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)")
Related
I'm trying to make a math app with different equations and formulas but I'm trying to circle sector but i just wanted to try to divide the input value by 360 but when I do that it only says 0 unless the value is over 360. I have tried using String, Double and Float with no luck I don't know what I'm doing is wrong but down here is the code. I'm thankful for help but I have been sitting a while and searched online for an answer with no result I might have been searching with the wrong search.
if graderna.text == ""{
}
else{
var myInt: Int? = Int(graderna.text!) // conversion of string to Int
var myInt2: Int? = Int(radien.text!)
let pi = 3.1415926
let lutning = 360
let result = (Double(myInt! / lutning) * Double(pi))
svar2.text = "\(result)"
}
Your code is performing integer division, taking the integer result and converting it to a double. Instead, you want to convert these individual integers to doubles and then do the division. So, instead of
let result = (Double(myInt! / lutning) * Double(pi))
You should
let result = Double(myInt!) / Double(lutning) * Double(pi)
Note, Double already has a .pi constant, so you can remove your pi constant, and simplify the above to:
let result = Double(myInt!) / Double(lutning) * .pi
Personally, I’d define myInt and lutning to be Double from the get go (and, while we’re at it, remove all of the forced unwrapping (with the !) of the optionals):
guard
let text = graderna.text,
let text2 = radien.text,
let value = Double(text),
let value2 = Double(text2)
else {
return
}
let lutning: Double = 360
let result = value / lutning * .pi
Or, you can use flatMap to safely unwrap those optional strings:
guard
let value = graderna.text.flatMap({ Double($0) }),
let value2 = radien.text.flatMap({ Double($0) })
else {
return
}
let lutning: Double = 360
let result = value / lutning * .pi
(By the way, if you’re converting between radians and degrees, it should be 2π/360, not π/360.)
You are dividing an Int by an Int.
Integer division rounds to the nearest integer towards zero. Therefore for example 359 / 360 is not a number close to 1, it is 0. 360 / 360 up to 719 / 360 equals 1. 720 / 360 to 1079 / 360 equals 2, and so on.
But your use of optionals is atrocious. I'd write
let myInt = Int(graderna.text!)
let myInt2 = Int(radien.text!)
if let realInt = myInt, realInt2 = myInt2 {
let pi = 3.1415926
let lutning = 360.0
let result = Double (realInt) * (pi / lutning)
svar2.text = "\(result)"
}
In the line let result = (Double(myInt! / lutning) * Double(pi)) you cast your type to double after dividing two integers so your result will always be zero. You have to make them doubles before division.
let result = (Double(myInt!) / Double(lutning)) * Double(pi))
If you want the value should be correct, then try as
let division = ((Float(V1) / Float(V2)) * Float(pi))
Let's say we have an NSDecimal constant called total and it contains the value of 3.33333333 divided by 10/3. 10 and 3 are both NSDecimalNumber.
We want the total to be 4 as an NSDecimalNumber in Swift.
let total = ten/three
// the ideal total would be rounded up whenever there is a floating
in the doc. we have
func NSDecimalRound(_ result: UnsafeMutablePointer<NSDecimal>,
_ number: UnsafePointer<NSDecimal>,
_ scale: Int,
_ roundingMode: NSRoundingMode)
maximumDecimalNumber()
Which one is the best fit for calculator with currency style? Please include an example each how to use them if you know so. Thank you .
let x = NSDecimalNumber(string:"10")
let y = NSDecimalNumber(string:"3")
let total = x.decimalNumberByDividingBy(y).decimalNumberByRoundingAccordingToBehavior( NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundUp, scale: 0, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: false)) // 4
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.
So I am trying to split a number in swift, I have tried searching for this on the internet but have had no success. So first I will start with a number like:
var number = 34.55
And from this number, I want to create two separate number by splitting from the dot. So the output should be something like:
var firstHalf = 34
var secondHalf = 55
Or the output can also be in an array of thats easier. How can I achieve this?
The easiest way would be to first cast the double to a string:
var d = 34.55
var b = "\(d)" // or String(d)
then split the string with the global split function:
var c = split(b) { $0 == "." } // [34, 55]
You can also bake this functionality into a double:
extension Double {
func splitAtDecimal() -> [Int] {
return (split("\(self)") { $0 == "." }).map({
return String($0).toInt()!
})
}
}
This would allow you to do the following:
var number = 34.55
print(number.splitAtDecimal()) // [34, 55]
Well, what you have there is a float, not a string. You can't really "split" it, and remember that a float is not strictly limited to 2 digits after the separator.
One solution is :
var firstHalf = Int(number)
var secondHalf = Int((number - firstHalf) * 100)
It's nasty but it'll do the right thing for your example (it will, however, fail when dealing with numbers that have more than two decimals of precision)
Alternatively, you could convert it into a string and then split that.
var stringified = NSString(format: "%.2f", number)
var parts = stringifed.componentsSeparatedByString(".")
Note that I'm explicitly calling the formatter here, to avoid unwanted behavior of standard float to string conversions.
Add the following extension:
extension Double {
func splitIntoParts(decimalPlaces: Int, round: Bool) -> (leftPart: Int, rightPart: Int) {
var number = self
if round {
//round to specified number of decimal places:
let divisor = pow(10.0, Double(decimalPlaces))
number = Darwin.round(self * divisor) / divisor
}
//convert to string and split on decimal point:
let parts = String(number).components(separatedBy: ".")
//extract left and right parts:
let leftPart = Int(parts[0]) ?? 0
let rightPart = Int(parts[1]) ?? 0
return(leftPart, rightPart)
}
Usage - Unrounded:
let number:Double = 95.99999999
let parts = number.splitIntoParts(decimalPlaces: 3, round: false)
print("LeftPart: \(parts.leftPart) RightPart: \(parts.rightPart)")
Outputs:
LeftPart: 95 RightPart: 999
Usage Rounded:
let number:Double = 95.199999999
let parts = number.splitIntoParts(decimalPlaces: 1, round: true)
Outputs:
LeftPart: 95 RightPart: 2
Actually, it's not possible to split a double by dot into two INTEGER numbers. All earlier offered solutions will produce a bug in nearly 10% of cases.
Here's why:
The breaking case will be numbers with decimal parts starting with one or more zeroes, for example: 1.05, 11.00698, etc.
Any decimal part that starts with one or more zeroes will have those zeroes discarded when converted to integers. The result of those conversions:
1.05 will become (1, 5)
11.00698 will become (11, 698)
An ugly and hard to find bug is guaranteed...
The only way to meaningfully split a decimal number is to convert it to (Int, Double). Below is a simple extension to Double that does that:
extension Double {
func splitAtDecimal() -> (Int, Double) {
let whole = Int(self)
let decimal = self - Darwin.floor(self)
return (whole, decimal)
}
}