Swift - Remove Trailing Zeros From Double

Question

What is the function that removes trailing zeros from doubles?

var double = 3.0
var double2 = 3.10

println(func(double)) // 3
println(func(double2)) // 3.1

Answer 1

In Swift 4 you can do it like that:

extension Double {
    func removeZerosFromEnd() -> String {
        let formatter = NumberFormatter()
        let number = NSNumber(value: self)
        formatter.minimumFractionDigits = 0
        formatter.maximumFractionDigits = 16 //maximum digits in Double after dot (maximum precision)
        return String(formatter.string(from: number) ?? "")
    }
}

example of use: print (Double("128834.567891000").removeZerosFromEnd()) result: 128834.567891

You can also count how many decimal digits has your string:

import Foundation

extension Double {
    func removeZerosFromEnd() -> String {
        let formatter = NumberFormatter()
        let number = NSNumber(value: self)
        formatter.minimumFractionDigits = 0
        formatter.maximumFractionDigits = (self.components(separatedBy: ".").last)!.count
        return String(formatter.string(from: number) ?? "")
    }
}

Answer 2

You can do it this way but it will return a string:

var double = 3.0
var double2 = 3.10

func forTrailingZero(temp: Double) -> String {
    var tempVar = String(format: "%g", temp)
    return tempVar
}

forTrailingZero(double)   //3
forTrailingZero(double2)  //3.1

Answer 3

Removing trailing zeros in output

This scenario is good when the default output precision is desired. We test the value for potential trailing zeros, and we use a different output format depending on it.

extension Double {
    var stringWithoutZeroFraction: String {
        return truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
    }
}

(works also with extension Float, but not Float80)

Output:

1.0 → "1"
0.1 → "0.1"
0.01 → "0.01"
0.001 → "0.001"
0.0001 → "0.0001"

Formatting with maximum fraction digits, without trailing zeros

This scenario is good when a custom output precision is desired. This solution seems roughly as fast as NumberFormatter + NSNumber solution from MirekE, but one benefit could be that we're avoiding NSObject here.

extension Double {
    func string(maximumFractionDigits: Int = 2) -> String {
        let s = String(format: "%.\(maximumFractionDigits)f", self)
        for i in stride(from: 0, to: -maximumFractionDigits, by: -1) {
            if s[s.index(s.endIndex, offsetBy: i - 1)] != "0" {
                return String(s[..<s.index(s.endIndex, offsetBy: i)])
            }
        }
        return String(s[..<s.index(s.endIndex, offsetBy: -maximumFractionDigits - 1)])
    }
}

(works also with extension Float, but not Float80)

Output for maximumFractionDigits: 2:

1.0 → "1"
0.12 → "0.12"
0.012 → "0.01"
0.0012 → "0"
0.00012 → "0"