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Swift - search for a character at multiple positions in a string

In Swift with the following line: "this is a string", how do I get an array of indices where the character "" (space) is present in the string?

Desired result: [4,7,9]

I tried:

let spaces: NSRange = full_string.rangeOfString(" ")

But this only returns 4, not all indices.

Any idea?

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2 answers


Here's a simple approach:

let string = "this is a string"

let indices = string
    .characters
    .enumerate()
    .filter { $0.element == " " }
    .map { $0.index }

print(indices) // [4, 7, 9]
  • characters

    returns CharacterView

    which is CollectionType

    (similar to an array of single characters)
  • enumerate

    converts this collection to SequenceType

    tuples containing index

    (0 to 15) and element

    (every single character)
  • filter

    removes tuples for which characters are not spaces
  • map

    converts an array of tuples to an array of indices only


This approach requires Swift 2 (in Xcode 7 beta). From the comments, the Swift 1.2 syntax is:

let indices = map(filter(enumerate(string), { $0.element == " " }), { $0.index } )

(tip for Martin R hat ).

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Solution for Swift 1.2 using Regex



func searchPattern(pattern : String, inString string : String) -> [Int]?
{
  let regex = NSRegularExpression(pattern: pattern, options: NSRegularExpressionOptions(), error: nil)
  return regex?.matchesInString(string, options: NSMatchingOptions(), range: NSRange(location:0, length:count(string)))
               .map { ($0 as! NSTextCheckingResult).range.location }
}

let string = "this is a string"

searchPattern("\\s\\S", inString : string) // [4, 7, 9]
searchPattern("i", inString : string) // [2, 5, 13]
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