = letters.index(letters.startIndex, offsetBy: letters.characters.count / 2)
for index in letters.characters.indices {
// to traverse to half the length of string
if index == middle { break } // s, t, r
print(letters[index]) // s, t, r, i, n, g
}
From the documentation in section Strings and Characters - Counting Characters:
Extended grapheme clusters can be composed of one or more Unicode scalars. This means that different characters—and different representations of the same character—can require different amounts of memory to store. Because of this, characters in Swift do not each take up the same amount of memory within a string’s representation. As a result, the number of characters in a string cannot be calculated without iterating through the string to determine its extended grapheme cluster boundaries.
emphasis is my own.
This will not work:
let secondChar = letters[1]
// error: subscript is unavailable, cannot subscript String with an Int
Another option is to use enumerated() e.g:
let string = "Hello World"
for (index, char) in string.characters.enumerated() {
print(char)
}
or for Swift 4 just use
let string = "Hello World"
for (index, char) in string.enumerated() {
print(char)
}
Use the following:
for i in s.characters.indices[s.startIndex..<s.endIndex] {
print(s[i])
}
Taken from Migrating to Swift 2.3 or Swift 3 from Swift 2.2
Iterating over characters in a string is cleaner in Swift 4:
let myString = "Hello World"
for char in myString {
print(char)
}
If you want to traverse over the characters of a String, then instead of explicitly accessing the indices of the String, you could simply work with the CharacterView of the String, which conforms to CollectionType, allowing you access to neat subsequencing methods such as prefix(_:) and so on.
/* traverse the characters of your string instance,
up to middle character of the string, where "middle"
will be rounded down for strings of an odd amount of
characters (e.g. 5 characters -> travers through 2) */
let m = "alpha"
for ch in m.characters.prefix(m.characters.count/2) {
print(ch, ch.dynamicType)
} /* a Character
l Character */
/* round odd division up instead */
for ch in m.characters.prefix((m.characters.count+1)/2) {
print(ch, ch.dynamicType)
} /* a Character
l Character
p Character */
If you'd like to treat the characters within the loop as strings, simply use String(ch) above.
With regard to your comment below: if you'd like to access a range of the CharacterView, you could easily implement your own extension of CollectionType (specified for when Generator.Element is Character) making use of both prefix(_:) and suffix(_:) to yield a sub-collection given e.g. a half-open (from..<to) range
/* for values to >= count, prefixed CharacterView will be suffixed until its end */
extension CollectionType where Generator.Element == Character {
func inHalfOpenRange(from: Int, to: Int) -> Self {
guard case let to = min(to, underestimateCount()) where from <= to else {
return self.prefix(0) as! Self
}
return self.prefix(to).suffix(to-from) as! Self
}
}
/* example */
let m = "0123456789"
for ch in m.characters.inHalfOpenRange(4, to: 8) {
print(ch) /* \ */
} /* 4 a (sub-collection) CharacterView
5
6
7 */
The best way to do this is :-
let name = "nick" // The String which we want to print.
for i in 0..<name.count
{
// Operation name[i] is not allowed in Swift, an alternative is
let index = name.index[name.startIndex, offsetBy: i]
print(name[index])
}
for more details visit here
Swift 4.2
Simply:
let m = "alpha"
for i in m.indices {
print(m[i])
}
Swift 4:
let mi: String = "hello how are you?"
for i in mi {
print(i)
}
To concretely demonstrate how to traverse through a range in a string in Swift 4, we can use the where filter in a for loop to filter its execution to the specified range:
func iterateStringByRange(_ sentence: String, from: Int, to: Int) {
let startIndex = sentence.index(sentence.startIndex, offsetBy: from)
let endIndex = sentence.index(sentence.startIndex, offsetBy: to)
for position in sentence.indices where (position >= startIndex && position < endIndex) {
let char = sentence[position]
print(char)
}
}
iterateStringByRange("string", from: 1, to: 3) will print t, r and i
When iterating over the indices of characters in a string, you seldom only need the index. You probably also need the character at the given index. As specified by Paulo (updated for Swift 4+), string.indices will give you the indices of the characters. zip can be used to combine index and character:
let string = "string"
// Define the range to conform to your needs
let range = string.startIndex..<string.index(string.startIndex, offsetBy: string.count / 2)
let substring = string[range]
// If the range is in the type "first x characters", like until the middle, you can use:
// let substring = string.prefix(string.count / 2)
for (index, char) in zip(substring.indices, substring) {
// index is the index in the substring
print(char)
}
Note that using enumerated() will produce a pair of index and character, but the index is not the index of the character in the string. It is the index in the enumeration, which can be different.