Parse a string of 4 numbers into a CGRect [closed] - ios

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I have this string (325, 140, 739, 979) . I want to parse it and take the four numbers separately.
This string represents a CGRect. I want to take the first number as x the second as y the third as width and the last as height.
What is the best way to achieve this ?

Given your string, you can fetch the numbers as an array of optional Int like this:
let str = "(325, 140, 739, 979)"
let nums = split(str) { contains("(), ", $0) }.map { $0.toInt() }
That split will remove any of the characters in the string passed to contains. You now have an array of optionals, which you can check for the correct contents:
let rect: CGRect
if nums.count == 4,
let x = nums[0], y = nums[1],
w = nums[2], h = nums[3]
{
rect = CGRect(x: x, y: y, width: w, height: h)
}
else {
// report an error, or default the values if you prefer
fatalError("Malformed input string")
}
Any extraneous characters in your input string will result in nil for one of the integers, or the wrong count in the array, so this should be safe against any garbage input.
For convenience, you could put all this in a failable initializer for CGRect:
extension CGRect {
init?(string: String) {
// note, since CGRect also has a contains method, need to specify Swift.contains
let nums = split(string) { Swift.contains("(), ", $0) }.map { $0.toInt() }
if nums.count == 4,
let x = nums[0], y = nums[1],
w = nums[2], h = nums[3]
{
self = CGRect(x: x, y: y, width: w, height: h)
}
else {
return nil
}
}
}
let rectangles = [
"(325, -140, 739, 979)", // valid
"(1,2,3,asdadaf)", // invalid (non-integer)
"1,2,3,4,", // valid
"(1,2,3,4,5)", // invalid (wrong count)
]
// returns an array of 2 valid CGRect and 2 nil
let cgrects = rectangles.map { CGRect(string: $0) }
Obviously there’s lots you could tweak here if you wanted to be more or less permissive in terms of the kind of input you’re willing to convert.

Here is my solution for your case:
func parse(str : String) -> [Int] {
var firstStepStr = str.stringByTrimmingCharactersInSet(NSCharacterSet(charactersInString: "()")) //remove ( and )
var secondStepArray = firstStepStr.componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: ", ")).filter{$0 != ""} //separated by , and ignore ""
return secondStepArray.map{$0.toInt() ?? 0} //convert to [Int], if cannot convert to Int, return 0
}
let cgrectString = "(325, 140, 739, 979)"
let intArray = parse(cgrectString)

First, we could get rid of the parentheses
let myString = "(325, 140, 739, 979)"
let myReplacementString = String(map(myString.generate()) {
$0 == "(" || $0 == ")" ? "-" : $0
})
You could also take a substring, which doesn't seem to be any nicer.
Then we could split the string into array
var myArray = myReplacementString.componentsSeparatedByString(", ")
And then use a loop to cast string to int
for item in myArray {
item.toInt()
}
or just use myArray[i].toInt() to give them straight to constructor etc.

This will get you an array of the four CGFloat values you are looking for:
func stringToCGFloatArray(string: String) -> [CGFloat] {
return string.componentsSeparatedByCharactersInSet(
NSCharacterSet(charactersInString: "(), ")).reduce([CGFloat]()) {
if let x = $1.toInt() {
return $0 + [CGFloat(x)]
} else {
return $0
}
}
}
You can then convert it to an optional tuple that can be provided directly to CGMakeRect:
func arrayToRectParameters(array: [CGFloat]) -> (CGFloat, CGFloat, CGFloat, CGFloat)? {
switch array.count {
case 4:
return (array[0], array[1], array[2], array[3])
default:
return nil
}
}
Now you can create a CGRect like this:
let str = "(325, 140, 739, 979)"
if let rectParameters = arrayToRectParameters(stringToCGFloatArray(str)) {
let myCGRect = CGRectMake(rectParameters)
println(myCGRect) // prints (325.0,140.0,739.0,979.0)
}

Related

Gradually and randomly visualize string

I am currently working on a simple program to gradually and randomly visualize a string in two iteration. Right now I have managed to get the first iteration but I'm not sure how to do the second one. If someone could give any example or advice I would be very grateful. My code looks like this:
let s = "Hello playground"
let factor = 0.25
let factor2 = 0.45
var n = s.filter({ $0 != " " }).count // # of non-space characters
var m = lrint(factor * Double(n)) // # of characters to display
let t = String(s.map { c -> Character in
if c == " " {
// Preserve space
return " "
} else if Int.random(in: 0..<n) < m {
// Replace
m -= 1
n -= 1
return c
} else {
// Keep
n -= 1
return "_"
}
})
print(t) // h_l__ _l_______d
To clarify, I want to use factor2 in the second iteration to print something that randomly add letters on top of t that looks something like this h_l_o pl_g_____d.
Replacing Characters
Starting from #MartinR's code, you should remember the indices that have been replaced. So, I am going to slightly change the code that replaces characters :
let s = "Hello playground"
let factor = 0.25
let factor2 = 0.45
var n = s.filter({ $0 != " " }).count // # of non-space characters
let nonSpaces = n
var m = lrint(factor * Double(n)) // # of characters to display
var indices = Array(s.indices)
var t = ""
for i in s.indices {
let c = s[i]
if c == " " {
// Preserve space
t.append(" ")
indices.removeAll(where: { $0 == i })
} else if Int.random(in: 0..<n) < m {
// Keep
m -= 1
n -= 1
t.append(c)
indices.removeAll(where: { $0 == i })
} else {
// Replace
n -= 1
t.append("_")
}
}
print(t) //For example: _e___ ______ou_d
Revealing Characters
In order to do that, we should calculate the number of characters that we want to reveal:
m = lrint((factor2 - factor) * Double(nonSpaces))
To pick three indices to reveal randomly, we shuffle indices and then replace the m first indices :
indices.shuffle()
var u = t
for i in 0..<m {
let index = indices[i]
u.replaceSubrange(index..<u.index(after: index), with: String(s[index]))
}
indices.removeSubrange(0..<m)
print(u) //For example: _e__o _l__g_ou_d
I wrote StringRevealer struct, that handle all revealing logic for you:
/// Hide all unicode letter characters as `_` symbol.
struct StringRevealer {
/// We need mapping between index of string character and his position in state array.
/// This struct represent one such record
private struct Symbol: Hashable {
let index: String.Index
let position: Int
}
private let originalString: String
private var currentState: [Character]
private let charactersCount: Int
private var revealed: Int
var revealedPercent: Double {
return Double(revealed) / Double(charactersCount)
}
private var unrevealedSymbols: Set<Symbol>
init(_ text: String) {
originalString = text
var state: [Character] = []
var symbols: [Symbol] = []
var count = 0
var index = originalString.startIndex
var i = 0
while index != originalString.endIndex {
let char = originalString[index]
if CharacterSet.letters.contains(char.unicodeScalars.first!) {
state.append("_")
symbols.append(Symbol(index: index, position: i))
count += 1
} else {
state.append(char)
}
index = originalString.index(after: index)
i += 1
}
currentState = state
charactersCount = count
revealed = 0
unrevealedSymbols = Set(symbols)
}
/// Current state of text. O(n) conplexity
func text() -> String {
return currentState.reduce(into: "") { $0.append($1) }
}
/// Reveal one random symbol in string
mutating func reveal() {
guard let symbol = unrevealedSymbols.randomElement() else { return }
unrevealedSymbols.remove(symbol)
currentState[symbol.position] = originalString[symbol.index]
revealed += 1
}
/// Reveal random symbols on string until `revealedPercent` > `percent`
mutating func reveal(until percent: Double) {
guard percent <= 1 else { return }
while revealedPercent < percent {
reveal()
}
}
}
var revealer = StringRevealer("Hello товарищ! 👋")
print(revealer.text())
print(revealer.revealedPercent)
for percent in [0.25, 0.45, 0.8] {
revealer.reveal(until: percent)
print(revealer.text())
print(revealer.revealedPercent)
}
It use CharacterSet.letters inside, so most of languages should be supported, emoji ignored and not-alphabetic characters as well.

Swift code to produce a number of possible anagrams from a selected word

I've attempted to research ways to take a given word and calculate the number of possible anagrams a user can make from that word eg an 8 letter word such as snowbanks has 5 eight letter possibilities, 25 seven letter possibilities, etc (those are made up numbers). My initial plan would be to iterate over a dictionary list and check each of the words to see if it is an anagram of the word in question as I've seen suggested in other places.
Rearrange Letters from Array and check if arrangement is in array
seemed very promising, except that it is in objective C and when I tried to convert it to Swift using Swiftify I couldn't get it to work as shown below:
func findAnagrams() -> Set<AnyHashable>? {
let nineCharacters = [unichar](repeating: 0, count: 8)
let anagramKey = self.anagramKey()
// make sure this word is not too long/short.
if anagramKey == nil {
return nil
}
(anagramKey as NSString?)?.getCharacters(nineCharacters, range: NSRange)
let middleCharPos = Int((anagramKey as NSString?)?.range(of: (self as NSString).substring(with: NSRange)).location ?? 0)
var anagrams = Set<AnyHashable>()
// 0x1ff means first 9 bits set: one for each character
for i in 0...0x1ff {
// skip permutations that do not contain the middle letter
if (i & (1 << middleCharPos)) == 0 {
continue
}
var length: Int = 0
var permutation = [unichar](repeating: 0, count: 9)
for bit in 0...9 {
if true {
permutation[length] = nineCharacters[bit]
length += 1
}
}
if length < 4 {
continue
}
let permutationString = String(permutation)
let matchingAnagrams = String.anagramMap()[permutationString] as? [Any]
for word: String? in matchingAnagrams {
anagrams.insert(word)
}
}
return anagrams
}
class func anagramMap() -> [AnyHashable: Any]? {
var anagramMap: [AnyHashable: Any]
if anagramMap != nil {
return anagramMap
}
// this file is present on Mac OS and other unix variants
let allWords = try? String(contentsOfFile: "/usr/share/dict/words", encoding: .utf8)
var map = [AnyHashable: Any]()
autoreleasepool {
allWords?.enumerateLines(invoking: {(_ word: String?, _ stop: UnsafeMutablePointer<ObjCBool>?) -> Void in
let key = word?.anagramKey()
if key == nil {
return
}
var keyWords = map[key] as? [AnyHashable]
if keyWords == nil {
keyWords = [AnyHashable]()
map[key] = keyWords
}
if let aWord = word {
keyWords?.append(aWord)
}
})
}
anagramMap = map
return anagramMap
}
func anagramKey() -> String? {
let lowercaseWord = word.lowercased()
// make sure to take the length *after* lowercase. it might change!
let length: Int = lowercaseWord.count
// in this case we're only interested in anagrams 4 - 9 characters long
if length < 3 || length > 9 {
return nil
}
let sortedWord = [unichar](repeating: 0, count: length)
(lowercaseWord as NSString).getCharacters(sortedWord, range: NSRange)
qsort_b(sortedWord, length, MemoryLayout<unichar>.size, {(_ aPtr: UnsafeRawPointer?, _ bPtr: UnsafeRawPointer?) -> Int in
let a = Int(unichar(aPtr))
let b = Int(unichar(bPtr))
return b - a
})
return String(describing: sortedWord)
}
func isReal(word: String) -> Bool {
let checker = UITextChecker()
let range = NSMakeRange(0, word.utf16.count)
let misspelledRange = checker.rangeOfMisspelledWord(in: word, range: range, startingAt: 0, wrap: false, language: "en")
return misspelledRange.location == NSNotFound
}
}
I've also tried the following in an attempt to just produce a list of words that I could iterate over to check for anagrams (I have working code that checks guesses vs the main word to check for anagrams) but I wasn't able to get them to work, possibly because they require a file to be copied to the app, since I was under the impression that the phone has a dictionary preloaded that I could use for words (although I may be mistaken):
var allTheWords = try? String(contentsOfFile: "/usr/share/dict/words", encoding: .utf8)
for line: String? in allTheWords?.components(separatedBy: "\n") ?? [String?]() {
print("\(line ?? "")")
print("Double Fail \(allTheWords)")
}
and
if let wordsFilePath = Bundle.main.path(forResource: "dict", ofType: nil) {
do {
let wordsString = try String(contentsOfFile: wordsFilePath)
let wordLines = wordsString.components(separatedBy: NSCharacterSet.newlines)
let randomLine = wordLines[Int(arc4random_uniform(UInt32(wordLines.count)))]
print(randomLine)
} catch { // contentsOfFile throws an error
print("Error: \(error)")
}
}
}
I looked at UIReferenceLibraryViewController as well in an attempt to use it to produce a list of words instead of defining a selected word, but the following isn't a valid option.
let words = UIReferenceLibraryViewController.enumerated
Any assistance would be greatly appreciated!

Swift Anagram checker

I am attempting to build an anagram checker for swift. This is my code. In case you don't know an anagram checker checks if two strings have the same characters in them but, order does not matter.
func checkForAnagram(#firstString: String, #secondString: String) -> Bool {
var firstStringArray: [Character] = []
var secondStringArray: [Character] = []
/* if case matters delete the next four lines
and make sure your variables are not constants */
var first = firstString
var second = secondString
first = first.lowercaseString
second = second.lowercaseString
for charactersOne in first {
firstStringArray += [charactersOne]
}
for charactersTwo in second {
secondStringArray += [charactersTwo]
}
if firstStringArray.count != secondStringArray.count {
return false
} else {
for elements in firstStringArray {
if secondStringArray.contains(elements){
return true
} else {
return false
}
}
}
}
var a = "Hello"
var b = "oellh"
var c = "World"
checkForAnagram(firstString: a, secondString: b)
I am getting an error message of.
'[Character]' does not have a member 'contains'
The accepted answer is compact and elegant, but very inefficient if compared to other solutions.
I'll now propose and discuss the implementation of a few variants of anagram checker. To measure performance, I'll use the different variants to find the anagrams of a given word out of an array of 50,000+ words.
// Variant 1: Sorting of Character
// Measured time: 30.46 s
func anagramCheck1(a: String, b: String) -> Bool {
return a.characters.sorted() == b.characters.sorted()
}
This is essentially the solution of the accepted answer, written in Swift 3 syntax. It's very slow because Swift's String, unlike NSString, is based on Character, which handles Unicode characters properly.
A more efficient solution exploits the NSCountedSet class, which allows us to represent a string as a set of characters, each with its own count. Two strings are anagrams if they map to the same NSCountedSet.
Note: checking string lengths as a precondition makes the implementation always more efficient.
// Variant 2: NSCountedSet of Character
// Measured time: 4.81 s
func anagramCheck2(a: String, b: String) -> Bool {
guard a.characters.count == b.characters.count else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for c in a.characters {
aSet.add(c)
}
for c in b.characters {
bSet.add(c)
}
return aSet == bSet
}
Better but not excellent. Here, one of the "culprits" is the use of the native Swift Character type (from Swift's String). Moving back to good old Objective-C types (NSString and unichar) makes things more efficient.
// Variant 3: NSCountedSet of unichar
// Measured time: 1.31 s
func anagramCheck3(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
bSet.add(bString.character(at: i))
}
return aSet == bSet
}
Using NSCountedSet is fine, but before we compare two NSCountedSet objects, we fully populate them. A useful alternative is to fully populate the NSCountedSet for only one of the two strings, and then, while we populate the NSCountedSet for the other string, we fail early if the other string contains a character that is not found in the NSCountedSet of the first string.
// Variant 4: NSCountedSet of unichar and early exit
// Measured time: 1.07 s
func anagramCheck4(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
}
for i in 0..<length {
let c = bString.character(at: i)
if bSet.count(for: c) >= aSet.count(for: c) {
return false
}
bSet.add(c)
}
return true
}
This is about the best timing we are going to get (with Swift). However, for completeness, let me discuss one more variant of this kind.
The next alternative exploits a Swift Dictionary of type [unichar: Int] to store the number of repetitions for each character instead of NSCountedSet. It's slightly slower than the previous two variants, but we can reuse it later to obtain a faster implementation.
// Variant 5: counting repetitions with [unichar:Int]
// Measured time: 1.36
func anagramCheck5(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
var aDic = [unichar:Int]()
var bDic = [unichar:Int]()
for i in 0..<length {
let c = aString.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
for i in 0..<length {
let c = bString.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
Note that a vanilla Objective-C implementation using NSCountedSet, corresponding to Variant 3, is faster than all the previous versions by a rather large margin.
// Variant 6: Objective-C and NSCountedSet
// Measured time: 0.65 s
- (BOOL)anagramChecker:(NSString *)a with:(NSString *)b {
if (a.length != b.length) {
return NO;
}
NSCountedSet *aSet = [[NSCountedSet alloc] init];
NSCountedSet *bSet = [[NSCountedSet alloc] init];
for (int i = 0; i < a.length; i++) {
[aSet addObject:#([a characterAtIndex:i])];
[bSet addObject:#([b characterAtIndex:i])];
}
return [aSet isEqual:bSet];
}
Another way we can improve upon the previous attempts is to observe that, if we need to find the anagram of a given word, we might as well consider that word as fixed, and thus we could build the corresponding structure (NSCountedSet, Dictionary, ...) for that word only once.
// Finding all the anagrams of word in words
// Variant 7: counting repetitions with [unichar:Int]
// Measured time: 0.58 s
func anagrams(word: String, from words: [String]) -> [String] {
let anagrammedWord = word as NSString
let length = anagrammedWord.length
var aDic = [unichar:Int]()
for i in 0..<length {
let c = anagrammedWord.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
let foundWords = words.filter {
let string = $0 as NSString
guard length == string.length else { return false }
var bDic = [unichar:Int]()
for i in 0..<length {
let c = string.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
return foundWords
}
Now, in the previous variant we have counted with a [unichar:Int] Dictionary. This proves slightly more efficient than using an NSCountedSet of unichar, either with early exit (0.60 s) or without (0.87 s).
You should try
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return firstString.lowercaseString.characters.sort() == secondString.lowercaseString.characters.sort()
}
func checkAnagrams(str1: String, str2: String) -> Bool {
guard str1.count == str2.count else { return false }
var dictionary = Dictionary<Character, Int>()
for index in 0..<str1.count {
let value1 = str1[str1.index(str1.startIndex, offsetBy: index)]
let value2 = str2[str2.index(str2.startIndex, offsetBy: index)]
dictionary[value1] = (dictionary[value1] ?? 0) + 1
dictionary[value2] = (dictionary[value2] ?? 0) - 1
}
return !dictionary.contains(where: {(_, value) in
return value != 0
})
}
Time complexity - O(n)
// Make sure name your variables correctly so you won't confuse
// Mutate the constants parameter, lowercase to handle capital letters and the sorted them to compare both. Finally check is there are equal return true or false.
func anagram(str1: String, srt2: String)->Bool{
let string1 = str1.lowercased().sorted()
let string2 = srt2.lowercased().sorted()
if string1 == string2 {
return true
}
return false
}
// This answer also would work
// Convert your parameters on Array, then sorted them and compare them
func ana(str1: String, str2: String)->Bool{
let a = Array(str1)
let b = Array(str2)
if a.sorted() == b.sorted() {
return true
}
return false
}
Don't forget whitespaces
func isAnagram(_ stringOne: String, stringTwo: String) -> Bool {
return stringOne.lowercased().sorted().filter { $0 != " "} stringTwo.lowercased().sorted().filter { $0 != " "}
}
Swift 4.1 Function will give you 3 questions answer for Anagram :-
1. Input Strings (a,b) are Anagram ? //Bool
2. If not an Anagram then number of count require to change Characters in strings(a,b) to make them anagram ? // Int
3. If not an Anagram then list of Characters needs to be change in strings(a,b) to make them anagram ? // [Character]
STEP 1:- Copy and Paste below function in to your required class:-
//MARK:- Anagram checker
func anagramChecker(a:String,b:String) -> (Bool,Int,[Character]) {
var aCharacters = Array(a)
var bCharacters = Array(b)
var count = 0
var isAnagram = true
var replacementRequiredWords:[Character] = [Character]()
if aCharacters.count == bCharacters.count {
let listA = aCharacters.filter { !bCharacters.contains($0) }
for i in 0 ..< listA.count {
if !replacementRequiredWords.contains(listA[i]) {
count = count + 1
replacementRequiredWords.append(listA[i])
isAnagram = false
}
}
let listB = bCharacters.filter { !aCharacters.contains($0) }
for i in 0 ..< listB.count {
if !replacementRequiredWords.contains(listB[i]) {
count = count + 1
replacementRequiredWords.append(listB[i])
isAnagram = false
}
}
}else{
//cant be an anagram
count = -1
}
return (isAnagram,count,replacementRequiredWords)
}
STEP 2 :- Make two Input Strings for test
// Input Strings
var a = "aeb"
var b = "abs"
STEP 3:- Print results :-
print("isAnagram : \(isAnagram(a: a, b: b).0)")
print("number of count require to change strings in anagram : \(isAnagram(a: a, b: b).1)")//-1 will come in case of cant be a Anagram
print("list of Characters needs to be change : \(isAnagram(a: a, b: b).2)")
Results of above exercise:-
isAnagram : false
number of count require to change strings in anagram : 2
list of Characters needs to be change : ["e", "s"]
Hope this 10 minutes exercise will give some support to my Swift
family for solving Anagram related problems easily. :)
We can use dictionary to construct a new data structure container. Then compare the value by key/character of the string.
func anagram(str1: String, str2 : String) -> Bool {
var dict1 = [Character: Int]()
var dict2 = [Character:Int]()
for i in str1 {
if let count = dict1[i] {
dict1[i] = count + 1
} else {
dict1[i] = 1
}
}
for j in str2 {
if let count = dict2[j] {
dict2[j] = count + 1
} else {
dict2[j] = 1
}
}
return dict1 == dict2 ? true : false
}
// input -> "anna", "aann"
// The count will look like:
// ["a": 2, "n": 2] & ["a": 2, "n": 2]
// then return true
Another easy that I just realise doing an Anagram function in Swift 5.X
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return !firstString.isEmpty && firstString.sorted() == secondString.sorted()
}
class Solution {
func isAnagram(_ s: String, _ t: String) -> Bool {
guard s.count == t.count else { return false }
let dictS = s.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
let dictT = t.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
for letter in s {
if let count = dictS[letter] {
guard count == dictT[letter] else { return false }
}
}
return true
}
}
Check two strings are anagram using inout method in Swift
func checkAnagramString(str1: inout String, str2: inout String)-> Bool{
var result:Bool = false
str1 = str1.lowercased().trimmingCharacters(in: .whitespace)
str2 = str2.lowercased().trimmingCharacters(in: .whitespaces)
if (str1.count != str2.count) {
return result
}
for c in str1 {
if str2.contains(c){
result = true
}
else{
result = false
return result
}
}
return result
}
Call function to check strings are anagram or not
var str1 = "tommarvoloriddle"
var str2 = "iamlordvoldemort"
print(checkAnagramString(str1: &str1, str2: &str2)) //Output = true.
func isAnagram(word1: String, word2: String) -> Bool {
let set1 = Set(word1)
let set2 = Set(word2)
return set1 == set2
}
or
func isAnagram(word1: String,word2: String) -> Bool {
return word1.lowercased().sorted() == word2.lowercased().sorted()
}

How to create a String with format?

I need to create a String with format which can convert Int, Int64, Double, etc types into String. Using Objective-C, I can do it by:
NSString *str = [NSString stringWithFormat:#"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE, STRING_VALUE];
How to do same but in Swift?
I think this could help you:
import Foundation
let timeNow = time(nil)
let aStr = String(format: "%#%x", "timeNow in hex: ", timeNow)
print(aStr)
Example result:
timeNow in hex: 5cdc9c8d
nothing special
let str = NSString(format:"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, LONG_VALUE, STRING_VALUE)
let str = "\(INT_VALUE), \(FLOAT_VALUE), \(DOUBLE_VALUE), \(STRING_VALUE)"
Update: I wrote this answer before Swift had String(format:) added to it's API. Use the method given by the top answer.
No NSString required!
String(format: "Value: %3.2f\tResult: %3.2f", arguments: [2.7, 99.8])
or
String(format:"Value: %3.2f\tResult: %3.2f", 2.7, 99.8)
I would argue that both
let str = String(format:"%d, %f, %ld", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE)
and
let str = "\(INT_VALUE), \(FLOAT_VALUE), \(DOUBLE_VALUE)"
are both acceptable since the user asked about formatting and both cases fit what they are asking for:
I need to create a string with format which can convert int, long, double etc. types into string.
Obviously the former allows finer control over the formatting than the latter, but that does not mean the latter is not an acceptable answer.
First read Official documentation for Swift language.
Answer should be
var str = "\(INT_VALUE) , \(FLOAT_VALUE) , \(DOUBLE_VALUE), \(STRING_VALUE)"
println(str)
Here
1) Any floating point value by default double
EX.
var myVal = 5.2 // its double by default;
-> If you want to display floating point value then you need to explicitly define such like a
EX.
var myVal:Float = 5.2 // now its float value;
This is far more clear.
let INT_VALUE=80
let FLOAT_VALUE:Double= 80.9999
let doubleValue=65.0
let DOUBLE_VALUE:Double= 65.56
let STRING_VALUE="Hello"
let str = NSString(format:"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE, STRING_VALUE);
println(str);
The accepted answer is definitely the best general solution for this (i.e., just use the String(format:_:) method from Foundation) but...
If you are running Swift ≥ 5, you can leverage the new StringInterpolationProtocol protocol to give yourself some very nice syntax sugar for common string formatting use cases in your app.
Here is how the official documentation summarizes this new protocol:
Represents the contents of a string literal with interpolations while it’s being built up.
Some quick examples:
extension String.StringInterpolation {
/// Quick formatting for *floating point* values.
mutating func appendInterpolation(float: Double, decimals: UInt = 2) {
let floatDescription = String(format: "%.\(decimals)f%", float)
appendLiteral(floatDescription)
}
/// Quick formatting for *hexadecimal* values.
mutating func appendInterpolation(hex: Int) {
let hexDescription = String(format: "0x%X", hex)
appendLiteral(hexDescription)
}
/// Quick formatting for *percents*.
mutating func appendInterpolation(percent: Double, decimals: UInt = 2) {
let percentDescription = String(format: "%.\(decimals)f%%", percent * 100)
appendLiteral(percentDescription)
}
/// Formats the *elapsed time* since the specified start time.
mutating func appendInterpolation(timeSince startTime: TimeInterval, decimals: UInt = 2) {
let elapsedTime = CACurrentMediaTime() - startTime
let elapsedTimeDescription = String(format: "%.\(decimals)fs", elapsedTime)
appendLiteral(elapsedTimeDescription)
}
}
which could be used as:
let number = 1.2345
"Float: \(float: number)" // "Float: 1.23"
"Float: \(float: number, decimals: 1)" // "Float: 1.2"
let integer = 255
"Hex: \(hex: integer)" // "Hex: 0xFF"
let rate = 0.15
"Percent: \(percent: rate)" // "Percent: 15.00%"
"Percent: \(percent: rate, decimals: 0)" // "Percent: 15%"
let startTime = CACurrentMediaTime()
Thread.sleep(forTimeInterval: 2.8)
"∆t was \(timeSince: startTime)" // "∆t was 2.80s"
"∆t was \(timeSince: startTime, decimals: 0)" // "∆t was 3s"
This was introduced by SE-0228, so please be sure to read the original proposal for a deeper understanding of this new feature. Finally, the protocol documentation is helpful as well.
I know a lot's of time has passed since this publish, but I've fallen in a similar situation and create a simples class to simplify my life.
public struct StringMaskFormatter {
public var pattern : String = ""
public var replecementChar : Character = "*"
public var allowNumbers : Bool = true
public var allowText : Bool = false
public init(pattern:String, replecementChar:Character="*", allowNumbers:Bool=true, allowText:Bool=true)
{
self.pattern = pattern
self.replecementChar = replecementChar
self.allowNumbers = allowNumbers
self.allowText = allowText
}
private func prepareString(string:String) -> String {
var charSet : NSCharacterSet!
if allowText && allowNumbers {
charSet = NSCharacterSet.alphanumericCharacterSet().invertedSet
}
else if allowText {
charSet = NSCharacterSet.letterCharacterSet().invertedSet
}
else if allowNumbers {
charSet = NSCharacterSet.decimalDigitCharacterSet().invertedSet
}
let result = string.componentsSeparatedByCharactersInSet(charSet)
return result.joinWithSeparator("")
}
public func createFormattedStringFrom(text:String) -> String
{
var resultString = ""
if text.characters.count > 0 && pattern.characters.count > 0
{
var finalText = ""
var stop = false
let tempString = prepareString(text)
var formatIndex = pattern.startIndex
var tempIndex = tempString.startIndex
while !stop
{
let formattingPatternRange = formatIndex ..< formatIndex.advancedBy(1)
if pattern.substringWithRange(formattingPatternRange) != String(replecementChar) {
finalText = finalText.stringByAppendingString(pattern.substringWithRange(formattingPatternRange))
}
else if tempString.characters.count > 0 {
let pureStringRange = tempIndex ..< tempIndex.advancedBy(1)
finalText = finalText.stringByAppendingString(tempString.substringWithRange(pureStringRange))
tempIndex = tempIndex.advancedBy(1)
}
formatIndex = formatIndex.advancedBy(1)
if formatIndex >= pattern.endIndex || tempIndex >= tempString.endIndex {
stop = true
}
resultString = finalText
}
}
return resultString
}
}
The follow link send to the complete source code:
https://gist.github.com/dedeexe/d9a43894081317e7c418b96d1d081b25
This solution was base on this article:
http://vojtastavik.com/2015/03/29/real-time-formatting-in-uitextfield-swift-basics/
There is a simple solution I learned with "We <3 Swift" if you can't either import Foundation, use round() and/or does not want a String:
var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0
result: 31.726
Use this following code:
let intVal=56
let floatval:Double=56.897898
let doubleValue=89.0
let explicitDaouble:Double=89.56
let stringValue:"Hello"
let stringValue="String:\(stringValue) Integer:\(intVal) Float:\(floatval) Double:\(doubleValue) ExplicitDouble:\(explicitDaouble) "
The beauty of String(format:) is that you can save a formatting string and then reuse it later in dozen of places. It also can be localized in this single place. Where as in case of the interpolation approach you must write it again and again.
Simple functionality is not included in Swift, expected because it's included in other languages, can often be quickly coded for reuse. Pro tip for programmers to create a bag of tricks file that contains all this reuse code.
So from my bag of tricks we first need string multiplication for use in indentation.
#inlinable func * (string: String, scalar: Int) -> String {
let array = [String](repeating: string, count: scalar)
return array.joined(separator: "")
}
and then the code to add commas.
extension Int {
#inlinable var withCommas:String {
var i = self
var retValue:[String] = []
while i >= 1000 {
retValue.append(String(format:"%03d",i%1000))
i /= 1000
}
retValue.append("\(i)")
return retValue.reversed().joined(separator: ",")
}
#inlinable func withCommas(_ count:Int = 0) -> String {
let retValue = self.withCommas
let indentation = count - retValue.count
let indent:String = indentation >= 0 ? " " * indentation : ""
return indent + retValue
}
}
I just wrote this last function so I could get the columns to line up.
The #inlinable is great because it takes small functions and reduces their functionality so they run faster.
You can use either the variable version or, to get a fixed column, use the function version. Lengths set less than the needed columns will just expand the field.
Now you have something that is pure Swift and does not rely on some old objective C routine for NSString.
Since String(format: "%s" ...) is crashing at run time, here is code to allow write something like "hello".center(42); "world".alignLeft(42):
extension String {
// note: symbol names match to nim std/strutils lib:
func align (_ boxsz: UInt) -> String {
self.withCString { String(format: "%\(boxsz)s", $0) }
}
func alignLeft (_ boxsz: UInt) -> String {
self.withCString { String(format: "%-\(boxsz)s", $0) }
}
func center (_ boxsz: UInt) -> String {
let n = self.count
guard boxsz > n else { return self }
let padding = boxsz - UInt(n)
let R = padding / 2
guard R > 0 else { return " " + self }
let L = (padding%2 == 0) ? R : (R+1)
return " ".withCString { String(format: "%\(L)s\(self)%\(R)s", $0,$0) }
}
}
Success to try it:
var letters:NSString = "abcdefghijkl"
var strRendom = NSMutableString.stringWithCapacity(strlength)
for var i=0; i<strlength; i++ {
let rndString = Int(arc4random() % 12)
//let strlk = NSString(format: <#NSString#>, <#CVarArg[]#>)
let strlk = NSString(format: "%c", letters.characterAtIndex(rndString))
strRendom.appendString(String(strlk))
}

How to convert AnyBase to Base10?

I found some code to encode a Base10-String with to a custom BaseString:
func stringToCustomBase(encode: Int, alphabet: String) -> String {
var base = alphabet.count, int = encode, result = ""
repeat {
let index = alphabet.index(alphabet.startIndex, offsetBy: (int % base))
result = [alphabet[index]] + result
int /= base
} while (int > 0)
return result
}
... calling it with this lines:
let encoded = stringToCustomBase(encode: 9291, alphabet: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789")
print(encoded)
The encoding above works pretty good. But what about decoding the encoded string?
So because I got no idea how to decode a (in this case Base62 [alphabet.count=62]) to a human readable string (in this case [Base10]) any help would be super appreciated.
PS: (A full code solution is not required, I can also come up with some kind of pseudo-code or maybe just a few-lines of code)
This is what I've tried so far:
func reVal(num: Int) -> Character {
if (num >= 0 && num <= 9) {
return Character("\(num)")
}
return Character("\(num - 10)A");
}
func convertBack() {
var index = 0;
let encoded = "w2RDn3"
var decoded = [Character]()
var inputNum = encoded.count
repeat {
index+=1
decoded[index] = reVal(num: inputNum % 62)
//encoded[index] = reVal(inputNum % 62);
inputNum /= 62;
} while (inputNum > 0)
print(decoded);
}
Based on the original algorithm you need to iterate through each character of the encoded string, find the location of that character within the alphabet, and calculate the new result.
Here are both methods and some test code:
func stringToCustomBase(encode: Int, alphabet: String) -> String {
var base = alphabet.count, string = encode, result = ""
repeat {
let index = alphabet.index(alphabet.startIndex, offsetBy: (string % base))
result = [alphabet[index]] + result
string /= base
} while (string > 0)
return result
}
func customBaseToInt(encoded: String, alphabet: String) -> Int? {
let base = alphabet.count
var result = 0
for ch in encoded {
if let index = alphabet.index(of: ch) {
let mult = result.multipliedReportingOverflow(by: base)
if (mult.overflow) {
return nil
} else {
let add = mult.partialValue.addingReportingOverflow(alphabet.distance(from: alphabet.startIndex, to: index))
if (add.overflow) {
return nil
} else {
result = add.partialValue
}
}
} else {
return nil
}
}
return result
}
let startNum = 234567
let alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let codedNum = stringToCustomBase(encode: startNum, alphabet: alphabet)
let origNun = customBaseToInt(encoded: codedNum, alphabet: alphabet)
I made the customBaseToInt method return an optional result in case there are characters in the encoded value that are not in the provided alphabet.
You can achieve this via reduce:
enum RadixDecodingError: Error {
case invalidCharacter
case overflowed
}
func customRadixToInt(str: String, alphabet: String) throws -> Int {
return try str.reduce(into: 0) {
guard let digitIndex = alphabet.index(of: $1) else {
throw RadixDecodingError.invalidCharacter
}
let multiplied = $0.multipliedReportingOverflow(by: alphabet.count)
guard !multiplied.overflow else {
throw RadixDecodingError.overflowed
}
let added = multiplied.partialValue.addingReportingOverflow(alphabet.distance(from: alphabet.startIndex, to: digitIndex))
guard !added.overflow else {
throw RadixDecodingError.overflowed
}
$0 = added.partialValue
}
}
I used the exception throwing mechanism so that the caller can distinguish between invalid characters or overflow errors.

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