When I got two numbers, like 5.085 and 70.085. My code rounds the first number to 5.09, but the second one it goes to 70.08. For some reason, when making let aux1 = aux * 100 the value goes to 7008.49999999. Any one have the solution to it?
Here is my code:
let aux = Double(value)!
let aux1 = aux * 100
let aux2 = (aux1).rounded()
let number = aux2 / 100
return formatter.string(from: NSNumber(value: number))!
If you want to format the Double by rounding it's fraction digits. Try't:
First, implement this method
func formatDouble(_ double: Double, withFractionDigits digits: Int) -> String{
let formatter = NumberFormatter()
formatter.maximumFractionDigits = digits
let string = formatter.string(from: (NSNumber(floatLiteral: double)))!
return string
/*if you want a Double instead of a String, change the return value and uncomment the bellow lines*/
//let number = formatter.number(from: string)!
//return number.doubleValue
}
after, you can call't that way
let roundedNumber = formatDouble(Double(value)!, withFractionDigits: 2)
Related
Can someone tell me why this happening?
let formatter = NumberFormatter.init()
formatter.numberStyle = .decimal
formatter.usesGroupingSeparator = false
formatter.roundingMode = .floor
formatter.minimumFractionDigits = 2
formatter.maximumFractionDigits = 2
let v = 36
let scale = 10
let float = formatter.string(from: NSNumber(value: Float(v) / Float(scale)))!
let double = formatter.string(from: NSNumber(value: Double(v) / Double(scale)))!
print(float) // 3.59
print(double) // 3.60
When I use Float the result is 3.59 (wrong result in my opinion) and when I use Double the result is 3.60.
I know it is something related to .floor roundingMode, but i don't fully understand the reason.
If you would like to preserve your fraction digits precision it is better to use Swift native Decimal type. That's what it is. You can use the Decimal init(sign: FloatingPointSign, exponent: Int, significand: Decimal) initializer and use your scale exponent and your value significand. Just make sure to negate its value:
extension SignedInteger {
var negated: Self { self * -1 }
}
let v = 36
let scale = 10
let sign: FloatingPointSign = v >= 0 ? .plus : .minus
let exponent = Decimal(scale).exponent.negated
let significand = Decimal(v).significand
let decimal = Decimal.init(sign: sign, exponent: exponent, significand: significand)
let formatted = formatter.string(for: decimal) // "3.60"
I need separate thouthands with space, and two digit after point.
let d1: Double = 20000000.0
let d2: Double = 1.2345
I want to view:
let s1 = String(format: "????", d1) //20 000 000.00
let s2 = String(format: "????", d2) //1.23
How to do it?
let d1: Double = 20000000.0
let d2: Double = 1.2345
let formatter = NumberFormatter()
formatter.groupingSeparator = " "
formatter.numberStyle = .decimal
formatter.minimumFractionDigits = 2
formatter.maximumFractionDigits = 2
formatter.decimalSeparator = "." // Default separator is dependent to the current local.
print(formatter.string(for: d1)) // 20 000 000.00
print(formatter.string(for: d2)) // 1.23
This question already has answers here:
How to truncate decimals to x places in Swift
(10 answers)
Closed 3 years ago.
I am new with swift and I need help. I want to get first two digits after the decimal point, for example -
1456.456214 -> 1456.45
35629.940812 -> 35629.94
without rounding the double to next one.
Try the below code
let num1 : Double = 1456.456214
let num2 : Double = 35629.940812
let numberFormatter = NumberFormatter()
numberFormatter.minimumFractionDigits = 2
numberFormatter.maximumFractionDigits = 2
numberFormatter.roundingMode = .down
let str = numberFormatter.string(from: NSNumber(value: num1))
let str2 = numberFormatter.string(from: NSNumber(value: num2))
print(str)
print(str2)
Output
1456.45
35629.94
To keep it a double you can do
let result = Double(Int(value * 100)) / 100.0
or, as #vacawama pointed out, use floor instead
let result = floor(value * 100) / 100
extension Double {
func truncate(places : Int)-> Double
{
return Double(floor(pow(10.0, Double(places)) * self)/pow(10.0, Double(places)))
}
}
and use this like as;
let ex: Double = 35629.940812
print(ex.truncate(places: 2)) //35629.94
let ex1: Double = 1456.456214
print(ex1.truncate(places: 2)) //1456.45
I'm printing out a number whose value I don't know. In most cases the number is whole or has a trailing .5. In some cases the number ends in .25 or .75, and very rarely the number goes to the thousandths place. How do I specifically detect that last case? Right now my code detects a whole number (0 decimal places), exactly .5 (1 decimal), and then reverts to 2 decimal spots in all other scenarios, but I need to go to 3 when it calls for that.
class func getFormattedNumber(number: Float) -> NSString {
var formattedNumber = NSString()
// Use the absolute value so it works even if number is negative
if (abs(number % 2) == 0) || (abs(number % 2) == 1) { // Whole number, even or odd
formattedNumber = NSString(format: "%.0f", number)
}
else if (abs(number % 2) == 0.5) || (abs(number % 2) == 1.5) {
formattedNumber = NSString(format: "%.1f", number)
}
else {
formattedNumber = NSString(format: "%.2f", number)
}
return formattedNumber
}
A Float uses a binary (IEEE 754) representation and cannot represent
all decimal fractions precisely. For example,
let x : Float = 123.456
stores in x the bytes 42f6e979, which is approximately
123.45600128173828. So does x have 3 or 14 fractional digits?
You can use NSNumberFormatter if you specify a maximum number
of decimal digits that should be presented:
let fmt = NSNumberFormatter()
fmt.locale = NSLocale(localeIdentifier: "en_US_POSIX")
fmt.maximumFractionDigits = 3
fmt.minimumFractionDigits = 0
println(fmt.stringFromNumber(123)!) // 123
println(fmt.stringFromNumber(123.4)!) // 123.4
println(fmt.stringFromNumber(123.45)!) // 123.45
println(fmt.stringFromNumber(123.456)!) // 123.456
println(fmt.stringFromNumber(123.4567)!) // 123.457
Swift 3/4 update:
let fmt = NumberFormatter()
fmt.locale = Locale(identifier: "en_US_POSIX")
fmt.maximumFractionDigits = 3
fmt.minimumFractionDigits = 0
print(fmt.string(for: 123.456)!) // 123.456
You can use %g to suppress trailing zeros. Then I think you do not need to go through the business of determining the number of places. Eg -
var num1:Double = 5.5
var x = String(format: "%g", num1) // "5.5"
var num2:Double = 5.75
var x = String(format: "%g", num2) // "5.75"
Or this variation where the number of places is specified. Eg -
var num3:Double = 5.123456789
var x = String(format: "%.5g", num3) // "5.1235"
My 2 cents ;) Swift 3 ready
Rounds the floating number and strips the trailing zeros to the required minimum/maximum fraction digits.
extension Double {
func toString(minimumFractionDigits: Int = 0, maximumFractionDigits: Int = 2) -> String {
let formatter = NumberFormatter()
formatter.locale = Locale(identifier: "en_US_POSIX")
formatter.minimumFractionDigits = minimumFractionDigits
formatter.maximumFractionDigits = maximumFractionDigits
return formatter.string(from: self as NSNumber)!
}
}
Usage:
Double(394.239).toString() // Output: 394.24
Double(394.239).toString(maximumFractionDigits: 1) // Output: 394.2
If you want to print a floating point number to 3 decimal places, you can use String(format: "%.3f"). This will round, so 0.10000001 becomes 0.100, 0.1009 becomes 0.101 etc.
But it sounds like you don’t want the trailing zeros, so you might want to trim them off. (is there a way to do this with format? edit: yes, g as #simons points out)
Finally, this really shouldn’t be a class function since it’s operating on primitive types. Better to either make it a free function, or perhaps extend Double/Float:
extension Double {
func toString(#decimalPlaces: Int)->String {
return String(format: "%.\(decimalPlaces)g", self)
}
}
let number = -0.3009
number.toString(decimalPlaces: 3) // -0.301
I'm using Swift
let myDouble = 8.5 as Double
let percentFormatter = NSNumberFormatter()
percentFormatter.numberStyle = NSNumberFormatterStyle.PercentStyle
percentFormatter.multiplier = 1.00
let myString = percentFormatter.stringFromNumber(myDouble)!
println(myString)
Outputs 8% and not 8.5%, how would I get it to output 8.5%? (But only up to 2 decimal places)
To set the number of fraction digits use:
percentFormatter.minimumFractionDigits = 1
percentFormatter.maximumFractionDigits = 1
Set minimum and maximum to your needs. Should be self-explanatory.
With Swift 5, NumberFormatter has an instance property called minimumFractionDigits. minimumFractionDigits has the following declaration:
var minimumFractionDigits: Int { get set }
The minimum number of digits after the decimal separator allowed as input and output by the receiver.
NumberFormatter also has an instance property called maximumFractionDigits. maximumFractionDigits has the following declaration:
var maximumFractionDigits: Int { get set }
The maximum number of digits after the decimal separator allowed as input and output by the receiver.
The following Playground code shows how to use minimumFractionDigits and maximumFractionDigits in order to set the number of digits after the decimal separator when using NumberFormatter:
import Foundation
let percentFormatter = NumberFormatter()
percentFormatter.numberStyle = NumberFormatter.Style.percent
percentFormatter.multiplier = 1
percentFormatter.minimumFractionDigits = 1
percentFormatter.maximumFractionDigits = 2
let myDouble1: Double = 8
let myString1 = percentFormatter.string(for: myDouble1)
print(String(describing: myString1)) // Optional("8.0%")
let myDouble2 = 8.5
let myString2 = percentFormatter.string(for: myDouble2)
print(String(describing: myString2)) // Optional("8.5%")
let myDouble3 = 8.5786
let myString3 = percentFormatter.string(for: myDouble3)
print(String(describing: myString3)) // Optional("8.58%")
When in doubt, look in apple documentation for minimum fraction digits and maximum fraction digits which will give you these lines you have to add before formatting your number:
numberFormatter.minimumFractionDigits = 1
numberFormatter.maximumFractionDigits = 2
Also notice, your input has to be 0.085 to get 8.5%. This is caused by the multiplier property, which is for percent style set to 100 by default.