Rounding a double value to x number of decimal pla

Question

Can anyone tell me how to round a double value to x number of decimal places in Swift?

I have:

var totalWorkTimeInHours = (totalWorkTime/60/60)

With totalWorkTime being an NSTimeInterval (double) in second.

totalWorkTimeInHours will give me the hours, but it gives me the amount of time in such a long precise number e.g. 1.543240952039......

How do I round this down to, say, 1.543 when I print totalWorkTimeInHours?

Answer 1

The best way to format a double property is to use the Apple predefined methods.

mutating func round(_ rule: FloatingPointRoundingRule)

FloatingPointRoundingRule is a enum which has following possibilities

Enumeration Cases:

case awayFromZero Round to the closest allowed value whose magnitude is greater than or equal to that of the source.

case down Round to the closest allowed value that is less than or equal to the source.

case toNearestOrAwayFromZero Round to the closest allowed value; if two values are equally close, the one with greater magnitude is chosen.

case toNearestOrEven Round to the closest allowed value; if two values are equally close, the even one is chosen.

case towardZero Round to the closest allowed value whose magnitude is less than or equal to that of the source.

case up Round to the closest allowed value that is greater than or equal to the source.

var aNumber : Double = 5.2
aNumber.rounded(.up) // 6.0

Answer 2

This is a sort of a long workaround, which may come in handy if your needs are a little more complex. You can use a number formatter in Swift.

let numberFormatter: NSNumberFormatter = {
    let nf = NSNumberFormatter()
    nf.numberStyle = .DecimalStyle
    nf.minimumFractionDigits = 0
    nf.maximumFractionDigits = 1
    return nf
}()

Suppose your variable you want to print is

var printVar = 3.567

This will make sure it is returned in the desired format:

numberFormatter.StringFromNumber(printVar)

The result here will thus be "3.6" (rounded). While this is not the most economic solution, I give it because the OP mentioned printing (in which case a String is not undesirable), and because this class allows for multiple parameters to be set.

Answer 3

The code for specific digits after decimals is:

var a = 1.543240952039
var roundedString = String(format: "%.3f", a)

Here the %.3f tells the swift to make this number rounded to 3 decimal places.and if you want double number, you may use this code:

// String to Double

var roundedString = Double(String(format: "%.3f", b))

Answer 4

Use the built in Foundation Darwin library

SWIFT 3

extension Double {

    func round(to places: Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return Darwin.round(self * divisor) / divisor
    }

}

Usage:

let number:Double = 12.987654321
print(number.round(to: 3)) 

Outputs: 12.988

Answer 5

round a double value to x number of decimal
NO. of digits after decimal

var x = 1.5657676754
var y = (x*10000).rounded()/10000
print(y)  // 1.5658 

---

var x = 1.5657676754 
var y = (x*100).rounded()/100
print(y)  // 1.57 

---

var x = 1.5657676754
var y = (x*10).rounded()/10
print(y)  // 1.6

Answer 6

Either:

1. Using String(format:):

   • Typecast Double to String with %.3f format specifier and then back to Double

     Double(String(format: "%.3f", 10.123546789))!
     

   • Or extend Double to handle N-Decimal places:

     extension Double {
         func rounded(toDecimalPlaces n: Int) -> Double {
             return Double(String(format: "%.\(n)f", self))!
         }
     }
     

2. By calculation

   • multiply with 10^3, round it and then divide by 10^3...

     (1000 * 10.123546789).rounded()/1000
     

   • Or extend Double to handle N-Decimal places:

     extension Double {    
         func rounded(toDecimalPlaces n: Int) -> Double {
             let multiplier = pow(10, Double(n))
             return (multiplier * self).rounded()/multiplier
         }
     }