How to Split String Using Regex Expressions

Question

I have a string "323 ECO Economics Course 451 ENG English Course 789 Mathematical Topography" I want to split this string using the regex expression [0-9][0-9][0-9][A-Z][A-Z][A-Z] so that the function returns the array:

Array = 
["323 ECO Economics Course ", "451 ENG English Course",  "789 Mathematical Topography"]

How would I go about doing this using swift?

Edit My question is different than the one linked to. I realize that you can split a string in swift using myString.components(separatedBy: "splitting string") The issue is that that question doesn't address how to make the splitting string a regex expression. I tried using mystring.components(separatedBy: "[0-9][0-9][0-9][A-Z][A-Z][A-Z]", options: .regularExpression) but that didn't work.

How can I make the separatedBy: portion a regular expression?

Perhaps you are looking at this wrong. Instead of trying to find a fancy way to "split" a string using a regex, why not simply use the `NSRegularExpression` class and its `matches` function to get all of the matches of your regex? — rmaddy, Feb 27 '17 at 01:55
The answer already done below is a great answer, however, after reading your question, I thought you might find this useful. This is a Regex class written in Swift that can be dropped into your project. I've used it in multiple projects with great ease and success. https://gist.github.com/ningsuhen/dc6e589be7f5a41e7794/ — Kyle, Feb 27 '17 at 02:06

Leo Dabus · Answer 1 · 2020-03-19T23:44:00.757

You can use regex "\\b[0-9]{1,}[a-zA-Z ]{1,}" and this extension from this answer to get all ranges of a string using literal, caseInsensitive or regularExpression search:

extension StringProtocol {
    func ranges<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Range<Index>] {
        var result: [Range<Index>] = []
        var startIndex = self.startIndex
        while startIndex < endIndex,
            let range = self[startIndex...].range(of: string, options: options) {
                result.append(range)
                startIndex = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
}

let inputString = "323 ECO Economics Course 451 ENG English Course 789 Mathematical Topography"

let courses = inputString.ranges(of: "\\b[0-9]{1,}[a-zA-Z ]{1,}", options: .regularExpression).map { inputString[$0].trimmingCharacters(in: .whitespaces) }

print(courses)   //   ["323 ECO Economics Course", "451 ENG English Course", "789 Mathematical Topography"]

If your courses codes always have 3 digits and your string have at least 3 characters, you can use regex `"\\b[0-9]{3}[a-zA-Z ]{3,}"` — Leo Dabus, Feb 27 '17 at 02:04
This is a nice clean solution. I like how you build an array of ranges and then use map to extract the substrings from the original string. Very elegant use of functional programming. (Voted) — Duncan C, Mar 07 '17 at 23:54

score 5 · Accepted Answer · answered Feb 27 '17 at 02:07

Swift doesn't have native regular expressions as of yet. But Foundation provides NSRegularExpression.

import Foundation

let toSearch = "323 ECO Economics Course 451 ENG English Course 789 MAT Mathematical Topography"

let pattern = "[0-9]{3} [A-Z]{3}"
let regex = try! NSRegularExpression(pattern: pattern, options: [])

// NSRegularExpression works with objective-c NSString, which are utf16 encoded
let matches = regex.matches(in: toSearch, range: NSMakeRange(0, toSearch.utf16.count))

// the combination of zip, dropFirst and map to optional here is a trick
// to be able to map on [(result1, result2), (result2, result3), (result3, nil)]
let results = zip(matches, matches.dropFirst().map { Optional.some($0) } + [nil]).map { current, next -> String in
  let range = current.rangeAt(0)
  let start = String.UTF16Index(range.location)
  // if there's a next, use it's starting location as the ending of our match
  // otherwise, go to the end of the searched string
  let end = next.map { $0.rangeAt(0) }.map { String.UTF16Index($0.location) } ?? String.UTF16Index(toSearch.utf16.count)

  return String(toSearch.utf16[start..<end])!
}

dump(results)

Running this will output

▿ 3 elements
  - "323 ECO Economics Course "
  - "451 ENG English Course "
  - "789 MAT Mathematical Topography"

Ikechukwu Eze · Answer 3 · 2021-09-20T08:52:22.363

I needed something like this and should work more like JS String.prototype.split(pat: RegExp) or Rust's String.splitn(pat: Pattern<'a>) but with Regex. I ended up with this

extension NSRegularExpression {
    convenience init(_ pattern: String) {...}
    
    
    /// An array of substring of the given string, separated by this regular expression, restricted to returning at most n items.
    /// If n substrings are returned, the last substring (the nth substring) will contain the remainder of the string.
    /// - Parameter str: String to be matched
    /// - Parameter n: If `n` is specified and n != -1, it will be split into n elements else split into all occurences of this pattern
    func splitn(_ str: String, _ n: Int = -1) -> [String] {
        let range = NSRange(location: 0, length: str.utf8.count)
        let matches = self.matches(in: str, range: range);
        
        var result = [String]()
        if (n != -1 && n < 2) || matches.isEmpty  { return [str] }
        
        if let first = matches.first?.range {
            if first.location == 0 { result.append("") }
            if first.location != 0 {
                let _range = NSRange(location: 0, length: first.location)
                result.append(String(str[Range(_range, in: str)!]))
            }
        }
        
        for (cur, next) in zip(matches, matches[1...]) {
            let loc = cur.range.location + cur.range.length
            if n != -1 && result.count + 1 == n {
                let _range = NSRange(location: loc, length: str.utf8.count - loc)
                result.append(String(str[Range(_range, in: str)!]))
                return result
                
            }
            let len = next.range.location - loc
            let _range = NSRange(location: loc, length: len)
            result.append(String(str[Range(_range, in: str)!]))
        }
        
        if let last = matches.last?.range, !(n != -1 && result.count >= n) {
            let lastIndex = last.length + last.location
            if lastIndex == str.utf8.count { result.append("") }
            if lastIndex < str.utf8.count {
                let _range = NSRange(location: lastIndex, length: str.utf8.count - lastIndex)
                result.append(String(str[Range(_range, in: str)!]))
            }
        }
        
        return result;
    }
    
}

Passes the following tests

func testRegexSplit() {
        XCTAssertEqual(NSRegularExpression("\\s*[.]\\s+").splitn("My . Love"), ["My", "Love"])
        XCTAssertEqual(NSRegularExpression("\\s*[.]\\s+").splitn("My . Love . "), ["My", "Love", ""])
        XCTAssertEqual(NSRegularExpression("\\s*[.]\\s+").splitn(" . My . Love"), ["", "My", "Love"])
        XCTAssertEqual(NSRegularExpression("\\s*[.]\\s+").splitn(" . My . Love . "), ["", "My", "Love", ""])
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX"), ["", "My", "", "Love", ""])
    }



func testRegexSplitWithN() {
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX", 1), ["xXMyxXxXLovexX"])
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX", -1), ["", "My", "", "Love", ""])
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX", 2), ["", "MyxXxXLovexX"])
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX", 3), ["", "My", "xXLovexX"])
        XCTAssertEqual(NSRegularExpression("xX").splitn("xXMyxXxXLovexX", 4), ["", "My", "", "LovexX"])
    }

func testNoMatches() {
        XCTAssertEqual(NSRegularExpression("xX").splitn("MyLove", 1), ["MyLove"])
        XCTAssertEqual(NSRegularExpression("xX").splitn("MyLove"), ["MyLove"])
        XCTAssertEqual(NSRegularExpression("xX").splitn("MyLove", 3), ["MyLove"])
    }

I found this crashed for me if I provided a string that had 0 matches to the pattern. I ended up using this instead: https://gist.github.com/hcrub/218e1d25f1659d00b7f77aebfcebf15a — Patrick, Sep 19 '21 at 08:50

score 0 · Answer 4 · answered Nov 01 '22 at 10:29

Update to @tomahh answer for latest Swift (5).

import Foundation

let toSearch = "323 ECO Economics Course 451 ENG English Course 789 MAT Mathematical Topography"

let pattern = "[0-9]{3} [A-Z]{3}"
let regex = try! NSRegularExpression(pattern: pattern, options: [])

let matches = regex.matches(in: toSearch, range: NSRange(toSearch.startIndex..<toSearch.endIndex, in: toSearch))

// the combination of zip, dropFirst and map to optional here is a trick
// to be able to map on [(result1, result2), (result2, result3), (result3, nil)]
let results = zip(matches, matches.dropFirst().map { Optional.some($0) } + [nil]).map { current, next -> String in
  let start = toSearch.index(toSearch.startIndex, offsetBy: current.range.lowerBound)
  let end = next.map(\.range).map { toSearch.index(toSearch.startIndex, offsetBy: $0.lowerBound) } ?? toSearch.endIndex
  return String(toSearch[start..<end])
}

dump(results)

▿ 3 elements
  - "323 ECO Economics Course "
  - "451 ENG English Course "
  - "789 MAT Mathematical Topography"

How to Split String Using Regex Expressions

4 Answers4

Linked