How to use regex to split string into groups of identical characters? - regex

Question

I got a string like this:
var string = "AAAAAAABBBCCCCCCDD"
and like to split the string into an array of this format (same characters --> same group) using regular expressions:
Array: "AAAAAAA", "BBB", "CCCCCC", "DD"
This Is what I got so far but tbh I can not really get it working.
var array = [String]()
var string = "AAAAAAABBBCCCCCCDD"
let pattern = "\\ b([1,][a-z])\\" // mistake?!
let regex = try! NSRegularExpression(pattern: pattern, options: [])
array = regex.matchesInString(string, options: [], range: NSRange(location: 0, length: string.count))

Answer 1

You can achieve that using this function from this answer:
func matches(for regex: String, in text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex)
let results = regex.matches(in: text,
range: NSRange(text.startIndex..., in: text))
return results.map {
String(text[Range($0.range, in: text)!])
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
Passing (.)\\1+ as regex and AAAAAAABBBCCCCCCDD as text like this:
let result = matches(for: "(.)\\1+", in: "AAAAAAABBBCCCCCCDD")
print(result) // ["AAAAAAA", "BBB", "CCCCCC", "DD"]

Answer 2

You can achieve that with a "back reference", compare
NSRegularExpression:
\n
Back Reference. Match whatever the nth capturing group matched. n must be a number ≥ 1 and ≤ total number of capture groups in the pattern.
Example (using the utility method from Swift extract regex matches):
let string = "AAAAAAABBBCCCCCCDDE"
let pattern = "(.)\\1*"
let array = matches(for: pattern, in: string)
print(array)
// ["AAAAAAA", "BBB", "CCCCCC", "DD", "E"]
The pattern matches an arbitrary character, followed by zero or more
occurrences of the same character. If you are only interested in
repeating word characters use
let pattern = "(\\w)\\1*"
instead.