CODEWITHSUNDEEP
swiftregex
John Aron
John Aron

Reputation: 41

Swift - Regex - Extract numbers in brackets divided by comma(1,1)(2,3)

I need to extract the numbers from a string ex. "(1,1)(3,3)(4,4)". I tried with "\d(,)\d)", but I get the comma too. How can I exclude the comma and make e tuple or an array of arrays [[1,1],[3,3]...

let str = "(1,1) (2,2) (3,3)"
let pattern = "\\d(,)\\d"
let regex = try! NSRegularExpression(pattern:pattern)
let results = regex.matches(in: str, options: [], range: range)

Upvotes: 3

Views: 324

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You can use

let s = "(1,1) (2,2) (3,3)"
print(s.replacingOccurrences(of: #"\)\s*\("#, with: "|",  options: .regularExpression)
  .trimmingCharacters(in: ["(", ")"])
  .components(separatedBy: "|")
  .map { $0.components(separatedBy: [",", "."]).map { Int($0)!} }
)

Output: [[1, 1], [2, 2], [3, 3]].

This will work even if you have a single number inside parentheses. If you have let s = "(1,1) (3,3)(4.4)(4)", the output will be [[1, 1], [3, 3], [4, 4], [4]].

Details

  • \)\s*\( matches ), 0+ whitespace chars, (, .replacingOccurrences(of: #"\)\s*\("#, with: "|", options: .regularExpression) replaces all these matches with |
  • .trimmingCharacters(in: ["(", ")"]) removes ( and ) at the start/end of the string
  • .components(separatedBy: "|") splits the resulting string with |
  • .map { $0.components(separatedBy: [",", "."]).map { Int($0)!} } splits each tuple into constituent numbers and then casts to Int.

Upvotes: 1

vadian
vadian

Reputation: 285069

A possible solution is to capture only the digits, replace each occurrence with "[$1,$2]", then insert commas between, wrap the expression in square brackets to get JSON and decode it

let str =  "(1,1)(3,3)(4.4)"
let str1 = str.replacingOccurrences(of: "\\((\\d+)[.,](\\d+)\\)", with: "[$1,$2]", options: .regularExpression).replacingOccurrences(of: "][", with: "],[")
let result = try! JSONDecoder().decode([[Int]].self, from: Data(("[" + str1 + "]").utf8))

The intermediate results are:

  • first replacingOccurrences: "[1,1][3,3][4,4]"
  • second replacingOccurrences: "[1,1],[3,3],[4,4]"
  • Argument of Data(... "[[1,1],[3,3],[4,4]]"

Upvotes: 2

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