I have a requirement to turn keyboard digits input into amount as cents, for example:
Typing 1 will output 0.01
Typing 10 will output 0.10
Typing 123 will output 1.23
Typing 12345 will output 123.45
My current solution involves multiplying by 0.01 and looking at the reminder:
// split into whole and reminder
let mod = modf(doubleAmount * 0.01)
// define a rounding behavior
let behavior = NSDecimalNumberHandler(roundingMode: .plain,
scale: Int16(places),
raiseOnExactness: false,
raiseOnOverflow: false,
raiseOnUnderflow: false,
raiseOnDivideByZero: true)
let whole = mod.0
let reminder = NSDecimalNumber(value: mod.1).rounding(accordingToBehavior: behavior)
// formatting...
let newnumber = NSNumber(value: whole + reminder.doubleValue)
This will work correct for most values but will break format when working with whole numbers like 10, 100, 1000 etc... The format will then be 0.1, 1.0, 10.0 while I need them to appear as 0.10, 1.00, 10.00 etc...
what is the best/correct way to achieve this?
The NumberFormatter type has a multiplier attribute that can be used for this scenario and then no further calculations will be needed
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.minimumFractionDigits = 2
formatter.maximumFractionDigits = 2
formatter.multiplier = 0.01
Related
I am receiving two kid of Doubles from JSON:
1.12 and 0.00007067999999
The second number switches automatically to scientific notation(
7.067e-05), so I'm using the function String(format:"%.8f", NUMBER) to make it 0.00007067, yes it works, but now my first number becomes 1.12000000.
How to clean trailing numbers?
I've tried with Swift - Remove Trailing Zeros From Double , but the %g format changes second number to scientific notation again, so %g is not an option. Any suggestions?
You can use NumberFormater and set minimum and maximum fraction digits:
let double1 = 1.12
let double2 = 0.00007067999999
let numberFormatter = NumberFormatter()
numberFormatter.minimumFractionDigits = 2
numberFormatter.maximumFractionDigits = 8
numberFormatter.minimumIntegerDigits = 1
numberFormatter.string(for: double1) ?? "" // "1.12"
numberFormatter.string(for: double2) ?? "" // "0.00007068"
if you would like to round the fraction digits down you can set the formatter rounding mode option to .down:
numberFormatter.roundingMode = .down
numberFormatter.string(for: double2) ?? "" // "0.00007067"
Is there any proper implementation for 0°00'00.00" format?
NSNumberFormatter supports only groupingSeparator and decimalSeparator.
Or not?
There is no degree-minute-second formatter in Foundation but it's not too hard to roll your own. Here's my crude attempt:
class DegreeFormatter : NumberFormatter {
func string(from degree: Double) -> String {
var remaining = degree
let degree = remaining.rounded(.towardZero)
remaining -= degree
remaining *= 60.0
let minute = remaining.rounded(.towardZero)
remaining -= minute
remaining *= 60
let seconds = remaining
return "\(Int(degree))°\(Int(minute))'\(self.string(from: seconds as NSNumber)!)"
}
}
let formatter = DegreeFormatter()
formatter.maximumFractionDigits = 2
print(formatter.string(from: 1.23456789)) // 1°14'4.44
You can customize it further to include custom degree symbols, whether to include the second, what units are allowed, etc.
I need to round stocks, indices and futures prices to the nearest tick. The first step is to look if the price is a multiple of the tick. Apple docs says "Unlike the remainder operator in C and Objective-C, Swift’s remainder operator can also operate on floating-point numbers".
If I write the following code in a playground or in a console app and I run it, I expect 0 as result but I get a remainder value equals to 0.00999999999999775:
var stringPrice = "17.66"
var price = Double(stringPrice)
var tickSize: Double = 0.01
let remainder = price! % ticksize
This problem breaks my rounding function when using values such 17.66 as aPrice and 0.01 as aTickSize:
func roundPriceToNearestTick(Price aPrice: Double, TickSize a TickSize: Double)-> Double{
let remainder = aPrice % aTickSize
let shouldRoundUp = remainder >= aTickSize/2 ? true : false
let multiple = floor(aPrice/aTickSize)
let returnPrice = !shouldRoundUp ? aTickSize*multiple : aTickSize*multiple + aTickSize
return returnPrice
}
What is the best way to fix this?
Following the comments about the broken floating point math and the need to avoid floats and doubles for all the operations concerning money I changed my code to perform the remainder operation using NSDecimalNumbers. This seems to solve the precision problem.
var stringPrice = "17.66"
var tickSizeDouble : Double = 0.01
var tickSizeDecimalNumber: NSDecimalNumber = 0.01
func decimalNumberRemainder(Dividend aDividend: NSDecimalNumber, Divisor aDivisor: NSDecimalNumber)->NSDecimalNumber{
let behaviour = NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundDown,
scale: 0,
raiseOnExactness: false ,
raiseOnOverflow: false,
raiseOnUnderflow: false,
raiseOnDivideByZero: false )
let quotient = aDividend.decimalNumberByDividingBy(aDivisor, withBehavior: behaviour)
let subtractAmount = quotient.decimalNumberByMultiplyingBy(aDivisor)
let remainder = aDividend.decimalNumberBySubtracting(subtractAmount)
return remainder
}
let doubleRemainder = Double(stringPrice)! % tickSizeDouble
let decimalRemainder = decimalNumberRemainder(Dividend: NSDecimalNumber(string: stringPrice), Divisor:tickSizeDecimalNumber)
print("Using Double: \(doubleRemainder)")
print("Using NSDecimalNumber: \(decimalRemainder)")
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.