Unlike string.components(separatedBy: ...) I want to keep the separators in place in the resulting array. Code is more explanatory
let input = "foo&bar|hello"
let output = string.tokenize(splitMarks: ["&", "|"])
let desiredResult = ["foo", "&", "bar", "|", "hello"]
Is there any function in the standard library which does this? If not how can I implement such a function?
For that you need to loop through the String and check its each characters that is it tokens or not. You can make extension of String for that like this.
extension String {
func stringTokens(splitMarks: Set<String>) -> [String] {
var string = ""
var desiredOutput = [String]()
for ch in self.characters {
if splitMarks.contains(String(ch)) {
if !string.isEmpty {
desiredOutput.append(string)
}
desiredOutput.append(String(ch))
string = ""
}
else {
string += String(ch)
}
}
if !string.isEmpty {
desiredOutput.append(string)
}
return desiredOutput
}
}
Now you can call this function like this way.
let input = "foo&bar|hello"
print(input.stringTokens(splitMarks: ["&", "|"]))
Output
["foo", "&", "bar", "|", "hello"]
You can use rangeOfCharacter(from: CharacterSet, ...) in a loop to
find the next occurrence of a split mark in the string, and then
append both the preceding part and the separator to an array:
extension String {
func tokenize(splitMarks: String) -> [Substring] {
let cs = CharacterSet(charactersIn: splitMarks)
var result = [Substring]()
var pos = startIndex
while let range = rangeOfCharacter(from: cs, range: pos..<endIndex) {
// Append string preceding the split mark:
if range.lowerBound != pos {
result.append(self[pos..<range.lowerBound])
}
// Append split mark:
result.append(self[range])
// Update position for next search:
pos = range.upperBound
}
// Append string following the last split mark:
if pos != endIndex {
result.append(self[pos..<endIndex])
}
return result
}
}
Example:
let input = "foo&bar|hello"
let output = input.tokenize(splitMarks: "&|")
print(output)
// ["foo", "&", "bar", "|", "hello"]
Related
Target: A string with a built-in separator shall be split in an int and another string. In the case that the separator sequence '###' occurs more than once, the string shall always be spliced at the last '###'.
Is there an operator like string.lastIndexOf("###"), like in C#?
This is how my parser looks like:
func parseTuple(from string: String) -> (String, Int)? {
let parsedString = string.components(separatedBy: "###")
if let tupleString = String(parsedString[0]), let tupleInt = Int(parsedString[1]) {
return (tupleString, tupleInt)
} else {
return nil
}
}
The range(of:...) method of String has a .backwards option
to find the last occurrence of a string.
Then substring(to:) and substring(from:) can be used with the
lower/upper bound of that range to extract the parts of the string
preceding/following the separator:
func parseTuple(from string: String) -> (String, Int)? {
if let theRange = string.range(of: "###", options: .backwards),
let i = Int(string.substring(from: theRange.upperBound)) {
return (string.substring(to: theRange.lowerBound), i)
} else {
return nil
}
}
Example:
if let tuple = parseTuple(from: "Connect###Four###Player###7") {
print(tuple)
// ("Connect###Four###Player", 7)
}
Swift 4 update:
func parseTuple(from string: String) -> (String, Int)? {
if let theRange = string.range(of: "###", options: .backwards),
let i = Int(string[theRange.upperBound...]) {
return (String(string[...theRange.lowerBound]), i)
} else {
return nil
}
}
let input = "Connect###Four###Player###7"
let seperator = "###"
// split components
var components = input.components(separatedBy: seperator)
// remove the last component ...
components = Array(components.dropLast())
// ... and re-join the remaining ones
let output = components.joined(separator: seperator)
print(output)
prints:
Connect###Four###Player
For a given String instance, I want to check whether the last three characters are numeric characters (0, 1, 2, ..., 9) or not.
For example, the string
let str1 = "SACH092"
should return true for such a query, whereas e.g.
let str2 = "SACHA92"
should return false for the query.
I am using Xcode 7.3.1.
(As pointed out by #NiravD, for pre Swift 3, use where to join parts of multi-clause conditions. For Swift 3, parts of multi-clause conditions are simply joined by ,. For both methods below, both Swift 2.2 and 3 versions are included)
Use pattern matching for numeric characters "0"..."9"
Swift 2.2
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3) where chars.count > 2 {
let numbersPattern = Character("0")..."9"
return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
}
return false
}
}
Swift 3
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3), chars.count > 2 {
let numbersPattern = Character("0")..."9"
return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
}
return false
}
}
/* example usage, common for both Swift 2.2/3 version */
let str1 = "SACH092"
let str2 = "SACH0B2"
print(str1.lastThreeLettersAreNumbers) // true
print(str2.lastThreeLettersAreNumbers) // false
Make use of nil-return Int by String initializer, with flatMap
You can make use of the fact that the Int by String initializer returns nil for strings that cannot be represented as integers.
Swift 2.2
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3) where chars.count > 2 {
return chars.flatMap{Int(String($0))}.count == 3
}
return false
}
}
Swift 3
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3), chars.count > 2 {
return chars.flatMap{Int(String($0))}.count == 3
}
return false
}
}
/* example usage, common for both Swift 2.2/3 version */
let str1 = "SACH092"
let str2 = "SACH0B2"
print(str1.lastThreeLettersAreNumbers) // true
print(str2.lastThreeLettersAreNumbers) // false
For getting last 3 characters,
let exampleString = "SACH092"
let last3Char = exampleString.substringFromIndex(exampleString.endIndex.advancedBy(-3))
Check if last3Char contains all Digits,
let badCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
if last3Char.rangeOfCharacterFromSet(badCharacters) == nil {
print("String contains all digits")
} else {
print("String contains non-digit characters")
}
The best way to achieve string control is to use Regular Expressions.
For you :
var str = "SACH092"
let pattern = "^.*[0-9]{3,3}$"
let regexp = try! NSRegularExpression(pattern: pattern, options: [])
let matches = regexp.matches(in: str, options: [], range: NSMakeRange(0, str.characters.count))
print("End with 3 numbers : \(matches.count > 0)")
You can use like this
let s : NSString = "SACH092"
let trimmedString: String = (s as NSString).substringFromIndex(max(s.length-3,0))
print(trimmedString.isNumeric) // return true or false
//Make Extension of String
extension String {
var isNumeric: Bool {
let nums: Set<Character> = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
return Set(self.characters).isSubsetOf(nums)
}
}
To get the last three letters of a string
let oldString = "yourString"
let newString = a.substringFromIndex(a.endIndex.advancedBy(-3))
To check those characters are numbers,
func isNumber(num: String) -> Bool {
let numberCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
return !num.isEmpty && num.rangeOfCharacterFromSet(numberCharacters) == nil
}
I have this extension:
extension String {
public func seperateHashtags(char : String) -> [String]{
var word : String = ""
var words : [String] = [String]()
for chararacter in self.characters {
if String(chararacter) == char && word != "" {
words.append(word)
word = char
}else {
word += String(chararacter)
}
}
words.append(word)
return words
}
}
and I want to get an array of hashtags from any given text.
I have a UITextView that contains the text:
write anything here... #one #two three #four
and I expect to have an output:
["one", "two", "four"]
but when I do this:
print(myTextView.text.seperateHashtags("#"))
I'm getting:
["write anything here... ", "#one ", "#two three ", "#four"]
how can I fix that?
You can use regex and get all the hashtag like this way:
var hashtags = [String]()
var str = "#one #two three #four"
let regex = try NSRegularExpression(pattern: "(#[A-Za-z0-9]*)", options: [])
let matches = regex.matchesInString(str, options:[], range:NSMakeRange(0, str.characters.count))
for match in matches {
print("match = \(match.range)")
hashtags.append(NSString(string: str).substringWithRange(NSRange(location:match.range.location + 1, length:match.range.length - 1)))
}
Output:
["one", "two", "four"]
write anything here...
Shouldn't you put this as a placeholder for the UITextField?
Pardon me as I am a newbie on this language.
Edit: Is there a way to reverse the position of a array element?
I am trying to create a function that test the given input if its a palindrome or not. I'm trying to avoid using functions using reversed()
let word = ["T","E","S","T"]
var temp = [String]()
let index_count = 3
for words in word{
var text:String = words
print(text)
temp.insert(text, atIndex:index_count)
index_count = index_count - 1
}
Your approach can be used to reverse an array. But you have to
insert each element of the original array at the start position
of the destination array (moving the other elements to the end):
// Swift 2.2:
let word = ["T", "E", "S", "T"]
var reversed = [String]()
for char in word {
reversed.insert(char, atIndex: 0)
}
print(reversed) // ["T", "S", "E", "T"]
// Swift 3:
let word = ["T", "E", "S", "T"]
var reversed = [String]()
for char in word {
reversed.insert(char, at: 0)
}
print(reversed) // ["T", "S", "E", "T"]
The same can be done on the characters of a string directly:
// Swift 2.2:
let word = "TEST"
var reversed = ""
for char in word.characters {
reversed.insert(char, atIndex: reversed.startIndex)
}
print(reversed) // "TSET"
// Swift 3:
let word = "TEST"
var reversed = ""
for char in word.characters {
reversed.insert(char, at: reversed.startIndex)
}
print(reversed)
Swift 5
extension String {
func invert() -> String {
var word = [Character]()
for char in self {
word.insert(char, at: 0)
}
return String(word)
}
}
var anadrome = "god"
anadrome.invert()
// "dog"
Here's my solution:
extension String {
func customReverse() -> String {
var chars = Array(self)
let count = chars.count
for i in 0 ..< count/2 {
chars.swapAt(i, count - 1 - i)
}
return String(chars)
}
}
let input = "abcdef"
let output = input.customReverse()
print(output)
You can try it here.
func reverse(_ str: String) -> String {
let arr = Array(str) // turn the string into an array of all of the letters
let reversed = ""
for char in arr {
reversed.insert(char, at: reversed.startIndex)
}
return reversed
}
To use it:
let word = "hola"
let wordReversed = reverse(word)
print(wordReversed) // prints aloh
Another solution for reversing:
var original : String = "Test"
var reversed : String = ""
var c = original.characters
for _ in 0..<c.count{
reversed.append(c.popLast()!)
}
It simply appends each element of the old string that is popped, starting at the last element and working towards the first
Solution 1
let word = "aabbaa"
let chars = word.characters
let half = chars.count / 2
let leftSide = Array(chars)[0..<half]
let rightSide = Array(chars.reverse())[0..<half]
let palindrome = leftSide == rightSide
Solution 2
var palindrome = true
let chars = Array(word.characters)
for i in 0 ..< (chars.count / 2) {
if chars[i] != chars[chars.count - 1 - i] {
palindrome = false
break
}
}
print(palindrome)
static func reverseString(str : String) {
var data = Array(str)
var i = 0// initial
var j = data.count // final
//either j or i for while , data.count/2 buz only half we need check
while j != data.count/2 {
print("befor i:\(i) j:\(j)" )
j = j-1
data.swapAt(i, j) // swapAt API avalible only for array in swift
i = i+1
}
print(String(data))
}
//Reverse String
let str = "Hello World"
var reverseStr = ""
for char in Array(str) {
print(char)
reverseStr.insert(char, at: str.startIndex)
}
print(reverseStr)
This finds the duplicates in the array, but i'm looking for something that finds the first non-repeating character in a string. I've been trying to figure out a way to do this and I cannot figure it out. This is the closest i've gotten.
var strArray = ["P","Q","R","S","T","P","R","A","T","B","C","P","P","P","P","P","C","P","P","J"]
println(strArray)
var filter = Dictionary<String,Int>()
var len = strArray.count
for var index = 0; index < len ;++index {
var value = strArray[index]
if (filter[value] != nil) {
strArray.removeAtIndex(index--)
len--
}else{
filter[value] = 1
}
}
println(strArray)
In order to tell if a character repeats itself, go through the entire array once, incrementing the count of occurrences in a dictionary:
let characters = ["P","Q","R","S","T","P","R","A","T","B","C","P","P","P","P","P","C","P","P","J"]
var counts: [String: Int] = [:]
for character in characters {
counts[character] = (counts[character] ?? 0) + 1
}
let nonRepeatingCharacters = characters.filter({counts[$0] == 1})
// ["Q", "S", "A", "B", "J"]
let firstNonRepeatingCharacter = nonRepeatingCharacters.first!
// "Q"
Here is a simple solution
let inputString = "PQRSTPRATBCPPPPPCPPJ"
func nonRepeat (_ input: String) -> String {
for char in input {
if input.firstIndex(of: char) == input.lastIndex(of: char) {
return String(char)
}
}
return ""
}
print (nonRepeat(inputString))
In the above example it would print "Q"
func firstNonRepeatedCharacter(input: String) -> Character?{
var characterCount : [Character : Int] = [:]
var uniqueCharacter: Character?
for character in input{
if let count = characterCount[character]{
characterCount[character] = count + 1
if(uniqueCharacter == character)
{
uniqueCharacter = nil
}
}
else{
characterCount[character] = 1
if(uniqueCharacter == nil){
uniqueCharacter = character
}
}
}
return uniqueCharacter
}
Without extra loop to find character from characterCount dictionary
Here is the way I have found to detect the first non-repeated character. It removes spaces and punctuation to find the actual letter or number that does not repeat.
extension String {
func removeNonAlphaNumChars() -> String {
let charSet = NSCharacterSet.alphanumericCharacterSet().invertedSet
return self
.componentsSeparatedByCharactersInSet(charSet)
.joinWithSeparator("")
}
var firstNonRepeatedCharacter: Character? {
let alphaNumString = self.removeNonAlphaNumChars()
let characters = alphaNumString.characters
let count = characters.count
guard count > 0 else { return nil }
// Find unique chars
var dict: [Character: Int?] = [:]
for (index, char) in characters.enumerate() {
if dict[char] != nil {
dict[char] = (nil as Int?)
}
else {
dict[char] = index
}
}
return dict.filter { $0.1 != nil }.sort { $0.1 < $1.1 }.first?.0
}
}
I totally wonder why the accepted answer was considered correct. They are using
.first
method of a dictionary and that according to documentation would return a random element in the dictionary and not the first element as a dictionary in swift is not ordered like an array.
please do find below an implementation that works
func firstNonRepeatingLetter(_ str: String) -> String{
var characterDict = [String : Int]()
for character in str{
let lower = character.lowercased()
if let count = characterDict[lower]{
characterDict[lower] = count + 1
}else{
characterDict[lower] = 1
}
}
let filtered = characterDict.filter { $0.value == 1}
for character in str{
let lower = character.lowercased()
if let _ = filtered[lower]{
return lower
}
}
return ""
}
firstNonRepeatingLetter("moonmen") would return "e".
We can iterate once and keep the letter counts inside a dictionary.
Then, iterate again and return first letter where we see it was encountered once only (or "_" if not found a non-repeating letter):
func firstNotRepeatingCharacter(s: String) -> Character {
var letterCounts: [String: Int] = [:]
var result: Character = "_"
for letter in s {
if let currentLetterCount = letterCounts[String(letter)] {
letterCounts[String(letter)] = currentLetterCount + 1
} else {
letterCounts[String(letter)] = 1
}
}
for letter in s {
if letterCounts[String(letter)] == 1 {
result = letter
break
}
}
return result
}
OrderedDictionary makes this easy for all Sequences of Hashables, not just Strings:
import struct OrderedCollections.OrderedDictionary
extension Sequence where Element: Hashable {
var firstUniqueElement: Element? {
OrderedDictionary(zip(self, true)) { _, _ in false }
.first(where: \.value)?
.key
}
}
/// `zip` a sequence with a single value, instead of another sequence.
public func zip<Sequence: Swift.Sequence, Constant>(
_ sequence: Sequence, _ constant: Constant
) -> LazyMapSequence<
LazySequence<Sequence>.Elements,
(LazySequence<Sequence>.Element, Constant)
> {
sequence.lazy.map { ($0, constant) }
}
func getFirstUniqueChar(string:String)->Character?{
var counts: [String: Int] = [:]
for character in string {
let charString = "\(character)"
counts[charString] = (counts[charString] ?? 0) + 1
}
let firstNonRepeatingCharacter = string.first {counts["\($0)"] == 1}
return firstNonRepeatingCharacter
}
print(getFirstUniqueChar(string: string))
import Foundation
import Glibc
var str:String = "aacbbcee"//your input string
var temp:String = ""
var dict:[Character:Int] = [:]
for char in str{
if let count = dict[char]{
dict[char] = count+1//storing values in dict and incrmenting counts o key
}
else{
dict[char] = 0
}
}
var arr:[Character] = []
for (key, value) in dict{
if value == 0{
arr.append(key)//filtering out, take characters which has value>0
} //int(arr)
}//print(arr.count)
if arr.count != 0{
outer:for char in str{//outer is labeling the loop
for i in arr{
if i == char{
print(i,"is first")//matching char with array elements if found break
break outer
}
else{
continue
}
}
}
}
else{
print("not found")
}
func firstNonRepeatedChar(string: String) -> Character {
var arr: [Character] = []
var dict: [Character : Int] = [:]
for character in string.description {
arr.append(character)
}
for character in arr {
dict[character] = (dict[character] ?? 0) + 1
}
let nonRepeatedArray = arr.filter { char in
if dict[char] == 1 {return true}
return false
}
let firstNonRepeatedChar = nonRepeatedArray.first
return firstNonRepeatedChar!
}
print(firstNonRepeatedChar(string: "strinstrig"))