Related
I'm trying to properly reduce scale, formatting a float value and returning it as a String in Swift.
For example:
let value: Float = 4.8962965
// formattedFalue should be 4.90 or 4,90 based on localization
let formattedValue = value.formatNumber()
Here is what I did:
extension Float {
func reduceScale(to places: Int) -> Float {
let multiplier = pow(10, Float(places))
let newDecimal = multiplier * self // move the decimal right
let truncated = Float(Int(newDecimal)) // drop the fraction
let originalDecimal = truncated / multiplier // move the decimal back return originalDecimal
}
func formatNumber() -> String {
let num = abs(self)
let numberFormatter = NumberFormatter()
numberFormatter.usesGroupingSeparator = true
numberFormatter.minimumFractionDigits = 0
numberFormatter.maximumFractionDigits = 2
numberFormatter.roundingMode = .up
numberFormatter.numberStyle = .decimal
numberFormatter.locale = // we take it from app settings
let formatted = num.reduceScale(to: 2)
let returningString = numberFormatter.string(from: NSNumber(value: formatted))!
return "\(returningString)"
}
}
But when I use this code I get 4.89 (or 4,89 depending on the localization) instead of 4.90 (or 4,90) as I expect.
Thanks in advance.
You get 4.89 because reduceScale(to:) turns the number into 4.89 (actually, probably 4.89000something because 4.89 cannot be expressed exactly as a binary floating point). When the number formatter truncates this to two decimal places, it naturally rounds it down.
In fact, you don't need reduceScale(to:) at all because the rounding function of the number formatter will do it for you.
Also the final string interpolation is unnecessary because the result of NumberFormatter.string(from:) is automatically bridged to a String?
Also (see comments below by Dávid Pásztor and Sulthan) you can use string(for:) to obviate the NSNumber conversion.
This is what you need
import Foundation
extension Float {
func formatNumber() -> String {
let num = abs(self)
let numberFormatter = NumberFormatter()
numberFormatter.usesGroupingSeparator = true
numberFormatter.minimumFractionDigits = 0
numberFormatter.maximumFractionDigits = 2
numberFormatter.roundingMode = .up
numberFormatter.numberStyle = .decimal
numberFormatter.locale = whatever
return numberFormatter.string(for: num)!
}
}
let value: Float = 4.8962965
// formattedFalue should be 4.90 or 4,90 based on localization
let formattedValue = value.formatNumber() // "4.9"
Solved by following Sulthan's comments:
remove that reduceScale method which is not necessary and it will probably work as expected. You are truncating the decimal to 4.89 which cannot be rounded any more (it is already rounded). – Sulthan 6 hours ago
That's because you have specified minimumFractionDigits = 0. If you always want to display two decimal digits, you will have to set minimumFractionDigits = 2. – Sulthan 5 hours ago
Foundation has a better API now.
In-place, you can use .number:
value.formatted(.number
.precision(.fractionLength(2))
.locale(locale)
)
But it's only available on specific types. For an extension for more than one floating-point type, you'll need to use the equivalent initializer instead:
extension BinaryFloatingPoint {
var formattedTo2Places: String {
formatted(FloatingPointFormatStyle()
.precision(.fractionLength(2))
.locale(locale)
)
}
}
let locale = Locale(identifier: "it")
(4.8962965 as Float).formattedTo2Places // 4,90
I'm trying to write a BMI program in swift language.
And I got this problem: how to convert a String to a Double?
In Objective-C, I can do like this:
double myDouble = [myString doubleValue];
But how can I achieve this in Swift language?
Swift 2 Update
There are new failable initializers that allow you to do this in more idiomatic and safe way (as many answers have noted, NSString's double value is not very safe because it returns 0 for non number values. This means that the doubleValue of "foo" and "0" are the same.)
let myDouble = Double(myString)
This returns an optional, so in cases like passing in "foo" where doubleValue would have returned 0, the failable intializer will return nil. You can use a guard, if-let, or map to handle the Optional<Double>
Original Post:
You don't need to use the NSString constructor like the accepted answer proposes. You can simply bridge it like this:
(swiftString as NSString).doubleValue
Swift 4.2+ String to Double
You should use the new type initializers to convert between String and numeric types (Double, Float, Int). It'll return an Optional type (Double?) which will have the correct value or nil if the String was not a number.
Note: The NSString doubleValue property is not recommended because it returns 0 if the value cannot be converted (i.e.: bad user input).
let lessPrecisePI = Float("3.14")
let morePrecisePI = Double("3.1415926536")
let invalidNumber = Float("alphabet") // nil, not a valid number
Unwrap the values to use them using if/let
if let cost = Double(textField.text!) {
print("The user entered a value price of \(cost)")
} else {
print("Not a valid number: \(textField.text!)")
}
You can convert formatted numbers and currency using the NumberFormatter class.
let formatter = NumberFormatter()
formatter.locale = Locale.current // USA: Locale(identifier: "en_US")
formatter.numberStyle = .decimal
let number = formatter.number(from: "9,999.99")
Currency formats
let usLocale = Locale(identifier: "en_US")
let frenchLocale = Locale(identifier: "fr_FR")
let germanLocale = Locale(identifier: "de_DE")
let englishUKLocale = Locale(identifier: "en_GB") // United Kingdom
formatter.numberStyle = .currency
formatter.locale = usLocale
let usCurrency = formatter.number(from: "$9,999.99")
formatter.locale = frenchLocale
let frenchCurrency = formatter.number(from: "9999,99€")
// Note: "9 999,99€" fails with grouping separator
// Note: "9999,99 €" fails with a space before the €
formatter.locale = germanLocale
let germanCurrency = formatter.number(from: "9999,99€")
// Note: "9.999,99€" fails with grouping separator
formatter.locale = englishUKLocale
let englishUKCurrency = formatter.number(from: "£9,999.99")
Read more on my blog post about converting String to Double types (and currency).
For a little more Swift feeling, using NSFormatter() avoids casting to NSString, and returns nil when the string does not contain a Double value (e.g. "test" will not return 0.0).
let double = NSNumberFormatter().numberFromString(myString)?.doubleValue
Alternatively, extending Swift's String type:
extension String {
func toDouble() -> Double? {
return NumberFormatter().number(from: self)?.doubleValue
}
}
and use it like toInt():
var myString = "4.2"
var myDouble = myString.toDouble()
This returns an optional Double? which has to be unwrapped.
Either with forced unwrapping:
println("The value is \(myDouble!)") // prints: The value is 4.2
or with an if let statement:
if let myDouble = myDouble {
println("The value is \(myDouble)") // prints: The value is 4.2
}
Update:
For localization, it is very easy to apply locales to the NSFormatter as follows:
let formatter = NSNumberFormatter()
formatter.locale = NSLocale(localeIdentifier: "fr_FR")
let double = formatter.numberFromString("100,25")
Finally, you can use NSNumberFormatterCurrencyStyle on the formatter if you are working with currencies where the string contains the currency symbol.
Another option here is converting this to an NSString and using that:
let string = NSString(string: mySwiftString)
string.doubleValue
Here's an extension method that allows you to simply call doubleValue() on a Swift string and get a double back (example output comes first)
println("543.29".doubleValue())
println("543".doubleValue())
println(".29".doubleValue())
println("0.29".doubleValue())
println("-543.29".doubleValue())
println("-543".doubleValue())
println("-.29".doubleValue())
println("-0.29".doubleValue())
//prints
543.29
543.0
0.29
0.29
-543.29
-543.0
-0.29
-0.29
Here's the extension method:
extension String {
func doubleValue() -> Double
{
let minusAscii: UInt8 = 45
let dotAscii: UInt8 = 46
let zeroAscii: UInt8 = 48
var res = 0.0
let ascii = self.utf8
var whole = [Double]()
var current = ascii.startIndex
let negative = current != ascii.endIndex && ascii[current] == minusAscii
if (negative)
{
current = current.successor()
}
while current != ascii.endIndex && ascii[current] != dotAscii
{
whole.append(Double(ascii[current] - zeroAscii))
current = current.successor()
}
//whole number
var factor: Double = 1
for var i = countElements(whole) - 1; i >= 0; i--
{
res += Double(whole[i]) * factor
factor *= 10
}
//mantissa
if current != ascii.endIndex
{
factor = 0.1
current = current.successor()
while current != ascii.endIndex
{
res += Double(ascii[current] - zeroAscii) * factor
factor *= 0.1
current = current.successor()
}
}
if (negative)
{
res *= -1;
}
return res
}
}
No error checking, but you can add it if you need it.
As of Swift 1.1, you can directly pass String to const char * parameter.
import Foundation
let str = "123.4567"
let num = atof(str) // -> 123.4567
atof("123.4567fubar") // -> 123.4567
If you don't like deprecated atof:
strtod("765.4321", nil) // -> 765.4321
One caveat: the behavior of conversion is different from NSString.doubleValue.
atof and strtod accept 0x prefixed hex string:
atof("0xffp-2") // -> 63.75
atof("12.3456e+2") // -> 1,234.56
atof("nan") // -> (not a number)
atof("inf") // -> (+infinity)
If you prefer .doubleValue behavior, we can still use CFString bridging:
let str = "0xff"
atof(str) // -> 255.0
strtod(str, nil) // -> 255.0
CFStringGetDoubleValue(str) // -> 0.0
(str as NSString).doubleValue // -> 0.0
I haven't seen the answer I was looking for.
I just post in here mine in case it can help anyone. This answer is valid only if you don't need a specific format.
Swift 3
extension String {
var toDouble: Double {
return Double(self) ?? 0.0
}
}
In Swift 2.0 the best way is to avoid thinking like an Objective-C developer. So you should not "convert a String to a Double" but you should "initialize a Double from a String". Apple doc over here:
https://developer.apple.com/library/ios//documentation/Swift/Reference/Swift_Double_Structure/index.html#//apple_ref/swift/structctr/Double/s:FSdcFMSdFSSGSqSd_
It's an optional init so you can use the nil coalescing operator (??) to set a default value. Example:
let myDouble = Double("1.1") ?? 0.0
On SWIFT 3, you can use:
if let myDouble = NumberFormatter().number(from: yourString)?.doubleValue {
print("My double: \(myDouble)")
}
Note:
- If a string contains any characters other than numerical digits or locale-appropriate group or decimal separators, parsing will fail.
- Any leading or trailing space separator characters in a string are ignored. For example, the strings “ 5”, “5 ”, and “5” all produce the number 5.
Taken from the documentation:
https://developer.apple.com/reference/foundation/numberformatter/1408845-number
Try this:
var myDouble = myString.bridgeToObjectiveC().doubleValue
println(myDouble)
NOTE
Removed in Beta 5. This no longer works ?
This is building upon the answer by #Ryu
His solution is great as long as you're in a country where dots are used as separators. By default NSNumberFormatter uses the devices locale. Therefore this will fail in all countries where a comma is used as the default separator (including France as #PeterK. mentioned) if the number uses dots as separators (which is normally the case). To set the locale of this NSNumberFormatter to be US and thus use dots as separators replace the line
return NSNumberFormatter().numberFromString(self)?.doubleValue
with
let numberFormatter = NSNumberFormatter()
numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
return numberFormatter.numberFromString(self)?.doubleValue
Therefore the full code becomes
extension String {
func toDouble() -> Double? {
let numberFormatter = NSNumberFormatter()
numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
return numberFormatter.numberFromString(self)?.doubleValue
}
}
To use this, just call "Your text goes here".toDouble()
This will return an optional Double?
As #Ryu mentioned you can either force unwrap:
println("The value is \(myDouble!)") // prints: The value is 4.2
or use an if let statement:
if let myDouble = myDouble {
println("The value is \(myDouble)") // prints: The value is 4.2
}
SWIFT 4
extension String {
func toDouble() -> Double? {
let numberFormatter = NumberFormatter()
numberFormatter.locale = Locale(identifier: "en_US_POSIX")
return numberFormatter.number(from: self)?.doubleValue
}
}
Swift 4.0
try this
let str:String = "111.11"
let tempString = (str as NSString).doubleValue
print("String:-",tempString)
Swift : 4 and 5
There are possibly two ways to do this:
String -> Int -> Double:
let strNumber = "314"
if let intFromString = Int(strNumber){
let dobleFromInt = Double(intFromString)
print(dobleFromInt)
}
String -> NSString -> Double
let strNumber1 = "314"
let NSstringFromString = NSString(string: strNumber1)
let doubleFromNSString = NSstringFromString.doubleValue
print(doubleFromNSString)
Use it anyway you like according to you need of the code.
Please check it on playground!
let sString = "236.86"
var dNumber = NSNumberFormatter().numberFromString(sString)
var nDouble = dNumber!
var eNumber = Double(nDouble) * 3.7
By the way in my Xcode
.toDouble() - doesn't exist
.doubleValue create value 0.0 from not numerical strings...
As already pointed out, the best way to achieve this is with direct casting:
(myString as NSString).doubleValue
Building from that, you can make a slick native Swift String extension:
extension String {
var doubleValue: Double {
return (self as NSString).doubleValue
}
}
This allows you to directly use:
myString.doubleValue
Which will perform the casting for you. If Apple does add a doubleValue to the native String you just need to remove the extension and the rest of your code will automatically compile fine!
1.
let strswift = "12"
let double = (strswift as NSString).doubleValue
2.
var strswift= "10.6"
var double : Double = NSString(string: strswift).doubleValue
May be this help for you.
Extension with optional locale
Swift 2.2
extension String {
func toDouble(locale: NSLocale? = nil) -> Double? {
let formatter = NSNumberFormatter()
if let locale = locale {
formatter.locale = locale
}
return formatter.numberFromString(self)?.doubleValue
}
}
Swift 3.1
extension String {
func toDouble(_ locale: Locale) -> Double {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.locale = locale
formatter.usesGroupingSeparator = true
if let result = formatter.number(from: self)?.doubleValue {
return result
} else {
return 0
}
}
}
SWIFT 3
To clear, nowadays there is a default method:
public init?(_ text: String)` of `Double` class.
It can be used for all classes.
let c = Double("-1.0")
let f = Double("0x1c.6")
let i = Double("inf")
, etc.
Or you could do:
var myDouble = Double((mySwiftString.text as NSString).doubleValue)
You can use StringEx. It extends String with string-to-number conversions including toDouble().
extension String {
func toDouble() -> Double?
}
It verifies the string and fails if it can't be converted to double.
Example:
import StringEx
let str = "123.45678"
if let num = str.toDouble() {
println("Number: \(num)")
} else {
println("Invalid string")
}
Swift 4
extension String {
func toDouble() -> Double {
let nsString = self as NSString
return nsString.doubleValue
}
}
What also works:
// Init default Double variable
var scanned: Double()
let scanner = NSScanner(string: "String to Scan")
scanner.scanDouble(&scanned)
// scanned has now the scanned value if something was found.
Using Scanner in some cases is a very convenient way of extracting numbers from a string. And it is almost as powerful as NumberFormatter when it comes to decoding and dealing with different number formats and locales. It can extract numbers and currencies with different decimal and group separators.
import Foundation
// The code below includes manual fix for whitespaces (for French case)
let strings = ["en_US": "My salary is $9,999.99",
"fr_FR": "Mon salaire est 9 999,99€",
"de_DE": "Mein Gehalt ist 9999,99€",
"en_GB": "My salary is £9,999.99" ]
// Just for referce
let allPossibleDecimalSeparators = Set(Locale.availableIdentifiers.compactMap({ Locale(identifier: $0).decimalSeparator}))
print(allPossibleDecimalSeparators)
for str in strings {
let locale = Locale(identifier: str.key)
let valStr = str.value.filter{!($0.isWhitespace || $0 == Character(locale.groupingSeparator ?? ""))}
print("Value String", valStr)
let sc = Scanner(string: valStr)
// we could do this more reliably with `filter` as well
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
However, there are issues with separators that could be conceived as word delimiters.
// This doesn't work. `Scanner` just ignores grouping separators because scanner tends to seek for multiple values
// It just refuses to ignore spaces or commas for example.
let strings = ["en_US": "$9,999.99", "fr_FR": "9999,99€", "de_DE": "9999,99€", "en_GB": "£9,999.99" ]
for str in strings {
let locale = Locale(identifier: str.key)
let sc = Scanner(string: str.value)
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted.union(CharacterSet(charactersIn: locale.groupingSeparator ?? ""))
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: \(locale.groupingSeparator ?? "") . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
// sc.scanDouble(representation: Scanner.NumberRepresentation) could help if there were .currency case
There is no problem to auto detect locale. Note that groupingSeparator in French locale in string "Mon salaire est 9 999,99€" is not a space, though it may render exactly as space (here it doesn't). Thats why the code below works fine without !$0.isWhitespace characters being filtered out.
let stringsArr = ["My salary is $9,999.99",
"Mon salaire est 9 999,99€",
"Mein Gehalt ist 9.999,99€",
"My salary is £9,999.99" ]
let tagger = NSLinguisticTagger(tagSchemes: [.language], options: Int(NSLinguisticTagger.Options.init().rawValue))
for str in stringsArr {
tagger.string = str
let locale = Locale(identifier: tagger.dominantLanguage ?? "en")
let valStr = str.filter{!($0 == Character(locale.groupingSeparator ?? ""))}
print("Value String", valStr)
let sc = Scanner(string: valStr)
// we could do this more reliably with `filter` as well
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
// Also will fail if groupingSeparator == decimalSeparator (but don't think it's possible)
Use this code in Swift 2.0
let strWithFloat = "78.65"
let floatFromString = Double(strWithFloat)
In the cases of strings contain other characters like: "27.8 °C", "52.523553 kM" or "Total: 349.0".
This works in Swift 4:
let anyString = "52.523553 kM"
let digitsCharacterSet = CharacterSet.init(charactersIn: "0123456789.")
let doubleResult = Double(anyString.components(separatedBy:digitsCharacterSet.inverted).joined())
Caution! This not working for strings contain multiple . like "27.8 °C 3.5 kM"
I find more readable to add an extension to String as follow:
extension String {
var doubleValue: Double {
return (self as NSString).doubleValue
}
}
and then you just could write your code:
myDouble = myString.doubleValue
my problem was comma so i solve it this way:
extension String {
var doubleValue: Double {
return Double((self.replacingOccurrences(of: ",", with: ".") as NSString).doubleValue)
}
}
var stringValue = "55"
var convertToDouble = Double((stringValue as NSString).doubleValue)
we can use CDouble value which will be obtained by myString.doubleValue
My function is converting a string to Decimal
func getDecimalFromString(_ strValue: String) -> NSDecimalNumber {
let formatter = NumberFormatter()
formatter.maximumFractionDigits = 1
formatter.generatesDecimalNumbers = true
return formatter.number(from: strValue) as? NSDecimalNumber ?? 0
}
But it is not working as per expectation. Sometimes it's returning like
Optional(8.300000000000001)
Optional(8.199999999999999)
instead of 8.3 or 8.2. In the string, I have value like "8.3" or "8.2" but the converted decimal is not as per my requirements. Any suggestion where I made mistake?
Setting generatesDecimalNumbers to true does not work as one might expect. The returned value is an instance of NSDecimalNumber (which can represent the value 8.3 exactly), but apparently the formatter converts the string to a binary floating number first (and that can not represent 8.3 exactly). Therefore the returned decimal value is only approximately correct.
That has also been reported as a bug:
NSDecimalNumbers from NSNumberFormatter are affected by binary approximation error
Note also that (contrary to the documentation), the maximumFractionDigits property has no effect when parsing a string
into a number.
There is a simple solution: Use
NSDecimalNumber(string: strValue) // or
NSDecimalNumber(string: strValue, locale: Locale.current)
instead, depending on whether the string is localized or not.
Or with the Swift 3 Decimal type:
Decimal(string: strValue) // or
Decimal(string: strValue, locale: .current)
Example:
if let d = Decimal(string: "8.2") {
print(d) // 8.2
}
I would probably be inclined to just use Decimal(string:locale:), but if you want to use NumberFormatter, you would just manually round it.
func getDecimalFromString(_ string: String) -> NSDecimalNumber {
let formatter = NumberFormatter()
formatter.generatesDecimalNumbers = true
let value = formatter.number(from: string) as? NSDecimalNumber ?? 0
return value.rounding(accordingToBehavior: RoundingBehavior(scale: 1))
}
Or if you want to return a Decimal:
func getDecimalFromString(_ string: String) -> Decimal {
let formatter = NumberFormatter()
formatter.generatesDecimalNumbers = true
let value = formatter.number(from: string) as? NSDecimalNumber ?? 0
return value.rounding(accordingToBehavior: RoundingBehavior(scale: 1)) as Decimal
}
Where
class RoundingBehavior: NSDecimalNumberBehaviors {
private let _scale: Int16
init(scale: Int16) {
_scale = scale
}
func roundingMode() -> NSDecimalNumber.RoundingMode {
.plain
}
func scale() -> Int16 {
_scale
}
func exceptionDuringOperation(_ operation: Selector, error: NSDecimalNumber.CalculationError, leftOperand: NSDecimalNumber, rightOperand: NSDecimalNumber?) -> NSDecimalNumber? {
.notANumber
}
}
Before I updated xCode 6, I had no problems casting a double to a string but now it gives me an error
var a: Double = 1.5
var b: String = String(a)
It gives me the error message "double is not convertible to string". Is there any other way to do it?
It is not casting, it is creating a string from a value with a format.
let a: Double = 1.5
let b: String = String(format: "%f", a)
print("b: \(b)") // b: 1.500000
With a different format:
let c: String = String(format: "%.1f", a)
print("c: \(c)") // c: 1.5
You can also omit the format property if no formatting is needed.
let double = 1.5
let string = double.description
update Xcode 7.1 • Swift 2.1:
Now Double is also convertible to String so you can simply use it as you wish:
let double = 1.5
let doubleString = String(double) // "1.5"
Swift 3 or later we can extend LosslessStringConvertible and make it generic
Xcode 11.3 • Swift 5.1 or later
extension LosslessStringConvertible {
var string: String { .init(self) }
}
let double = 1.5
let string = double.string // "1.5"
For a fixed number of fraction digits we can extend FloatingPoint protocol:
extension FloatingPoint where Self: CVarArg {
func fixedFraction(digits: Int) -> String {
.init(format: "%.*f", digits, self)
}
}
If you need more control over your number format (minimum and maximum fraction digits and rounding mode) you can use NumberFormatter:
extension Formatter {
static let number = NumberFormatter()
}
extension FloatingPoint {
func fractionDigits(min: Int = 2, max: Int = 2, roundingMode: NumberFormatter.RoundingMode = .halfEven) -> String {
Formatter.number.minimumFractionDigits = min
Formatter.number.maximumFractionDigits = max
Formatter.number.roundingMode = roundingMode
Formatter.number.numberStyle = .decimal
return Formatter.number.string(for: self) ?? ""
}
}
2.12345.fractionDigits() // "2.12"
2.12345.fractionDigits(min: 3, max: 3, roundingMode: .up) // "2.124"
In addition to #Zaph's answer, you can create an extension on Double:
extension Double {
func toString() -> String {
return String(format: "%.1f",self)
}
}
Usage:
var a:Double = 1.5
println("output: \(a.toString())") // output: 1.5
Swift 3+: Try these line of code
let num: Double = 1.5
let str = String(format: "%.2f", num)
to make anything a string in swift except maybe enum values simply do what you do in the println() method
for example:
var stringOfDBL = "\(myDouble)"
There are many answers here that suggest a variety of techniques. But when presenting numbers in the UI, you invariably want to use a NumberFormatter so that the results are properly formatted, rounded, and localized:
let value = 10000.5
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
guard let string = formatter.string(for: value) else { return }
print(string) // 10,000.5
If you want fixed number of decimal places, e.g. for currency values
let value = 10000.5
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.maximumFractionDigits = 2
formatter.minimumFractionDigits = 2
guard let string = formatter.string(for: value) else { return }
print(string) // 10,000.50
But the beauty of this approach, is that it will be properly localized, resulting in 10,000.50 in the US but 10.000,50 in Germany. Different locales have different preferred formats for numbers, and we should let NumberFormatter use the format preferred by the end user when presenting numeric values within the UI.
Needless to say, while NumberFormatter is essential when preparing string representations within the UI, it should not be used if writing numeric values as strings for persistent storage, interface with web services, etc.
Swift 4:
Use following code
let number = 2.4
let string = String(format: "%.2f", number)
This function will let you specify the number of decimal places to show:
func doubleToString(number:Double, numberOfDecimalPlaces:Int) -> String {
return String(format:"%."+numberOfDecimalPlaces.description+"f", number)
}
Usage:
let numberString = doubleToStringDecimalPlacesWithDouble(number: x, numberOfDecimalPlaces: 2)
In swift 3:
var a: Double = 1.5
var b: String = String(a)
In swift 3 it is simple as given below
let stringDouble = String(describing: double)
I would prefer NSNumber and NumberFormatter approach (where need), also u can use extension to avoid bloating code
extension Double {
var toString: String {
return NSNumber(value: self).stringValue
}
}
U can also need reverse approach
extension String {
var toDouble: Double {
return Double(self) ?? .nan
}
}
var b = String(stringInterpolationSegment: a)
This works for me. You may have a try
In Swift 4 if you like to modify and use a Double in the UI as a textLabel "String" you can add this in the end of your file:
extension Double {
func roundToInt() -> Int{
return Int(Darwin.round(self))
}
}
And use it like this if you like to have it in a textlabel:
currentTemp.text = "\(weatherData.tempCelsius.roundToInt())"
Or print it as an Int:
print(weatherData.tempCelsius.roundToInt())
Swift 5:
Use following code
extension Double {
func getStringValue(withFloatingPoints points: Int = 0) -> String {
let valDouble = modf(self)
let fractionalVal = (valDouble.1)
if fractionalVal > 0 {
return String(format: "%.*f", points, self)
}
return String(format: "%.0f", self)
}
}
You shouldn't really ever cast a double to a string, the most common reason for casting a float to a string is to present it to a user, but floats are not real number and can only approximate lots of values, similar to how ⅓ can not be represented as a decimal number with a finite number of decimal places. Instead keep you values as float for all their use, then when you want to present them to the user, use something like NumberFormater to convert them for your. This stage of converting for user presentation is what something like your viewModel should do.
Use this.
Text(String(format: "%.2f", doubleValue))
I'm trying to write a BMI program in swift language.
And I got this problem: how to convert a String to a Double?
In Objective-C, I can do like this:
double myDouble = [myString doubleValue];
But how can I achieve this in Swift language?
Swift 2 Update
There are new failable initializers that allow you to do this in more idiomatic and safe way (as many answers have noted, NSString's double value is not very safe because it returns 0 for non number values. This means that the doubleValue of "foo" and "0" are the same.)
let myDouble = Double(myString)
This returns an optional, so in cases like passing in "foo" where doubleValue would have returned 0, the failable intializer will return nil. You can use a guard, if-let, or map to handle the Optional<Double>
Original Post:
You don't need to use the NSString constructor like the accepted answer proposes. You can simply bridge it like this:
(swiftString as NSString).doubleValue
Swift 4.2+ String to Double
You should use the new type initializers to convert between String and numeric types (Double, Float, Int). It'll return an Optional type (Double?) which will have the correct value or nil if the String was not a number.
Note: The NSString doubleValue property is not recommended because it returns 0 if the value cannot be converted (i.e.: bad user input).
let lessPrecisePI = Float("3.14")
let morePrecisePI = Double("3.1415926536")
let invalidNumber = Float("alphabet") // nil, not a valid number
Unwrap the values to use them using if/let
if let cost = Double(textField.text!) {
print("The user entered a value price of \(cost)")
} else {
print("Not a valid number: \(textField.text!)")
}
You can convert formatted numbers and currency using the NumberFormatter class.
let formatter = NumberFormatter()
formatter.locale = Locale.current // USA: Locale(identifier: "en_US")
formatter.numberStyle = .decimal
let number = formatter.number(from: "9,999.99")
Currency formats
let usLocale = Locale(identifier: "en_US")
let frenchLocale = Locale(identifier: "fr_FR")
let germanLocale = Locale(identifier: "de_DE")
let englishUKLocale = Locale(identifier: "en_GB") // United Kingdom
formatter.numberStyle = .currency
formatter.locale = usLocale
let usCurrency = formatter.number(from: "$9,999.99")
formatter.locale = frenchLocale
let frenchCurrency = formatter.number(from: "9999,99€")
// Note: "9 999,99€" fails with grouping separator
// Note: "9999,99 €" fails with a space before the €
formatter.locale = germanLocale
let germanCurrency = formatter.number(from: "9999,99€")
// Note: "9.999,99€" fails with grouping separator
formatter.locale = englishUKLocale
let englishUKCurrency = formatter.number(from: "£9,999.99")
Read more on my blog post about converting String to Double types (and currency).
For a little more Swift feeling, using NSFormatter() avoids casting to NSString, and returns nil when the string does not contain a Double value (e.g. "test" will not return 0.0).
let double = NSNumberFormatter().numberFromString(myString)?.doubleValue
Alternatively, extending Swift's String type:
extension String {
func toDouble() -> Double? {
return NumberFormatter().number(from: self)?.doubleValue
}
}
and use it like toInt():
var myString = "4.2"
var myDouble = myString.toDouble()
This returns an optional Double? which has to be unwrapped.
Either with forced unwrapping:
println("The value is \(myDouble!)") // prints: The value is 4.2
or with an if let statement:
if let myDouble = myDouble {
println("The value is \(myDouble)") // prints: The value is 4.2
}
Update:
For localization, it is very easy to apply locales to the NSFormatter as follows:
let formatter = NSNumberFormatter()
formatter.locale = NSLocale(localeIdentifier: "fr_FR")
let double = formatter.numberFromString("100,25")
Finally, you can use NSNumberFormatterCurrencyStyle on the formatter if you are working with currencies where the string contains the currency symbol.
Another option here is converting this to an NSString and using that:
let string = NSString(string: mySwiftString)
string.doubleValue
Here's an extension method that allows you to simply call doubleValue() on a Swift string and get a double back (example output comes first)
println("543.29".doubleValue())
println("543".doubleValue())
println(".29".doubleValue())
println("0.29".doubleValue())
println("-543.29".doubleValue())
println("-543".doubleValue())
println("-.29".doubleValue())
println("-0.29".doubleValue())
//prints
543.29
543.0
0.29
0.29
-543.29
-543.0
-0.29
-0.29
Here's the extension method:
extension String {
func doubleValue() -> Double
{
let minusAscii: UInt8 = 45
let dotAscii: UInt8 = 46
let zeroAscii: UInt8 = 48
var res = 0.0
let ascii = self.utf8
var whole = [Double]()
var current = ascii.startIndex
let negative = current != ascii.endIndex && ascii[current] == minusAscii
if (negative)
{
current = current.successor()
}
while current != ascii.endIndex && ascii[current] != dotAscii
{
whole.append(Double(ascii[current] - zeroAscii))
current = current.successor()
}
//whole number
var factor: Double = 1
for var i = countElements(whole) - 1; i >= 0; i--
{
res += Double(whole[i]) * factor
factor *= 10
}
//mantissa
if current != ascii.endIndex
{
factor = 0.1
current = current.successor()
while current != ascii.endIndex
{
res += Double(ascii[current] - zeroAscii) * factor
factor *= 0.1
current = current.successor()
}
}
if (negative)
{
res *= -1;
}
return res
}
}
No error checking, but you can add it if you need it.
As of Swift 1.1, you can directly pass String to const char * parameter.
import Foundation
let str = "123.4567"
let num = atof(str) // -> 123.4567
atof("123.4567fubar") // -> 123.4567
If you don't like deprecated atof:
strtod("765.4321", nil) // -> 765.4321
One caveat: the behavior of conversion is different from NSString.doubleValue.
atof and strtod accept 0x prefixed hex string:
atof("0xffp-2") // -> 63.75
atof("12.3456e+2") // -> 1,234.56
atof("nan") // -> (not a number)
atof("inf") // -> (+infinity)
If you prefer .doubleValue behavior, we can still use CFString bridging:
let str = "0xff"
atof(str) // -> 255.0
strtod(str, nil) // -> 255.0
CFStringGetDoubleValue(str) // -> 0.0
(str as NSString).doubleValue // -> 0.0
I haven't seen the answer I was looking for.
I just post in here mine in case it can help anyone. This answer is valid only if you don't need a specific format.
Swift 3
extension String {
var toDouble: Double {
return Double(self) ?? 0.0
}
}
In Swift 2.0 the best way is to avoid thinking like an Objective-C developer. So you should not "convert a String to a Double" but you should "initialize a Double from a String". Apple doc over here:
https://developer.apple.com/library/ios//documentation/Swift/Reference/Swift_Double_Structure/index.html#//apple_ref/swift/structctr/Double/s:FSdcFMSdFSSGSqSd_
It's an optional init so you can use the nil coalescing operator (??) to set a default value. Example:
let myDouble = Double("1.1") ?? 0.0
On SWIFT 3, you can use:
if let myDouble = NumberFormatter().number(from: yourString)?.doubleValue {
print("My double: \(myDouble)")
}
Note:
- If a string contains any characters other than numerical digits or locale-appropriate group or decimal separators, parsing will fail.
- Any leading or trailing space separator characters in a string are ignored. For example, the strings “ 5”, “5 ”, and “5” all produce the number 5.
Taken from the documentation:
https://developer.apple.com/reference/foundation/numberformatter/1408845-number
Try this:
var myDouble = myString.bridgeToObjectiveC().doubleValue
println(myDouble)
NOTE
Removed in Beta 5. This no longer works ?
This is building upon the answer by #Ryu
His solution is great as long as you're in a country where dots are used as separators. By default NSNumberFormatter uses the devices locale. Therefore this will fail in all countries where a comma is used as the default separator (including France as #PeterK. mentioned) if the number uses dots as separators (which is normally the case). To set the locale of this NSNumberFormatter to be US and thus use dots as separators replace the line
return NSNumberFormatter().numberFromString(self)?.doubleValue
with
let numberFormatter = NSNumberFormatter()
numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
return numberFormatter.numberFromString(self)?.doubleValue
Therefore the full code becomes
extension String {
func toDouble() -> Double? {
let numberFormatter = NSNumberFormatter()
numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
return numberFormatter.numberFromString(self)?.doubleValue
}
}
To use this, just call "Your text goes here".toDouble()
This will return an optional Double?
As #Ryu mentioned you can either force unwrap:
println("The value is \(myDouble!)") // prints: The value is 4.2
or use an if let statement:
if let myDouble = myDouble {
println("The value is \(myDouble)") // prints: The value is 4.2
}
SWIFT 4
extension String {
func toDouble() -> Double? {
let numberFormatter = NumberFormatter()
numberFormatter.locale = Locale(identifier: "en_US_POSIX")
return numberFormatter.number(from: self)?.doubleValue
}
}
Swift 4.0
try this
let str:String = "111.11"
let tempString = (str as NSString).doubleValue
print("String:-",tempString)
Swift : 4 and 5
There are possibly two ways to do this:
String -> Int -> Double:
let strNumber = "314"
if let intFromString = Int(strNumber){
let dobleFromInt = Double(intFromString)
print(dobleFromInt)
}
String -> NSString -> Double
let strNumber1 = "314"
let NSstringFromString = NSString(string: strNumber1)
let doubleFromNSString = NSstringFromString.doubleValue
print(doubleFromNSString)
Use it anyway you like according to you need of the code.
Please check it on playground!
let sString = "236.86"
var dNumber = NSNumberFormatter().numberFromString(sString)
var nDouble = dNumber!
var eNumber = Double(nDouble) * 3.7
By the way in my Xcode
.toDouble() - doesn't exist
.doubleValue create value 0.0 from not numerical strings...
As already pointed out, the best way to achieve this is with direct casting:
(myString as NSString).doubleValue
Building from that, you can make a slick native Swift String extension:
extension String {
var doubleValue: Double {
return (self as NSString).doubleValue
}
}
This allows you to directly use:
myString.doubleValue
Which will perform the casting for you. If Apple does add a doubleValue to the native String you just need to remove the extension and the rest of your code will automatically compile fine!
1.
let strswift = "12"
let double = (strswift as NSString).doubleValue
2.
var strswift= "10.6"
var double : Double = NSString(string: strswift).doubleValue
May be this help for you.
Extension with optional locale
Swift 2.2
extension String {
func toDouble(locale: NSLocale? = nil) -> Double? {
let formatter = NSNumberFormatter()
if let locale = locale {
formatter.locale = locale
}
return formatter.numberFromString(self)?.doubleValue
}
}
Swift 3.1
extension String {
func toDouble(_ locale: Locale) -> Double {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.locale = locale
formatter.usesGroupingSeparator = true
if let result = formatter.number(from: self)?.doubleValue {
return result
} else {
return 0
}
}
}
SWIFT 3
To clear, nowadays there is a default method:
public init?(_ text: String)` of `Double` class.
It can be used for all classes.
let c = Double("-1.0")
let f = Double("0x1c.6")
let i = Double("inf")
, etc.
Or you could do:
var myDouble = Double((mySwiftString.text as NSString).doubleValue)
You can use StringEx. It extends String with string-to-number conversions including toDouble().
extension String {
func toDouble() -> Double?
}
It verifies the string and fails if it can't be converted to double.
Example:
import StringEx
let str = "123.45678"
if let num = str.toDouble() {
println("Number: \(num)")
} else {
println("Invalid string")
}
Swift 4
extension String {
func toDouble() -> Double {
let nsString = self as NSString
return nsString.doubleValue
}
}
What also works:
// Init default Double variable
var scanned: Double()
let scanner = NSScanner(string: "String to Scan")
scanner.scanDouble(&scanned)
// scanned has now the scanned value if something was found.
Using Scanner in some cases is a very convenient way of extracting numbers from a string. And it is almost as powerful as NumberFormatter when it comes to decoding and dealing with different number formats and locales. It can extract numbers and currencies with different decimal and group separators.
import Foundation
// The code below includes manual fix for whitespaces (for French case)
let strings = ["en_US": "My salary is $9,999.99",
"fr_FR": "Mon salaire est 9 999,99€",
"de_DE": "Mein Gehalt ist 9999,99€",
"en_GB": "My salary is £9,999.99" ]
// Just for referce
let allPossibleDecimalSeparators = Set(Locale.availableIdentifiers.compactMap({ Locale(identifier: $0).decimalSeparator}))
print(allPossibleDecimalSeparators)
for str in strings {
let locale = Locale(identifier: str.key)
let valStr = str.value.filter{!($0.isWhitespace || $0 == Character(locale.groupingSeparator ?? ""))}
print("Value String", valStr)
let sc = Scanner(string: valStr)
// we could do this more reliably with `filter` as well
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
However, there are issues with separators that could be conceived as word delimiters.
// This doesn't work. `Scanner` just ignores grouping separators because scanner tends to seek for multiple values
// It just refuses to ignore spaces or commas for example.
let strings = ["en_US": "$9,999.99", "fr_FR": "9999,99€", "de_DE": "9999,99€", "en_GB": "£9,999.99" ]
for str in strings {
let locale = Locale(identifier: str.key)
let sc = Scanner(string: str.value)
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted.union(CharacterSet(charactersIn: locale.groupingSeparator ?? ""))
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: \(locale.groupingSeparator ?? "") . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
// sc.scanDouble(representation: Scanner.NumberRepresentation) could help if there were .currency case
There is no problem to auto detect locale. Note that groupingSeparator in French locale in string "Mon salaire est 9 999,99€" is not a space, though it may render exactly as space (here it doesn't). Thats why the code below works fine without !$0.isWhitespace characters being filtered out.
let stringsArr = ["My salary is $9,999.99",
"Mon salaire est 9 999,99€",
"Mein Gehalt ist 9.999,99€",
"My salary is £9,999.99" ]
let tagger = NSLinguisticTagger(tagSchemes: [.language], options: Int(NSLinguisticTagger.Options.init().rawValue))
for str in stringsArr {
tagger.string = str
let locale = Locale(identifier: tagger.dominantLanguage ?? "en")
let valStr = str.filter{!($0 == Character(locale.groupingSeparator ?? ""))}
print("Value String", valStr)
let sc = Scanner(string: valStr)
// we could do this more reliably with `filter` as well
sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
sc.locale = locale
print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
while !(sc.isAtEnd) {
if let val = sc.scanDouble() {
print(val)
}
}
}
// Also will fail if groupingSeparator == decimalSeparator (but don't think it's possible)
Use this code in Swift 2.0
let strWithFloat = "78.65"
let floatFromString = Double(strWithFloat)
In the cases of strings contain other characters like: "27.8 °C", "52.523553 kM" or "Total: 349.0".
This works in Swift 4:
let anyString = "52.523553 kM"
let digitsCharacterSet = CharacterSet.init(charactersIn: "0123456789.")
let doubleResult = Double(anyString.components(separatedBy:digitsCharacterSet.inverted).joined())
Caution! This not working for strings contain multiple . like "27.8 °C 3.5 kM"
I find more readable to add an extension to String as follow:
extension String {
var doubleValue: Double {
return (self as NSString).doubleValue
}
}
and then you just could write your code:
myDouble = myString.doubleValue
my problem was comma so i solve it this way:
extension String {
var doubleValue: Double {
return Double((self.replacingOccurrences(of: ",", with: ".") as NSString).doubleValue)
}
}
var stringValue = "55"
var convertToDouble = Double((stringValue as NSString).doubleValue)
we can use CDouble value which will be obtained by myString.doubleValue