Return results from regular expression pattern matching - swift

I have a string (HTML in this example case) which contains the same pattern for displaying the results of sports games. So, the HTML tags are known, but the values for each game are not.
In Perl, we can do this:
if ( $content =~ /\<\/a\>\<br\>(\d+)\<\/span\>\<br\>(\d+)\-(\d+).+\<\/a\>\<br\>(\d+)\<\/span\>\<br\>(\d+)\-(\d+)/) {
$visitingTeamScore = $1; // $1 is the 1st matched digit
$visitingTeamWins = $2; // $2 is the 2nd matched digit
$visitingTeamLosses = $3; // Etc
$homeTeamScore = $4;
$homeTeamWins = $5;
$homeTeamLosses = $6;
}
which returns the digits inside the parentheses, in this case 6 total integers of varying digit lengths. We can then assign those matches to variables.
From an answer in this question: Swift Get string between 2 strings in a string, I have the following Swift code:
extension String {
func sliceFrom(start: String, to: String) -> String? {
return (rangeOfString(start)?.endIndex).flatMap { sInd in
(rangeOfString(to, range: sInd..<endIndex)?.startIndex).map { eInd in
substringWithRange(sInd..<eInd)
}
}
}
}
let firstMatch = content?.sliceFrom("</a><br>", to: "</span>") // The first integer in the string
The problem comes in when getting the 4th integer which is also between </a\><br> and </span> so the resulting match will be the first digit again.
I can manually count the characters (which itself isn't a perfect science because the digits in each integer can differ) to do something ugly like:
let newRawHTML = content![content!.startIndex.advancedBy(15)...content!.startIndex.advancedBy(5)]
Another possibility is to remove anything matched already from the string, making it shorter for each subsequent search (which I'm not sure how to implement.) What's the way to do this? Is there any way in Swift to "pluck out" the matches?

The code you have shown as a Perl example, uses regular expression.
And in case the pattern is getting a little bit complex, you'd better use NSRegularExpression directly.
let pattern = "</a><br>(\\d+)</span><br>(\\d+)-(\\d+).+</a><br>(\\d+)</span><br>(\\d+)-(\\d+)"
let regex = try! NSRegularExpression(pattern: pattern, options: [])
if let match = regex.firstMatchInString(content, options: [], range: NSRange(0..<content.utf16.count)) {
let visitingTeamScore = (content as NSString).substringWithRange(match.rangeAtIndex(1))
let visitingTeamWins = (content as NSString).substringWithRange(match.rangeAtIndex(2))
let visitingTeamLosses = (content as NSString).substringWithRange(match.rangeAtIndex(3))
let homeTeamScore = (content as NSString).substringWithRange(match.rangeAtIndex(4))
let homeTeamWins = (content as NSString).substringWithRange(match.rangeAtIndex(5))
let homeTeamLosses = (content as NSString).substringWithRange(match.rangeAtIndex(6))
//...use the values
}

Related

Split String or Substring with Regex pattern in Swift

First let me point out... I want to split a String or Substring with any character that is not an alphabet, a number, # or #. That means, I want to split with whitespaces(spaces & line breaks) and special characters or symbols excluding # and #
In Android Java, I am able to achieve this with:
String[] textArr = text.split("[^\\w_##]");
Now, I want to do the same in Swift. I added an extension to String and Substring classes
extension String {}
extension Substring {}
In both extensions, I added a method that returns an array of Substring
func splitWithRegex(by regexStr: String) -> [Substring] {
//let string = self (for String extension) | String(self) (for Substring extension)
let regex = try! NSRegularExpression(pattern: regexStr)
let range = NSRange(string.startIndex..., in: string)
return regex.matches(in: string, options: .anchored, range: range)
.map { match -> Substring in
let range = Range(match.range(at: 1), in: string)!
return string[range]
}
}
And when I tried to use it, (Only tested with a Substring, but I also think String will give me the same result)
let textArray = substring.splitWithRegex(by: "[^\\w_##]")
print("substring: \(substring)")
print("textArray: \(textArray)")
This is the out put:
substring: This,is a #random #text written for debugging
textArray: []
Please can Someone help me. I don't know if the problem if from my regex [^\\w_##] or from splitWithRegex method
The main reason why the code doesn't work is range(at: 1) which returns the content of the first captured group, but the pattern does not capture anything.
With just range the regex returns the ranges of the found matches, but I suppose you want the characters between.
To accomplish that you need a dynamic index starting at the first character. In the map closure return the string from the current index to the lowerBound of the found range and set the index to its upperBound. Finally you have to add manually the string from the upperBound of the last match to the end.
The Substring type is a helper type for slicing strings. It should not be used beyond a temporary scope.
extension String {
func splitWithRegex(by regexStr: String) -> [String] {
guard let regex = try? NSRegularExpression(pattern: regexStr) else { return [] }
let range = NSRange(startIndex..., in: self)
var index = startIndex
var array = regex.matches(in: self, range: range)
.map { match -> String in
let range = Range(match.range, in: self)!
let result = self[index..<range.lowerBound]
index = range.upperBound
return String(result)
}
array.append(String(self[index...]))
return array
}
}
let text = "This,is a #random #text written for debugging"
let textArray = text.splitWithRegex(by: "[^\\w_##]")
print(textArray) // ["This", "is", "a", "#random", "#text", "written", "for", "debugging"]
However in macOS 13 and iOS 16 there is a new API quite similar to the java API
let text = "This,is a #random #text written for debugging"
let textArray = Array(text.split(separator: /[^\w_##]/))
print(textArray)
The forward slashes indicate a regex literal

Swift String Tokenizer / Parser [closed]

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Hello there fellow Swift devs!
I am a junior dev, and I'm trying to figure out a best way to tokenize / parse Swift String as an exercise.
What I have is a string which looks like this:
let string = "This is a {B}string{/B} and this is a substring."
What I would like to do is, tokenize the string, and change the "strings / tokens" inside the tags you see.
I can see using NSRegularExpression and it's matches, but it feels too generic. I would like to have only say 2 of these tags, that change the text. What would be the best approach in Swift 5.2^?
if let regex = try? NSRegularExpression(pattern: "\\{[a-z0-9]+\}", options: .caseInsensitive) {
let string = self as NSString
return regex.matches(in: self, options: [], range: NSRange(location: 0, length: string.length)).map {
// now $0 is the result? but it won't work for enclosing the tags :/
}
}
If the option of using html tags instead of {B}{/B} is acceptable, then you can use the StringEx library that I wrote for this purpose.
You can select a substring inside the html tag and replace it with another string like this:
let string = "This is a <b>string</b> and this is a substring."
let ex = string.ex
ex[.tag("b")].replace(with: "some value")
print(ex.rawString) // This is a <b>some value</b> and this is a substring.
print(ex.string) // This is a some value and this is a substring.
if necessary, you can also style the selected substrings and get NSAttributedString:
ex[.tag("b")].style([
.font(.boldSystemFont(ofSize: 16)),
.color(.black)
])
myLabel.attributedText = ex.attributedString
Not sure if you have solved it with NLTokenizer or not, but you can certainly solve it with Regx here is how (I have implemented it as generic, in future if you have to handle different kinds of tags and substite different string for them small tweak to the logic should do the job )
override func viewDidLoad() {
super.viewDidLoad()
let regexStr = "(\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\})"
let regex = try! NSRegularExpression(pattern: regexStr)
var string = "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
var foundRanges = [NSRange]()
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
debugPrint(replacedString)
}
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
Input: "Sandeep {B}Bhandaari{/B} is here"
Output: Sandeep abcd is here
Input: "Sandeep {B}Bhandaari{/B} is here{B}Sandeep{/B}"
Output: Sandeep abcd is hereabcd
Look at the edge case handling Longer strings replaced by smaller substitute strings and vice versa also detection of string enclosed in tag with / without space
EDIT 1:
Regx (\\{B\\}(\\s*\\w+\\s*)*\\{\\/B\\}) should be self explanatory, incase you need help with understanding it use cheat sheet
regex.enumerateMatches(in: string, options: [], range: NSMakeRange(0, string.count)) { (match, flag, stop) in
if let matchRange = match?.range(at: 1) {
foundRanges.append(matchRange)
}
}
I could have modified substring here itself, but if you have more than one match and if you mutate string evaluated ranges will be corrupted hence am saving all found ranges into an array and apply replace on each one of them later
let substituteString = "abcd"
var replacedString = string as NSString
let foundRangesCount = foundRanges.count
var currentRange = 0
while foundRangesCount > currentRange {
let range = foundRanges[currentRange]
replacedString = replacedString.replacingCharacters(in: range, with: substituteString) as NSString
reEvaluateAllRanges(ranges: &foundRanges, byOffset: range.length - substituteString.count)
currentRange += 1
}
Here am iterating through all found match ranges and replace character in range with substitute string, you can always have a switch / if else ladder inside while loop to look for different types of tags and pass different substitute strings for each tags
func reEvaluateAllRanges(ranges: inout [NSRange], byOffset: Int) {
var newFoundRange = [NSRange]()
for range in ranges {
newFoundRange.append(NSMakeRange(range.location - byOffset, range.length))
}
ranges = newFoundRange
}
This function modifies all the ranges in array using the offset, remember you need to only modify range's location, length remains same
One bit of optimisation you can do is probably get rid of ranges from array for which you have already applied substitute strings

I have problems with ł, ó, ą type characters in Swift

I have a piece of code in my iOS app that should go through a word and check if a character is in it. When it finds at least one, it should change a string full of "_" of the same length as the word to one with the character in the right place:
wordToGuess = six
letterGuessed = i
wordAsUnderscores = _i_
The code works. But I start to have problems when I type in characters like: "ć", "ł", "ą", etc. From using character.utf8.count I saw that Swift thinks those are not 1 but 2 characters. So I get something like this:
wordToGuess = cześć
letterGuessed = ś
wordAsUnderscores = _ _ ś (place filled with empty char) _
It takes up 2 places.
I was at it for 6 hours and didn't figure out how to fix it, so I'm asking you guys for help.
Code that is supposed to do that:
let characterGuessed = Character(letterGuessed)
for index in wordToGuess.indices {
if (wordToGuess[index] == characterGuessed) {
let endIndex = wordToGuess.index(after: index)
let charRange = index..<endIndex
wordAsUnderscores = wordAsUnderscores.replacingCharacters(in: charRange, with: letterGuessed)
wordToGuessLabel.text = wordAsUnderscores
}
}
I would like the code to treat "ć", "ł", "ą" characters the same as "i", "a" and so on. I don't want them to be treated as 2.
The reason is that you cannot use indices from one string (wordToGuess) for subscripting another string (wordAsUnderscores). Generally, indices of one collection must not be used with a different collection. (There are exception like Array though).
Here is a working variant:
let wordToGuess = "cześć"
let letterGuessed: Character = "ś"
var wordAsUnderscores = "c____"
wordAsUnderscores = String(zip(wordToGuess, wordAsUnderscores)
.map { $0 == letterGuessed ? letterGuessed : $1 })
print(wordAsUnderscores) // c__ś_
The strings are traversed in parallel, and for each correctly guessed character in wordToGuess the corresponding character in wordAsUnderscores is replaced by that character.

Regular expressions in swift

I'm bit confused by NSRegularExpression in swift, can any one help me?
task:1 given ("name","john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets.
task:2 given ("name"," john","name of john")
then I should get ["name","john","name of john"]. Here I should avoid the brackets and extra spaces and finally get array of strings.
task:3 given key = value // comment
then I should get ["key","value","comment"]. Here I should get only strings in the line by avoiding = and //
I have tried below code for task 1 but not passed.
let string = "(name,john,string for user name)"
let pattern = "(?:\\w.*)"
do {
let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive)
let matches = regex.matches(in: string, options: [], range: NSRange(location: 0, length: string.utf16.count))
for match in matches {
if let range = Range(match.range, in: string) {
let name = string[range]
print(name)
}
}
} catch {
print("Regex was bad!")
}
Thanks in advance.
RegEx in Swift
These posts might help you to explore regular expressions in swift:
Does a string match a pattern?
Swift extract regex matches
How can I use String slicing subscripts in Swift 4?
How to use regex with Swift?
Swift 3 - How do I extract captured groups in regular expressions?
How to group search regular expressions using swift?
Task 1 & 2
This expression might help you to match your desired outputs for both Task 1 and 2:
"(\s+)?([a-z\s]+?)(\s+)?"
Based on Rob's advice, you could much reduce the boundaries, such as the char list [a-z\s]. For example, here, we can also use:
"(\s+)?(.*?)(\s+)?"
or
"(\s+)?(.+?)(\s+)?"
to simply pass everything in between two " and/or space.
RegEx
If this wasn't your desired expression, you can modify/change your expressions in regex101.com.
RegEx Circuit
You can also visualize your expressions in jex.im:
JavaScript Demo
const regex = /"(\s+)?([a-z\s]+?)(\s+)?"/gm;
const str = `"name","john","name of john"
"name"," john","name of john"
" name "," john","name of john "
" name "," john"," name of john "`;
const subst = `\n$2`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Task 3
This expression might help you to design an expression for the third task:
(.*?)([a-z\s]+)(.*?)
const regex = /(.*?)([a-z\s]+)(.*?)/gm;
const str = `key = value // comment
key = value with some text // comment`;
const subst = `$2,`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
Separate the string by non alpha numeric characters except white spaces. Then trim the elements with white spaces.
extension String {
func words() -> [String] {
return self.components(separatedBy: CharacterSet.alphanumerics.inverted.subtracting(.whitespaces))
.filter({ !$0.isEmpty })
.map({ $0.trimmingCharacters(in: .whitespaces) })
}
}
let string1 = "(name,john,string for user name)"
let string2 = "(name, john,name of john)"
let string3 = "key = value // comment"
print(string1.words())//["name", "john", "string for user name"]
print(string2.words())//["name", "john", "name of john"]
print(string3.words())//["key", "value", "comment"]
Here I have done with after understanding all of above comments.
let text = """
Capturing and non-capturing groups are somewhat advanced topics. You’ll encounter examples of capturing and non-capturing groups later on in the tutorial
"""
extension String {
func rex (_ expr : String)->[String] {
return try! NSRegularExpression(pattern: expr, options: [.caseInsensitive])
.matches(in: self, options: [], range: NSRange(location: 0, length: self.count))
.map {
String(self[Range($0.range, in: self)!])
}
}
}
let r = text.rex("(?:\\w+-\\w+)") // pass any rex
A single pattern, works for test:1...3, in Swift.
let string =
//"(name,john,string for user name)" //test:1
//#"("name"," john","name of john")"# //test:2
"key = value // comment" //test:3
let pattern = #"(?:\w+)(?:\s+\w+)*"# //Swift 5+ only
//let pattern = "(?:\\w+)(?:\\s+\\w+)*"
do {
let regex = try NSRegularExpression(pattern: pattern)
let matches = regex.matches(in: string, range: NSRange(0..<string.utf16.count))
let matchingWords = matches.map {
String(string[Range($0.range, in: string)!])
}
print(matchingWords) //(test:3)->["key", "value", "comment"]
} catch {
print("Regex was bad!")
}
Let’s consider:
let string = "(name,José,name is José)"
I’d suggest a regex that looks for strings where:
It’s the substring either after the ( at the start of the full string or after a comma, i.e., look behind assertion of (?<=^\(|,);
It’s the substring that does not contain , within it, i.e., [^,]+?;
It’s the substring that is terminated by either a comma or ) at the end of the full string, i.e., look ahead assertion of (?=,|\)$), and
If you want to have it skip white space before and after the substrings, throw in the \s*+, too.
Thus:
let pattern = #"(?<=^\(|,)\s*+([^,]+?)\s*+(?=,|\)$)"#
let regex = try! NSRegularExpression(pattern: pattern)
regex.enumerateMatches(in: string, range: NSRange(string.startIndex..., in: string)) { match, _, _ in
if let nsRange = match?.range(at: 1), let range = Range(nsRange, in: string) {
let substring = String(string[range])
// do something with `substring` here
}
}
Note, I’m using the Swift 5 extended string delimiters (starting with #" and ending with "#) so that I don’t have to escape my backslashes within the string. If you’re using Swift 4 or earlier, you’ll want to escape those back slashes:
let pattern = "(?<=^\\(|,)\\s*+([^,]+?)\\s*+(?=,|\\)$)"

Get numbers characters from a string [duplicate]

This question already has answers here:
Filter non-digits from string
(12 answers)
Closed 6 years ago.
How to get numbers characters from a string? I don't want to convert in Int.
var string = "string_1"
var string2 = "string_20_certified"
My result have to be formatted like this:
newString = "1"
newString2 = "20"
Pattern matching a String's unicode scalars against Western Arabic Numerals
You could pattern match the unicodeScalars view of a String to a given UnicodeScalar pattern (covering e.g. Western Arabic numerals).
extension String {
var westernArabicNumeralsOnly: String {
let pattern = UnicodeScalar("0")..."9"
return String(unicodeScalars
.flatMap { pattern ~= $0 ? Character($0) : nil })
}
}
Example usage:
let str1 = "string_1"
let str2 = "string_20_certified"
let str3 = "a_1_b_2_3_c34"
let newStr1 = str1.westernArabicNumeralsOnly
let newStr2 = str2.westernArabicNumeralsOnly
let newStr3 = str3.westernArabicNumeralsOnly
print(newStr1) // 1
print(newStr2) // 20
print(newStr3) // 12334
Extending to matching any of several given patterns
The unicode scalar pattern matching approach above is particularly useful extending it to matching any of a several given patterns, e.g. patterns describing different variations of Eastern Arabic numerals:
extension String {
var easternArabicNumeralsOnly: String {
let patterns = [UnicodeScalar("\u{0660}")..."\u{0669}", // Eastern Arabic
"\u{06F0}"..."\u{06F9}"] // Perso-Arabic variant
return String(unicodeScalars
.flatMap { uc in patterns.contains{ $0 ~= uc } ? Character(uc) : nil })
}
}
This could be used in practice e.g. if writing an Emoji filter, as ranges of unicode scalars that cover emojis can readily be added to the patterns array in the Eastern Arabic example above.
Why use the UnicodeScalar patterns approach over Character ones?
A Character in Swift contains of an extended grapheme cluster, which is made up of one or more Unicode scalar values. This means that Character instances in Swift does not have a fixed size in the memory, which means random access to a character within a collection of sequentially (/contiguously) stored character will not be available at O(1), but rather, O(n).
Unicode scalars in Swift, on the other hand, are stored in fixed sized UTF-32 code units, which should allow O(1) random access. Now, I'm not entirely sure if this is a fact, or a reason for what follows: but a fact is that if benchmarking the methods above vs equivalent method using the CharacterView (.characters property) for some test String instances, its very apparent that the UnicodeScalar approach is faster than the Character approach; naive testing showed a factor 10-25 difference in execution times, steadily growing for growing String size.
Knowing the limitations of working with Unicode scalars vs Characters in Swift
Now, there are drawbacks using the UnicodeScalar approach, however; namely when working with characters that cannot represented by a single unicode scalar, but where one of its unicode scalars are contained in the pattern to which we want to match.
E.g., consider a string holding the four characters "Café". The last character, "é", is represented by two unicode scalars, "e" and "\u{301}". If we were to implement pattern matching against, say, UnicodeScalar("a")...e, the filtering method as applied above would allow one of the two unicode scalars to pass.
extension String {
var onlyLowercaseLettersAthroughE: String {
let patterns = [UnicodeScalar("1")..."e"]
return String(unicodeScalars
.flatMap { uc in patterns.contains{ $0 ~= uc } ? Character(uc) : nil })
}
}
let str = "Cafe\u{301}"
print(str) // Café
print(str.onlyLowercaseLettersAthroughE) // Cae
/* possibly we'd want "Ca" or "Caé"
as result here */
In the particular use case queried by from the OP in this Q&A, the above is not an issue, but depending on the use case, it will sometimes be more appropriate to work with Character pattern matching over UnicodeScalar.
Edit: Updated for Swift 4 & 5
Here's a straightforward method that doesn't require Foundation:
let newstring = string.filter { "0"..."9" ~= $0 }
or borrowing from #dfri's idea to make it a String extension:
extension String {
var numbers: String {
return filter { "0"..."9" ~= $0 }
}
}
print("3 little pigs".numbers) // "3"
print("1, 2, and 3".numbers) // "123"
import Foundation
let string = "a_1_b_2_3_c34"
let result = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined(separator: "")
print(result)
Output:
12334
Here is a Swift 2 example:
let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
NSCharacterSet
.decimalDigitCharacterSet()
.invertedSet)
.joinWithSeparator("") // Return a string with all the numbers
This method iterate through the string characters and appends the numbers to a new string:
class func getNumberFrom(string: String) -> String {
var number: String = ""
for var c : Character in string.characters {
if let n: Int = Int(String(c)) {
if n >= Int("0")! && n < Int("9")! {
number.append(c)
}
}
}
return number
}
For example with regular expression
let text = "string_20_certified"
let pattern = "\\d+"
let regex = try! NSRegularExpression(pattern: pattern, options: [])
if let match = regex.firstMatch(in: text, options: [], range: NSRange(location: 0, length: text.characters.count)) {
let newString = (text as NSString).substring(with: match.range)
print(newString)
}
If there are multiple occurrences of the pattern use matches(in..
let matches = regex.matches(in: text, options: [], range: NSRange(location: 0, length: text.characters.count))
for match in matches {
let newString = (text as NSString).substring(with: match.range)
print(newString)
}