Easily truncating a Double - swift - swift

Question

I want to be able to take a longitude/latitude value of 19.3563443 and turn it into 19.35634 (five decimal places). Is there any easy way to do this?

Answer 1

You can use NumberFormatter:
extension Formatter {
static let number = NumberFormatter()
}
extension FloatingPoint {
func fractionDigitsRounded(to digits: Int, roundingMode: NumberFormatter.RoundingMode = .halfEven) -> String {
Formatter.number.roundingMode = roundingMode
Formatter.number.minimumFractionDigits = digits
Formatter.number.maximumFractionDigits = digits
return Formatter.number.string(for: self) ?? ""
}
}
let value = 19.3563443
let roundedToFive = value.fractionDigitsRounded(to: 5) // "19.35634"

Answer 2

Round, get double value back.
NSString(format: "%.5f", myfloat).doubleValue
This won't truncate, but may be accurate enough for long/lat to 5 decimal places and is quick.

Answer 3

This is a math problem, not a coding problem. Converting between String and Double is not very attractive. In mathematics, truncating digits to the right of the decimal is called reducing scale. To truncate a non-integer, all you need to do is move the decimal to the right however many places you want it truncated (multiply it by 10 to move it one, multiply it by 100 to move it two, etc.), convert it into an integer (which drops everything after the new decimal), and then divide it back by the original multiplier to return the decimal to its original position.
To reduce pi's scale to 3:
move the decimal right: 3.1415927 * 10^3 = 3141.5927
zero the fraction = 3141.0
move the decimal back: 3141 / 10^3 = 3.141
Add it as an extension to Double for convenience:
extension Double {
/// Reduces a double's scale (truncates digits right of the decimal).
func reduceScale(to places: Int) -> Double {
guard !self.isNaN,
!self.isInfinite else {
return 0 // if the double is not a number or infinite, this func will throw an exception error
}
let multiplier = pow(10, Double(places))
let newDecimal = multiplier * self // move the decimal right
let truncated = Double(Int(newDecimal)) // drop the fraction
let originalDecimal = truncated / multiplier // move the decimal back
return originalDecimal
}
}
let dirty = 3.1415927
let clean = dirty.reduceScale(to: 2) // 3.14

Answer 4

I suggest using NSDecimalNumber. I don't know if this is actually better or worse than #Leonardo's answer. But when dealing with numbers I prefer to stick to numbers, and not convert to/from strings. The code below generates a function for a requested number of digits that can be reused as needed.
This truncates by rounding down. To get normal arithmetic rounding, change it to use .RoundPlain instead of .RoundDown.
func makeRounder(#digits: Int16) -> (Double) -> Double {
class RoundHandler : NSObject, NSDecimalNumberBehaviors {
var roundToDigits : Int16
init(roundToDigits: Int16) {
self.roundToDigits = roundToDigits
}
#objc func scale() -> Int16 {
println("Digits: \(digits)")
return roundToDigits
}
#objc func roundingMode() -> NSRoundingMode {
return NSRoundingMode.RoundDown
}
#objc func exceptionDuringOperation(operation: Selector, error: NSCalculationError, leftOperand: NSDecimalNumber, rightOperand: NSDecimalNumber) -> NSDecimalNumber? {
return nil
}
}
let roundHandler = RoundHandler(roundToDigits:digits)
func roundDigits(original:Double) -> Double {
let decimal = NSDecimalNumber(double: original)
let rounded = decimal.decimalNumberByRoundingAccordingToBehavior(roundHandler)
return rounded.doubleValue
}
println("Digits: \(digits)")
return roundDigits
}
let roundTo5 = makeRounder(digits:Int16(5))
roundTo5(19.3563443)
roundTo5(1.23456789)

Answer 5

I found a different answer on a different site and wanted to share it:
let x = 129.88888888777
let y = Double(round(1000*x)/1000)
y returns 129.888
thanks to swift's "round" function.

Answer 6

let x = (long*100000).rounded(.towardZero)/100000