How to remove spaces from a string in Swift?

Question

I have the need to remove leading and trailing spaces around a punctuation character.

For example: Hello , World ... I 'm a newbie iOS Developer.

And I'd like to have: > Hello, World... I'm a newbie iOS Developer.

How can I do this? I tried to get components of the string and enumerate it by sentences. But that is not what I need

Answer 1

Simplest answer (to my knowledge) to remove all spaces in a string:

originalString.filter { $0 != " " }

Strings being arrays, we can use all Array functions on them.

Answer 2

Swift 4, 4.2 and 5

let str = "  Akbar Code  "
let trimmedString = str.trimmingCharacters(in: .whitespaces)

Answer 3

For Swift 3 or 4 you can use :

let trimmedString = string.trimmingCharacters(in: .whitespaces)

Answer 4

This is a really wonderful problem and a shame that it isn't easier to do in Swift today (someday it will be, but not today).

I kind of hate this code, but I'm getting on a plane for 20 hours, and don't have time to make it nicer. This may at least get you started using NSMutableString. It'd be nice to work in String, and Swift hates regular expressions, so this is kind of hideous, but at least it's a start.

import Foundation

let input = "Hello,  World ... I 'm a newbie iOS Developer."

let adjustments = [
    (pattern: "\\s*(\\.\\.\\.|\\.|,)\\s*", replacement: "$1 "), // elipsis or period or comma has trailing space
    (pattern: "\\s*'\\s*", replacement: "'"), // apostrophe has no extra space
    (pattern: "^\\s+|\\s+$", replacement: ""), // remove leading or trailing space
]

let mutableString = NSMutableString(string: input)

for (pattern, replacement) in adjustments {
    let re = try! NSRegularExpression(pattern: pattern)
    re.replaceMatches(in: mutableString,
                      options: [],
                      range: NSRange(location: 0, length: mutableString.length),
                      withTemplate: replacement)
}
mutableString // "Hello, World... I'm a newbie iOS Developer."

Regular expressions can be very confusing when you first encounter them. A few hints at reading these:

• The specific language Foundation uses is described by ICU.

• Backslash (\) means "the next character is special" for a regex. But inside a Swift string, backslash means "the next character is special" of the string. So you have to double them all.

• \s means "a whitespace character"

• \s* means "zero or more whitespace characters"

• \s+ means "one or more whitespace characters"

• $1 means "the thing we matched in parentheses"

• | means "or"

• ^ means "start of string"

• $ means "end of string"

• . means "any character" so to mean "an actual dot" you have to type "\\." in a Swift string.

Notice that I check for both "..." and "." in the same regular expression. You kind of have to do something like that, or else the "." will match three times inside the "...". Another approach would be to first replace "..." with "…" (the single ellipsis character, typed on a Mac by pressing Opt-;). Then "…" is a one-character punctuation. (You could also decide to re-expand all ellipsis back to dot-dot-dot at the end of the process.)

Something like this is probably how I'd do it in real life, get it done and shipped, but it may be worth the pain/practice to try to build this as a character-by-character state machine, walking one character at a time, and keeping track of your current state.

Answer 5

If you were doing this manually by processing characters arrays, you would merely need to check the previous and next characters around spaces. You can achieve the same result using functional style programming with zip, filter and map:

let testInput = "Hello , World ... I 'm a newbie iOS Developer."

let punctuation    = Set(".\',")
let previousNext   = zip( [" "] + testInput, String(testInput.dropFirst()) + [" "] )
let filteredChars  = zip(Array(previousNext),testInput)
                    .filter{  $1 != " "
                              || !($0.0 != " " && punctuation.contains($0.1))
                           }
let filteredInput = String(filteredChars.map{$1})

print(testInput)     // Hello , World ... I 'm a newbie iOS Developer.
print(filteredInput) // Hello, World... I'm a newbie iOS Developer.

Answer 6

Interesting problem; here's my stab at a non-Regex approach:

func correct(input: String) -> String {
    typealias Correction = (punctuation: String, replacement: String)

    let corrections: [Correction] = [
        (punctuation: "...", replacement: "... "),
        (punctuation: "'", replacement: "'"),
        (punctuation: ",", replacement: ", "),
    ]

    var transformed = input
    for correction in corrections {
        transformed = transformed
            .components(separatedBy: correction.punctuation)
            .map({ $0.trimmingCharacters(in: .whitespaces) })
            .joined(separator: correction.replacement)
    }

    return transformed
}

let testInput = "Hello , World ... I 'm a newbie iOS Developer."
let testOutput = correct(input: testInput)

// Hello, World... I'm a newbie iOS Developer.

Answer 7

Rob's answer is great, but you can trim it down quite a lot by taking advantage of the \p{Po} regular expression class. Getting rid of the spaces around punctuation then becomes a single regular expression replace:

import Foundation

let input = "Hello ,  World ... I 'm a newbie iOS Developer."
let result = input.replacingOccurrences(of: "\\s*(\\p{Po}\\s?)\\s*",
                                        with: "$1",
                                        options: [.regularExpression])
print(result) // "Hello, World... I'm a newbie iOS Developer."

Rob's answer also tries to trim leading/trailing spaces, but your input doesn't have any of those. If you do care about that you can just call result.trimmingCharacters(in: .whitespacesAndNewlines) on the result.

---

Here's an explanation for the regular expression. Removing the double-escapes it looks like

\s*(\p{Po}\s?)\s*

This is comprised of the following components:

• \s* - Match zero or more whitespace characters (and throw them away)
• (…) - Capturing group. Anything inside this group is preserved by the replacement (the $1 in the replacement refers to this group).
  • \p{Po} - Match a single character in the "Other_Punctuation" unicode category. This includes things like ., ', and …, but excludes things like ( or -.
  • \s? - Match a single optional whitespace character. This preserves the space after periods (or ellipses).
• \s* - Once again, match zero or more whitespace characters (and throw them away). This is what turns your , World into , World.

Answer 8

You can try something like string.replacingOccurrences(of: " ,", with: ",") for every punctuation...