How to do Inverse Regex match [regex negation] in Swift? - ios

Question

Is there a way to do inverse regular expression match and retrieve the un-matching string as return value in iOS Swift?
Let's say,
Input string: "232#$%4lion"
Regex Pattern: "[a-z]{4}"
Normal match Output: "lion"
Inverse match output: "232#$%4" (Expected result)
Please find the normal regex matching swift code below.
func regexMatch() {
let str = "232#$%4lion"
let regex = try! NSRegularExpression(pattern: "[a-z]{4}", options: .caseInsensitive)
let firstMatch = regex.firstMatch(in: str, options: .reportProgress, range: NSMakeRange(0, str.count))
guard let matches = firstMatch else {
print("No Match")
return
}
if matches.numberOfRanges > 0 {
let outputRange = matches.range(at: 0)
let startIndex = str.index(str.startIndex, offsetBy: outputRange.lowerBound)
let endIndex = str.index(str.startIndex, offsetBy: outputRange.upperBound)
print("Matched String: \(str[startIndex..<endIndex])")
}
}
I can somehow manipulate the matching result and then, can retrieve the inverse matching string by manipulating range operations. Instead of doing that,
I want to know if inverse matching can be done using regex pattern itself in Swift and directly retrieve un-matching pattern from the input string.
Above input and output values are for reference purpose only. Just to keep the question simple. Actual values in real time cases can be complex. Please answer with a generic solution.

Answer 1

You may use your regex to remove (replace with an empty string) the matches found:
let result = str.replacingOccurrences(of: "[a-z]{4}", with: "", options: .regularExpression)
Swift test:
let str = "232#$%4lion"
let result = str.replacingOccurrences(of: "[a-z]{4}", with: "", options: .regularExpression)
print(result)
Output: 232#$%4.

Answer 2

There is no need to use a regular expression for that. You can just use filter.
RangeReplaceableCollection has a filter instance method that returns Self, String conforms to RangeReplaceableCollection, so filter when used with a String returns another String.
You can combine it with the new Character property isLetter (Swift5) and create a predicate negating that property.
let str = "232#$%4lion"
let result = str.filter { !$0.isLetter }
print(result) // "232#$%4"

Answer 3

Either use this pattern, however it doesn't invert {4} because it's not predictable
"[^a-z]+"
or create a new mutable string from str and remove the found range
func invertedRegexMatch() {
let str = "232#$%4lion"
let regex = try! NSRegularExpression(pattern: "[a-z]{4}", options: .caseInsensitive)
let firstMatch = regex.firstMatch(in: str, options: .reportProgress, range: NSRange(str.startIndex..., in: str))
guard let matches = firstMatch else {
print("No Match")
return
}
if matches.numberOfRanges > 0 {
let outputRange = Range(matches.range(at: 0), in: str)!
var invertedString = str
invertedString.removeSubrange(outputRange)
print("Matched String: \(invertedString)")
}
}
Note: Use always the dedicated API to create NSRange from Range<String.Index> and vice versa
But if you just want to remove all letters there is a much simpler way
var str = "232#$%4lion"
str.removeAll{$0.isLetter}