Split text into array while maintaining the punctuation in Swift - ios

Question

I want to split the text into an array, maintaining the punctuation separated by the rest of the words, so a string like:
Hello, I am Albert Einstein.
should turn into an array like this:
["Hello", ",", "I", "am", "Albert", "Einstein", "."]
I have tried with sting.components(separatedBy: CharacterSet.init(charactersIn: " ,;;:")) but this method deletes all punctuations, and returns an array like this:
["Hello", "I", "am", "Albert", "Einstein"]
So, how can I get an array like my first example?

Answer 1

It's not beautiful as solution but you can try with:
var str = "Hello, I am Albert Einstein."
var list = [String]()
var currentSubString = "";
//enumerate to get all characters including ".", ",", ";", " "
str.enumerateSubstrings(in: str.startIndex..<str.endIndex, options: String.EnumerationOptions.byComposedCharacterSequences) { (substring, substringRange, enclosingRange, value) in
if let _subString = substring {
if (!currentSubString.isEmpty &&
(_subString.compare(" ") == .orderedSame
|| _subString.compare(",") == .orderedSame
|| _subString.compare(".") == .orderedSame
|| _subString.compare(";") == .orderedSame
)
) {
//create word if see any of those character and currentSubString is not empty
list.append(currentSubString)
currentSubString = _subString.trimmingCharacters(in: CharacterSet.whitespaces )
} else {
//add to current sub string if current character is not space.
if (_subString.compare(" ") != .orderedSame) {
currentSubString += _subString
}
}
}
}
//last word
if (!currentSubString.isEmpty) {
list.append(currentSubString)
}
In Swift3:
var str = "Hello, I am Albert Einstein."
var list = [String]()
var currentSubString = "";
//enumerate to get all characters including ".", ",", ";", " "
str.enumerateSubstrings(in: str.startIndex..<str.endIndex, options: String.EnumerationOptions.byComposedCharacterSequences) { (substring, substringRange, enclosingRange, value) in
if let _subString = substring {
if (!currentSubString.isEmpty &&
(_subString.compare(" ") == .orderedSame
|| _subString.compare(",") == .orderedSame
|| _subString.compare(".") == .orderedSame
|| _subString.compare(";") == .orderedSame
)
) {
//create word if see any of those character and currentSubString is not empty
list.append(currentSubString)
currentSubString = _subString.trimmingCharacters(in: CharacterSet.whitespaces )
} else {
//add to current sub string if current character is not space.
if (_subString.compare(" ") != .orderedSame) {
currentSubString += _subString
}
}
}
}
//last word
if (!currentSubString.isEmpty) {
list.append(currentSubString)
}
The idea is to loop for all character and create word in same time. A word is a group of consecutive character that is not , ,, . or ;. So, during the creation of word in loop, we finish the current word if we see one of those character, and the current word in construction is not empty.
To break down steps with your input:
get H (not space nor other terminal character)
-> currentSubString = "H"
get e (not space nor other terminal character)
-> currentSubString = "He"
get l (not space nor other terminal character)
-> currentSubString = "Hel"
get l (not space nor other terminal character)
-> currentSubString = "Hell"
get o (not space nor other terminal character)
-> currentSubString = "Hello"
get . (is terminal character)
-> as currentSubString is not empty, add to list and restart the construction for next word, then list = ["Hello"]
-> currentSubString = "." (the reason that I used trimming is just to remove if I get this character. but for other terminal character, we have to keep for next word.
get (is space character)
-> as currentSubString is not empty, add to listand restart the construction -> list = ["Hello", "."]
-> currentSubString = "" (trimmed).
... and so on.

Answer 2

To explain from my comment... Think of regular expressions as a way to nicely find patterns within Strings. In your case, the pattern is words (groups of letters) with other possible symbols (punctuation marks) in between.
Take the regex in my comment (which I've expanded a bit here), for example: ([,\.\:\"])*([A-Za-z0-9\']*)([,\.\:\"])*
In there, we have 3 groups. The first searches for any symbols (such as a leading quotation mark). The second is searching for letters, numbers, and an apostrophe (because people like to concatenate words, like "I'm"). and the third group searches for any trailing punctuation marks.
Edit to note: groups in the above are denoted by parentheses ( and ), while the [ and ] brackets denote acceptable characters for a search. So, for example, [A-Z] says that all upper case letters from A-Z are acceptable. [A-Za-z] lets you get both upper and lower, while [A-Za-z0-9] includes all letters and numbers from 0-9. Granted, there are shorthand versions to writing this, but those you'll discover down the road.
So now we have a way to separate all the words and punctuation marks, now you need to actually use it, doing something along the lines of:
func find(value: NSString) throws -> [NSString] {
let regex = try NSRegularExpression(pattern: "([,\\.\\:\\\"])*([A-Za-z0-9\\']*)([,\\.\\:\\\"])*") // Notice you have to escape the values in code
let results = regex.matches(in: value, range: NSRange(location: 0, length: nsString.length))
return results.map({ value.substring(with: $0.range) }).filter({ $0 != nil })
}
That should give you each non-nil group found within the String value you supply to the method.
Granted, that last filter method may not be necessary, but I'm not familiar enough with how Swift handles regex to know for sure.
But that should definitely point you in the right direction...
Cheers~