Function that separates sentences and questions in swift - ios

I want the function below to separate the sentences into an array and the questions into an array and inserting "," where the "." and "?" belong. At the moment it's printing both in the same array. Any ideas on how to fix this?
func separateAllSentences() {
// needs to print just the sentences
func separateDeclarations() { // AKA Separate sentences that end in "."
if userInput.range(of: ".") != nil { // Notice how lowercased() wasn't used
numSentencesBefore = userInput.components(separatedBy: ".") // Hasn't subtracted 1 yet
numSentencesAfter = numSentencesBefore.count - 1
separateSentencesArray = Array(numSentencesBefore)
print("# Of Sentences = \(numSentencesAfter)")
print(separateSentencesArray)
} else {
print("There are no declarations found.")
}
}
// needs to print just the questions
func separateQuestions() { // Pretty Self Explanitory
if userInput.range(of: "?") != nil {
numQuestionsBefore = userInput.components(separatedBy: "?")
numQuestionsAfter = numQuestionsBefore.count - 1
separateQuestionsArray = Array(numQuestionsBefore)
print("# Of Questions = \(numQuestionsAfter)")
print(separateQuestionsArray)
} else {
print("There are no questions found. I have nothing to solve. Please rephrase the work to solve as a question.")
}
}
// TODO: - Separate Commas
func separateCommas() {
}
separateDeclarations()
separateQuestions()
}
Console Prints Out:
Ned rode his bike 7 miles to the library.
He took a shortcut on the way home which was only 5 miles long.
How many miles did Ned ride altogether?
[# Of Sentences = 2]
["Ned rode his bike 7 miles to the library", "\nHe took a shortcut on the way home which was only 5 miles long", "\nHow many miles did Ned ride altogether?\n"]
[# Of Questions = 1]
["Ned rode his bike 7 miles to the library.\nHe took a shortcut on the way home which was only 5 miles long.\nHow many miles did Ned ride altogether", "\n"]
Ned rode his bike 7 miles to the library.
He took a shortcut on the way home which was only 5 miles long.
How many miles did Ned ride altogether?
It Should Print Out
[# Of Sentences = 2]
[# Of Questions = 1]
Sentences: ["Ned rode his bike 7 miles to the library. He took a shortcut on the way home which was only 5 miles long."]
Questions: ["How many miles did Ned ride altogether?"]

I would suggest not separating based upon the presence of a character, but rather enumate using the .bySentences option (which can more gracefully handle punctuation which is not terminating a sentence). Then iterate once through your string, appending to the appropriate array, e.g. in Swift 3:
var questions = [String]()
var statements = [String]()
var unknown = [String]()
let string = "Ned deployed version 1.0 of his app. He wrote very elegant code. How much money did Ned make?"
string.enumerateSubstrings(in: string.startIndex ..< string.endIndex, options: .bySentences) { string, _, _, _ in
if let sentence = string?.trimmingCharacters(in: .whitespacesAndNewlines), let lastCharacter = sentence.characters.last {
switch lastCharacter {
case ".":
statements.append(sentence)
case "?":
questions.append(sentence)
default:
unknown.append(sentence)
}
}
}
print("questions: \(questions)")
print("statements: \(statements)")
print("unknown: \(unknown)")

This snippet could use some refactoring to replace the common code but it works as is.
let punctuation = CharacterSet(charactersIn: ".?")
let sentences = userInput.components(separatedBy: punctuation)
let questions = sentences.filter {
guard let range = userInput.range(of: $0) else { return false }
let start = range.upperBound
let end = userInput.index(after: start)
let punctuation = userInput.substring(with: Range(uncheckedBounds: (start, end)))
return punctuation == "?"
}
let statements = sentences.filter {
guard let range = userInput.range(of: $0) else { return false }
let start = range.upperBound
let end = userInput.index(after: start)
let punctuation = userInput.substring(with: Range(uncheckedBounds: (start, end)))
return punctuation == "."
}
Looking at the closure first, the range variable contains the indices of the sentence in the user input. We want to get the punctuation trailing that particular sentence so we start with its upper bound and find the next index past it. Using substring, we extract the punctuation and compare it to either . or ?.
Now that we have code that will return true or false whether we have a question or statement sentence, we use filter to iterate over the array of all sentences and return only an array of questions or statements.

i make a simplest solution
func separateDeclarations(userInput: String) { // AKA Separate sentences that end in "."
let array = userInput.components(separatedBy: ".")
let arrayQ = userInput.components(separatedBy: "?")
arrayQ.map { (string) -> String in
if let question = string.components(separatedBy: ".").last {
return question
} else {
return ""
}
}
array.map { (string) -> String in
if let question = string.components(separatedBy: "?").last {
return question
} else {
return ""
}
}
}

Related

Get elements separate from array

What is the best way or elegant way to get all elements whithout spaces and ignore "DOMICILIO" and the next elements for example :
"LOAIZA"
"HERRERA"
"JESUS" (This is my expected output)
In this case I have one string with 2 elements ("LOAIZA\nHERRERA")
["LOAIZA HERRERA", "JESUS", "DOMICILIO", "CALLE1", "CALLE2"]
var dataID = ["LOAIZA HERRERA", "JESUS", "DOMICILIO"]
for i in dataID {
if i.contains(" "){
print(i) // LOAIZA HERRERA
let dataSeparate = i.components(separatedBy: " ")
print(dataSeparate) // ["LOAIZA", "HERRERA"]
}
}
Separate the terms by " " and flatMap into a new array. Then, find the index of "DOMICILIO" if it exists and use the segment of the array up to that point.
func findResult(dataID: [String]) -> Array<String> {
let terms = dataID.flatMap { $0.components(separatedBy: " ")}
let indexOfDom = terms.firstIndex(of: "DOMICILIO")
if let indexOfDom = indexOfDom, indexOfDom > 0 {
return Array(terms[0...(indexOfDom - 1)])
} else {
return terms
}
}

How to split string as English and non English using Swift 4?

I have a string which contains English and Arabic together. I am using an API, that is why I cannot set an indicator in it.
What I want to get is: the Arabic and English split into tow parts. Here is a sample String:
"بِاسْمِكَ رَبِّي وَضَعْتُ جَنْبِي، وَبِكَ أَرْفَعُهُ، فَإِنْ أَمْسَكْتَ نَفْسِي فَارْحَمْهَا، وَإِنْ أَرْسَلْتَهَا فَاحْفَظْهَا، بِمَا تَحْفَظُ بِهِ عِبَادَكَ الصَّالِحِينَ.Bismika rabbee wadaAAtu janbee wabika arfaAAuh, fa-in amsakta nafsee farhamha, wa-in arsaltaha fahfathha bima tahfathu bihi AAibadakas-saliheen. In Your name my Lord, I lie down and in Your name I rise, so if You should take my soul then have mercy upon it, and if You should return my soul then protect it in the manner You do so with Your righteous servants.",
I cannot find how to split it into 2 parts that I get Arabic and English into two different parts.
What I want:
so there can be any language, my problem is to only take out English or Arabic language and show them in respective fields.
How can I achieve it?
You can use a Natural Language Tagger, which would work even if both scripts are intermingled:
import NaturalLanguage
let str = "¿como? بداية start وسط middle начать средний конец نهاية end. 從中間開始. "
let tagger = NLTagger(tagSchemes: [.script])
tagger.string = str
var index = str.startIndex
var dictionary = [String: String]()
var lastScript = "other"
while index < str.endIndex {
let res = tagger.tag(at: index, unit: .word, scheme: .script)
let range = res.1
let script = res.0?.rawValue
switch script {
case .some(let s):
lastScript = s
dictionary[s, default: ""] += dictionary["other", default: ""] + str[range]
dictionary.removeValue(forKey: "other")
default:
dictionary[lastScript, default: ""] += str[range]
}
index = range.upperBound
}
print(dictionary)
and print the result if you'd like:
for entry in dictionary {
print(entry.key, ":", entry.value)
}
yielding :
Hant : 從中間開始.
Cyrl : начать средний конец
Arab : بداية وسط نهاية
Latn : ¿como? start middle end.
This is still not perfect since the language tagger only checks to which script the most number of letters in a word belong to. For example, in the string you're working with, the tagger would consider الصَّالِحِينَ.Bismika as one word. To overcome this, we could use two pointers and traverse the original string and check the script of words individually. Words are defined as contiguous letters:
let str = "بِاسْمِكَ رَبِّي وَضَعْتُ جَنْبِي، وَبِكَ أَرْفَعُهُ، فَإِنْ أَمْسَكْتَ نَفْسِي فَارْحَمْهَا، وَإِنْ أَرْسَلْتَهَا فَاحْفَظْهَا، بِمَا تَحْفَظُ بِهِ عِبَادَكَ الصَّالِحِينَ.Bismika rabbee wadaAAtu janbee wabika arfaAAuh, fa-in amsakta nafsee farhamha, wa-in arsaltaha fahfathha bima tahfathu bihi AAibadakas-saliheen. In Your name my Lord, I lie down and in Your name I rise, so if You should take my soul then have mercy upon it, and if You should return my soul then protect it in the manner You do so with Your righteous servants."
let tagger = NLTagger(tagSchemes: [.script])
var i = str.startIndex
var dictionary = [String: String]()
var lastScript = "glyphs"
while i < str.endIndex {
var j = i
while j < str.endIndex,
CharacterSet.letters.inverted.isSuperset(of: CharacterSet(charactersIn: String(str[j]))) {
j = str.index(after: j)
}
if i != j { dictionary[lastScript, default: ""] += str[i..<j] }
if j < str.endIndex { i = j } else { break }
while j < str.endIndex,
CharacterSet.letters.isSuperset(of: CharacterSet(charactersIn: String(str[j]))) {
j = str.index(after: j)
}
let tempo = String(str[i..<j])
tagger.string = tempo
let res = tagger.tag(at: tempo.startIndex, unit: .word, scheme: .script)
if let s = res.0?.rawValue {
lastScript = s
dictionary[s, default: ""] += dictionary["glyphs", default: ""] + tempo
dictionary.removeValue(forKey: "glyphs")
}
else { dictionary["other", default: ""] += tempo }
i = j
}
You can use the NaturalLanguageTagger as answered by #ielyamani but the only limitation is that it is iOS 12+
If you are trying to do this on earlier iOS versions, you can take a look at NSCharacterSet
You can create your own characterset to check whether a string has english characters and numbers
extension String {
func containsLatinCharacters() -> Bool {
var charSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890")
charSet = charSet.invertedSet
let range = (self as NSString).rangeOfCharacterFromSet(charSet)
if range.location != NSNotFound {
return false
}
return true
}
}
Another option is to use the charactersets already available:
let nonLatinString = string.trimmingCharacters(in: .alphanumerics)//symbols will still get through
let latinString = string.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)//symbols and non-latin characters wont get through
With these you can get the strings you want quite easily. But if these are not good enough, you can look to create your own characterset, use union, intersect etc to filter out the wanted and the unwanted characters.
Step 1:
You have to split whole string into an array by "." as I can see there are "." between sentence.
Step 2:
Pass each sentence to determine its language and append into different string.
Final Code
//add in your viewController
enum Language : String {
case arabic = "ar"
case english = "en"
}
override func viewDidLoad() {
super.viewDidLoad()
//make array of string
let kalmaArray = "بِاسْمِكَ رَبِّي وَضَعْتُ جَنْبِي، وَبِكَ أَرْفَعُهُ، فَإِنْ أَمْسَكْتَ نَفْسِي فَارْحَمْهَا، وَإِنْ أَرْسَلْتَهَا فَاحْفَظْهَا، بِمَا تَحْفَظُ بِهِ عِبَادَكَ الصَّالِحِينَ.Bismika rabbee wadaAAtu janbee wabika arfaAAuh, fa-in amsakta nafsee farhamha, wa-in arsaltaha fahfathha bima tahfathu bihi AAibadakas-saliheen. In Your name my Lord, I lie down and in Your name I rise, so if You should take my soul then have mercy upon it, and if You should return my soul then protect it in the manner You do so with Your righteous servants.".components(separatedBy: ".")
splitInLanguages(kalmaArray: kalmaArray)
}
private func splitInLanguages(kalmaArray: [String]){
var englishText = ""
var arabicText = ""
for kalma in kalmaArray {
if kalma.count > 0 {
if let language = NSLinguisticTagger.dominantLanguage(for: kalma) {
switch language {
case Language.arabic.rawValue:
arabicText.append(kalma)
arabicText.append(".")
break
default: // English
englishText.append(kalma)
englishText.append(".")
break
}
} else {
print("Unknown language")
}
}
}
debugPrint("Arabic: ", arabicText)
debugPrint("English: ", englishText)
}
I hope it will help you to split the string in two language. Let me know if you are still having any issue.

Comparing quantities of unique elements contained in large data sets [duplicate]

This question already has answers here:
Refactored Solution In Swift
(2 answers)
Closed 6 years ago.
I'm attempting to solve HackerRank's Hash Table Ransom Note challenge. There are 19 test cases and I'm passing all but two of time due to timeout on larger data sets (10,000-30,000 entries).
I'm given:
1) an array of words contained in a magazine and
2) an array of words for a ransom note. My objective is to determine if the words in the magazine can be used to construct a ransom note.
I need to have enough unique elements in the magazineWords to satisfy the quantity needed by noteWords.
I'm using this code to make that determination...and it takes FOREVER...
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
What is a faster way to accomplish this task?
Below is my complete code for the challenge:
import Foundation
var magazineWords = // Array of 1 to 30,000 strings
var noteWords = // Array of 1 to 30,000 strings
enum RegexString: String {
// Letters a to z, A to Z, 1 to 5 characters long
case wordCanBeUsed = "([a-zA-Z]{1,5})"
}
func matches(for regexString: String, in text: String) -> [String] {
// Hat tip MartinR for this
do {
let regex = try NSRegularExpression(pattern: regexString)
let nsString = text as NSString
let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range)}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
func canCreateRansomNote(from magazineWords: [String], for noteWords: [String]) -> String {
// figure out what's unique
let magazineWordsSet = Set(magazineWords)
let noteWordsSet = Set(noteWords)
let intersectingValuesSet = magazineWordsSet.intersection(noteWordsSet)
// constraints specified in challenge
guard magazineWords.count >= 1, noteWords.count >= 1 else { return "No" }
guard magazineWords.count <= 30000, noteWords.count <= 30000 else { return "No" }
// make sure there are enough individual words to work with
guard magazineWordsSet.count >= noteWordsSet.count else { return "No" }
guard intersectingValuesSet.count == noteWordsSet.count else { return "No" }
// check if all the words can be used. assume the regex method works perfectly
guard noteWords.count == matches(for: RegexString.wordCanBeUsed.rawValue, in: noteWords.joined(separator: " ")).count else { return "No" }
// FIXME: this is a processor hog. I'm timing out when I get to this point
// need to make sure there are enough magazine words to write the note
// compare quantity of word in magazine with quantity of word in note
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
return "Yes"
}
print(canCreateRansomNote(from: magazineWords, for: noteWords))
I don't know how to read from the test case on the contest website or what frameworks you are allowed. If Foundation is allowed, you can use NSCountedSet
import Foundation
let fileContent = try! String(contentsOf: URL(fileURLWithPath: "/path/to/file.txt"))
let scanner = Scanner(string: fileContent)
var m = 0
var n = 0
scanner.scanInt(&m)
scanner.scanInt(&n)
var magazineWords = NSCountedSet(capacity: m)
var ransomWords = NSCountedSet(capacity: n)
for i in 0..<(m+n) {
var word: NSString? = nil
scanner.scanUpToCharacters(from: .whitespacesAndNewlines, into: &word)
if i < m {
magazineWords.add(word!)
} else {
ransomWords.add(word!)
}
}
var canCreate = true
for w in ransomWords {
if ransomWords.count(for: w) > magazineWords.count(for: w) {
canCreate = false
break
}
}
print(canCreate ? "Yes" : "No")
It works by going through the input file one word at a time, counting how many times that word appears in the magazine and in the ransom note. Then if any word appear more frequently in the ransom note than in the magazine, it fails the test immediately. Run the 30,000 words test case in less than 1 second on my iMac 2012.

Refactored Solution In Swift

I've been studying for a coding exam by doing the HackerRank test cases, for the most part I've been doing well, but I get hung up on some easy cases and you all help me when I can't see the solution. I'm working on this problem:
https://www.hackerrank.com/challenges/ctci-ransom-note
A kidnapper wrote a ransom note but is worried it will be traced back to him. He found a magazine and wants to know if he can cut out whole words from it and use them to create an untraceable replica of his ransom note. The words in his note are case-sensitive and he must use whole words available in the magazine, meaning he cannot use substrings or concatenation to create the words he needs.
Given the words in the magazine and the words in the ransom note, print Yes if he can replicate his ransom note exactly using whole words from the magazine; otherwise, print No.
Input Format
The first line contains two space-separated integers describing the respective values of (the number of words in the magazine) and (the number of words in the ransom note).
The second line contains space-separated strings denoting the words present in the magazine.
The third line contains space-separated strings denoting the words present in the ransom note.
Each word consists of English alphabetic letters (i.e., to and to ).
The words in the note and magazine are case-sensitive.
Output Format
Print Yes if he can use the magazine to create an untraceable replica of his ransom note; otherwise, print No.
Sample Input
6 4
give me one grand today night
give one grand today
Sample Output
Yes
Explanation
All four words needed to write an untraceable replica of the ransom note are present in the magazine, so we print Yes as our answer.
And here is my solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
// Get list of elements that intersect in each array
let filtered = Set(a).intersection(Set(b))
// Map set to set of Boolean where true means set a has enough words to satisfy set b's needs
let checkB = filtered.map{ word in reduceSet(b, word: word) <= reduceSet(a, word: word) }
// If mapped set does not contain false, answer is Yes, else No
return !checkB.contains(false) ? "Yes" : "No"
}
func reduceSet(_ a: [String], word: String) -> Int {
return (a.reduce(0){ $0 + ($1 == word ? 1 : 0)})
}
print(main())
I always time out on three of the 20 test-cases with this solution. So the solution seems to solve all the test cases, but not within their required time constraints. These are great practice, but it's so extremely frustrating when you get stuck like this.
I should note that I use Sets and the Set(a).intersection(Set(b)) because when I tried mapping an array of Strings, half the test-cases timed out.
Any cleaner, or more efficient solutions will be greatly appreciated! Thank you!
Thanks to #Alexander - I was able to solve this issue using NSCountedSet instead of my custom reduce method. It's much cleaner and more efficient. Here is the solution:
import Foundation
func main() -> String {
let v = readLine()!.components(separatedBy: " ").map{Int($0)!}
var a = [String](); var b = [String]()
if v[0] < v[1] { return "No"}
for i in 0 ..< 2 {
if i == 0 {
a = (readLine()!).components(separatedBy: " ")
} else { b = (readLine()!).components(separatedBy: " ") }
}
let countA = NSCountedSet(array: a)
let countB = NSCountedSet(array: b)
let intersect = Set(a).intersection(Set(b))
let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
return !check.contains(false) ? "Yes" : "No"
}
print(main())
Many thanks!
I took the leisure of making some improvements on your code. I put comments to explain the changes:
import Foundation
func main() -> String {
// Give more meaningful variable names
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
// a guard reads more like an assertion, stating the affirmative, as opposed to denying the negation.
// it also
guard magazineWordCount > ransomNoteWordCount else { return "No" }
// Don't use a for loop if it only does 2 iterations, which are themselves hardcoded in.
// Just write the statements in order.
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ") //You don't need ( ) around readLine()!
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
// intersect is a verb. you're looking for the noun, "intersection"
// let intersection = Set(a).intersection(Set(b))
// let check = intersect.map{ countB.count(for: $0) <= countA.count(for: $0) }
// You don't actually care for the intersection of the two sets.
// You only need to worry about exactly the set of words that
// exists in the ransom note. Just check them directly.
let hasWordWithShortage = ransomNoteWordCounts.contains(where: { word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
})
// Don't negate the condition of a conditional expression. Just flip the order of the last 2 operands.
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
with the comments removed:
import Foundation
func main() -> String {
let firstLine = readLine()!.components(separatedBy: " ").map{Int($0)!}
let (magazineWordCount, ransomNoteWordCount) = (firstLine[0], firstLine[1])
guard magazineWordCount > ransomNoteWordCount else { return "No" }
let magazineWords = readLine()!.components(separatedBy: " ")
let ransomNoteWords = readLine()!.components(separatedBy: " ")
let magazineWordCounts = NSCountedSet(array: magazineWords)
let ransomNoteWordCounts = NSCountedSet(array: ransomNoteWords)
let hasWordWithShortage = ransomNoteWordCounts.contains{ word in
magazineWordCounts.count(for: word) < ransomNoteWordCounts.count(for: word)
}
return hasWordWithShortage ? "No" : "Yes"
}
print(main())
It's simpler, and much easier to follow. :)

Replace part of string with lower case letters - Swift

I have a Swift based iOS app and one of the features allows you to comment on a post. Anyway, users can add "#mentions" in their posts to tag other people. However I want to stop the user from adding a username with a capital letter.
Is there anyway I can convert a string, so that the #usernames are all in lowercase?
For example:
I really enjoy sightseeing with #uSerABC (not allowed)
I really enjoy sightseeing with #userabc (allowed)
I know there is a property for the string in swift called .lowercaseString - but the problem with that, is that it makes the entire string lowercase and thats not what I want. I only want the #username to be in lower case.
Is there any way around this with having to use the .lowercase property.
Thanks for your time, Dan.
This comes from a code I use to detect hashtags, I've modified to detect mentions:
func detectMentionsInText(text: String) -> [NSRange]? {
let mentionsDetector = try? NSRegularExpression(pattern: "#(\\w+)", options: NSRegularExpressionOptions.CaseInsensitive)
let results = mentionsDetector?.matchesInString(text, options: NSMatchingOptions.WithoutAnchoringBounds, range: NSMakeRange(0, text.utf16.count)).map { $0 }
return results?.map{$0.rangeAtIndex(0)}
}
It detects all the mentions in a string by using a regex and returns an NSRange array, by using a range you have the beginning and the end of the "mention" and you can easily replace them with a lower case version.
Split the string into two using the following command -
let arr = myString.componentsSeparatedByString("#")
//Convert arr[1] to lower case
//Append to arr[0]
//Enjoy
Thanks to everyone for their help. In the end I couldn't get any of the solutions to work and after a lot of testing, I came up with this solution:
func correctStringWithUsernames(inputString: String, completion: (correctString: String) -> Void) {
// Create the final string and get all
// the seperate strings from the data.
var finalString: String!
var commentSegments: NSArray!
commentSegments = inputString.componentsSeparatedByString(" ")
if (commentSegments.count > 0) {
for (var loop = 0; loop < commentSegments.count; loop++) {
// Check the username to ensure that there
// are no capital letters in the string.
let currentString = commentSegments[loop] as! String
let capitalLetterRegEx = ".*[A-Z]+.*"
let textData = NSPredicate(format:"SELF MATCHES %#", capitalLetterRegEx)
let capitalResult = textData.evaluateWithObject(currentString)
// Check if the current loop string
// is a #user mention string or not.
if (currentString.containsString("#")) {
// If we are in the first loop then set the
// string otherwise concatenate the string.
if (loop == 0) {
if (capitalResult == true) {
// The username contains capital letters
// so change it to a lower case version.
finalString = currentString.lowercaseString
}
else {
// The username does not contain capital letters.
finalString = currentString
}
}
else {
if (capitalResult == true) {
// The username contains capital letters
// so change it to a lower case version.
finalString = "\(finalString) \(currentString.lowercaseString)"
}
else {
// The username does not contain capital letters.
finalString = "\(finalString) \(currentString)"
}
}
}
else {
// The current string is NOT a #user mention
// so simply set or concatenate the finalString.
if (loop == 0) {
finalString = currentString
}
else {
finalString = "\(finalString) \(currentString)"
}
}
}
}
else {
// No issues pass back the string.
finalString = inputString
}
// Pass back the correct username string.
completion(correctString: finalString)
}
Its certainly not the most elegant or efficient solution around but it does work. If there are any ways of improving it, please leave a comment.

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