Related
I want to split string by chunks of given size 2
Example :
String "1234567" and output should be ["12", "34", "56","7"]
You can group your collection elements (in this case Characters) every n elements as follow:
extension Collection {
func unfoldSubSequences(limitedTo maxLength: Int) -> UnfoldSequence<SubSequence,Index> {
sequence(state: startIndex) { start in
guard start < self.endIndex else { return nil }
let end = self.index(start, offsetBy: maxLength, limitedBy: self.endIndex) ?? self.endIndex
defer { start = end }
return self[start..<end]
}
}
func subSequences(of n: Int) -> [SubSequence] {
.init(unfoldSubSequences(limitedTo: n))
}
}
let numbers = "1234567"
let subSequences = numbers.subSequences(of: 2)
print(subSequences) // ["12", "34", "56", "7"]
edit/update:
If you would like to append the exceeding characters to the last group:
extension Collection {
func unfoldSubSequencesWithTail(lenght: Int) -> UnfoldSequence<SubSequence,Index> {
let n = count / lenght
var counter = 0
return sequence(state: startIndex) { start in
guard start < endIndex else { return nil }
let end = index(start, offsetBy: lenght, limitedBy: endIndex) ?? endIndex
counter += 1
if counter == n {
defer { start = endIndex }
return self[start...]
} else {
defer { start = end }
return self[start..<end]
}
}
}
func subSequencesWithTail(n: Int) -> [SubSequence] {
.init(unfoldSubSequencesWithTail(lenght: n))
}
}
let numbers = "1234567"
let subSequencesWithTail = numbers.subSequencesWithTail(n: 2)
print(subSequencesWithTail) // ["12", "34", "567"]
var testString = "abcdefghijklmnopqrstu"
var startingPoint: Int = 0
var substringLength: Int = 1
var substringArray = [AnyHashable]()
for i in 0..<(testString.count ?? 0) / substringLength {
var substring: String = (testString as NSString).substring(with: NSRange(location: startingPoint, length: substringLength))
substringArray.append(substring)
startingPoint += substringLength
}
print("\(substringArray)")
OutPut :
(
a,
b,
c,
d,
e,
f,
g,
h,
i,
j,
k,
l,
m,
n,
o,
p,
q,
r,
s,
t,
u
)
try this
func SplitString(stringToBeSplitted:String, By:Int) -> [String]
{
var newArray = [String]()
var newStr = String()
for char in stringToBeSplitted
{
newStr += String(char)
if newStr.count == By
{
newArray.append(newStr)
newStr = ""
}
}
return newArray
}
Swift 5
extension Array {
func chunks(size: Int) -> [[Element]] {
return stride(from: 0, to: count, by: size).map {
Array(self[$0 ..< Swift.min($0 + size, count)])
}
}
}
extension String {
func chunks(size: Int) -> [String] {
map { $0 }.chunks(size: size).compactMap { String($0) }
}
}
let s = "1234567"
print(s.chunks(size: 2)) // ["12", "34", "56", "7"]
extension String {
func split(len: Int) -> [String] {
var currentIndex = 0
var array = [String]()
let length = self.characters.count
while currentIndex < length {
let startIndex = self.startIndex.advancedBy(currentIndex)
let endIndex = startIndex.advancedBy(len, limit: self.endIndex)
let substr = self.substringWithRange(Range(start: startIndex, end: endIndex))
array.append(substr)
currentIndex += len
}
return array
}
}
"123456789".split(2)
//output: ["12", "34", "56", "78", "9"]
I have write one method in objective c as below,
-(NSMutableArray*)splitString : (NSString*)str withRange : (int)range{
NSMutableArray *arr = [[NSMutableArray alloc]init];
NSMutableString *mutableStr = [[NSMutableString alloc]initWithString:str];
int j = 0;
int counter = 0;
for (int i = 0; i < str.length; i++) {
j++;
if (range == j) {
j = 0;
if (!(i == str.length - 1)) {
[mutableStr insertString:#"$" atIndex:i+1+counter];
}
counter++;
}
}
arr = (NSMutableArray*)[mutableStr componentsSeparatedByString:#"$"];
NSLog(#"%#",arr);
return arr;
}
You can call this method like,
[self splitString:#"123456" withRange:2];
and result will be,
(
12,
34,
56
)
You can also try below code:
var arrStr: [Substring] = []
let str = "1234567"
var i = 0
while i < str.count - 1 {
let index = str.index(str.startIndex, offsetBy: i)
//Below line gets current index and advances by 2
let substring = str[index..<str.index(index, offsetBy: 2)]
arrStr.append(substring)
i += 2
}
if str.count % 2 == 1 {
arrStr.append(str.suffix(1))
}
print(arrStr)
There's a stupid way, you can think about the rules of the data model.
var strOld = "123456"
print("The original string:\(strOld)")
strOld.insert("、", at: strOld.index(before: strOld.index(strOld.startIndex, offsetBy: 3)))
strOld.insert("、", at: strOld.index(before: strOld.index(strOld.startIndex, offsetBy: 6)))
print("After inserting:\(strOld)")
let str = strOld
let splitedArray = str.components(separatedBy: "、")
print("After the split of the array:\(splitedArray)")
let splitedArrayOther = str.split{$0 == "、"}.map(String.init)
print("After break up the array (method 2):\(splitedArrayOther)")
The results:
The original string:123456
After inserting:12、34、56
After the split of the array:["12", "34", "56"]
After break up the array (method 2):["12", "34", "56"]
Here's a short (and clean) solution, thanks to recursion:
extension Collection {
func chunks(ofSize size: Int) -> [SubSequence] {
// replace this by `guard count >= size else { return [] }`
// if you want to omit incomplete chunks
guard !isEmpty else { return [] }
return [prefix(size)] + dropFirst(size).chunks(ofSize: size)
}
}
The recursion should not pose a performance problem, as Swift has support for tail call optimization.
Also if Swift arrays are really fast when it comes to prepending or appending elements (like the Objective-C ones are), then the array operations should be fast.
Thus you get both fast and readable code (assuming my array assumptions are true).
How to parse string value like 12:02:21.3213 to NSTimeInterval? NSDateComponentsFormatter, available since iOS8, supports only formatting, not parsing.
Here is how you can do it in Swift,
It works for values like
2:12:12,
02:01:23.123213
Swift 5 (by #Youstanzr):
extension String {
func convertToTimeInterval() -> TimeInterval {
guard self != "" else {
return 0
}
var interval:Double = 0
let parts = self.components(separatedBy: ":")
for (index, part) in parts.reversed().enumerated() {
interval += (Double(part) ?? 0) * pow(Double(60), Double(index))
}
return interval
}
}
Swift 3 (by #Torre Lasley)
func parseDuration(_ timeString:String) -> TimeInterval {
guard !timeString.isEmpty else {
return 0
}
var interval:Double = 0
let parts = timeString.components(separatedBy: ":")
for (index, part) in parts.reversed().enumerated() {
interval += (Double(part) ?? 0) * pow(Double(60), Double(index))
}
return interval
}
Swift 2
func parseDuration(timeString:String) -> NSTimeInterval {
guard !timeString.isEmpty else {
return 0
}
var interval:Double = 0
let parts = timeString.componentsSeparatedByString(":")
for (index, part) in parts.reverse().enumerate() {
interval += (Double(part) ?? 0) * pow(Double(60), Double(index))
}
return interval
}
The solution provided by Bartosz Hernas worked for me, thank you!
For convenience, here it is for Swift 3:
func parseDuration(_ timeString:String) -> TimeInterval {
guard !timeString.isEmpty else {
return 0
}
var interval:Double = 0
let parts = timeString.components(separatedBy: ":")
for (index, part) in parts.reversed().enumerated() {
interval += (Double(part) ?? 0) * pow(Double(60), Double(index))
}
return interval
}
Here is the Swift 5 version that I've made of #Bartosz answer
extension String {
func convertToTimeInterval() -> TimeInterval {
guard self != "" else {
return 0
}
var interval:Double = 0
let parts = self.components(separatedBy: ":")
for (index, part) in parts.reversed().enumerated() {
interval += (Double(part) ?? 0) * pow(Double(60), Double(index))
}
return interval
}
}
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()
}
I'm displaying a distance with one decimal, and I would like to remove this decimal in case it is equal to 0 (ex: 1200.0Km), how could I do that in swift?
I'm displaying this number like this:
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: "%.1f", distanceFloat) + "Km"
Swift 3/4:
var distanceFloat1: Float = 5.0
var distanceFloat2: Float = 5.540
var distanceFloat3: Float = 5.03
extension Float {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
print("Value \(distanceFloat1.clean)") // 5
print("Value \(distanceFloat2.clean)") // 5.54
print("Value \(distanceFloat3.clean)") // 5.03
Swift 2 (Original answer)
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: distanceFloat == floor(distanceFloat) ? “%.0f" : "%.1f", distanceFloat) + "Km"
Or as an extension:
extension Float {
var clean: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Use NSNumberFormatter:
let formatter = NumberFormatter()
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = 2
// Avoid not getting a zero on numbers lower than 1
// Eg: .5, .67, etc...
formatter.numberStyle = .decimal
let nums = [3.0, 5.1, 7.21, 9.311, 600.0, 0.5677, 0.6988]
for num in nums {
print(formatter.string(from: num as NSNumber) ?? "n/a")
}
Returns:
3
5.1
7.21
9.31
600
0.57
0.7
extension is the powerful way to do it.
Extension:
Code for Swift 2 (not Swift 3 or newer):
extension Float {
var cleanValue: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Usage:
var sampleValue: Float = 3.234
print(sampleValue.cleanValue)
3.234
sampleValue = 3.0
print(sampleValue.cleanValue)
3
sampleValue = 3
print(sampleValue.cleanValue)
3
Sample Playground file is here.
Update of accepted answer for swift 3:
extension Float {
var cleanValue: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
usage would just be:
let someValue: Float = 3.0
print(someValue.cleanValue) //prints 3
To format it to String, follow this pattern
let aFloat: Float = 1.123
let aString: String = String(format: "%.0f", aFloat) // "1"
let aString: String = String(format: "%.1f", aFloat) // "1.1"
let aString: String = String(format: "%.2f", aFloat) // "1.12"
let aString: String = String(format: "%.3f", aFloat) // "1.123"
To cast it to Int, follow this pattern
let aInt: Int = Int(aFloat) // "1"
When you use String(format: initializer, Swift will automatically round the final digit as needed based on the following number.
You can use an extension as already mentioned, this solution is a little shorter though:
extension Float {
var shortValue: String {
return String(format: "%g", self)
}
}
Example usage:
var sample: Float = 3.234
print(sample.shortValue)
Swift 5
for Double it's same as #Frankie's answer for float
var dec: Double = 1.0
dec.clean // 1
for the extension
extension Double {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
Swift 5.5 makes it easy
Just use the new formatted() api with a default FloatingPointFormatStyle:
let values: [Double] = [1.0, 4.5, 100.0, 7]
for value in values {
print(value.formatted(FloatingPointFormatStyle()))
}
// prints "1, 4.5, 100, 7"
In Swift 4 try this.
extension CGFloat{
var cleanValue: String{
//return String(format: 1 == floor(self) ? "%.0f" : "%.2f", self)
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)//
}
}
//How to use - if you enter more then two-character after (.)point, it's automatically cropping the last character and only display two characters after the point.
let strValue = "32.12"
print(\(CGFloat(strValue).cleanValue)
Formatting with maximum fraction digits, without trailing zeros
This scenario is good when a custom output precision is desired.
This solution seems roughly as fast as NumberFormatter + NSNumber solution from MirekE, but one benefit could be that we're avoiding NSObject here.
extension Double {
func string(maximumFractionDigits: Int = 2) -> String {
let s = String(format: "%.\(maximumFractionDigits)f", self)
var offset = -maximumFractionDigits - 1
for i in stride(from: 0, to: -maximumFractionDigits, by: -1) {
if s[s.index(s.endIndex, offsetBy: i - 1)] != "0" {
offset = i
break
}
}
return String(s[..<s.index(s.endIndex, offsetBy: offset)])
}
}
(works also with extension Float, but not the macOS-only type Float80)
Usage: myNumericValue.string(maximumFractionDigits: 2) or myNumericValue.string()
Output for maximumFractionDigits: 2:
1.0 → "1"
0.12 → "0.12"
0.012 → "0.01"
0.0012 → "0"
0.00012 → "0"
Simple :
Int(floor(myFloatValue))
NSNumberFormatter is your friend
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
let numberFormatter = NSNumberFormatter()
numberFormatter.positiveFormat = "###0.##"
let distance = numberFormatter.stringFromNumber(NSNumber(float: distanceFloat))!
distanceLabel.text = distance + " Km"
Here's the full code.
let numberA: Float = 123.456
let numberB: Float = 789.000
func displayNumber(number: Float) {
if number - Float(Int(number)) == 0 {
println("\(Int(number))")
} else {
println("\(number)")
}
}
displayNumber(numberA) // console output: 123.456
displayNumber(numberB) // console output: 789
Here's the most important line in-depth.
func displayNumber(number: Float) {
Strips the float's decimal digits with Int(number).
Returns the stripped number back to float to do an operation with Float(Int(number)).
Gets the decimal-digit value with number - Float(Int(number))
Checks the decimal-digit value is empty with if number - Float(Int(number)) == 0
The contents within the if and else statements doesn't need explaining.
This might be helpful too.
extension Float {
func cleanValue() -> String {
let intValue = Int(self)
if self == 0 {return "0"}
if self / Float (intValue) == 1 { return "\(intValue)" }
return "\(self)"
}
}
Usage:
let number:Float = 45.23230000
number.cleanValue()
Maybe stringByReplacingOccurrencesOfString could help you :)
let aFloat: Float = 1.000
let aString: String = String(format: "%.1f", aFloat) // "1.0"
let wantedString: String = aString.stringByReplacingOccurrencesOfString(".0", withString: "") // "1"
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))
}