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
Use a DecimalFormat object with a format string of "0. #".
By using round(_:) , ceil(_:) , and floor(_:) you can round Double and Float values to any number of decimal places in Swift.
To convert a string to a double, we can use the built-in Double() initializer syntax in Swift. The Double() initializer takes the string as an input and returns the double instance.
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
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) ?? "")
}
}
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"
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"
Note that it performs a rounding (same as MirekE solution):
0.9950000 → "0.99"
0.9950001 → "1"
In case you're looking how to remove trailing zeros from a string:
string.replacingOccurrences(of: "^([\d,]+)$|^([\d,]+)\.0*$|^([\d,]+\.[0-9]*?)0*$", with: "$1$2$3", options: .regularExpression)
This will transform strings like "0.123000000" into "0.123"
All the answers i found was good but all of them had some problems like producing decimal numbers without the 0
in the beginning ( like .123
instead of 0.123
). but these two will do the job with no problem :
extension Double {
func formatNumberWithFixedFraction(maximumFraction: Int = 8) -> String {
let stringFloatNumber = String(format: "%.\(maximumFraction)f", self)
return stringFloatNumber
}
func formatNumber(maximumFraction: Int = 8) -> String {
let formatter = NumberFormatter()
let number = NSNumber(value: self)
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = maximumFraction
formatter.numberStyle = .decimal
formatter.allowsFloats = true
let formattedNumber = formatter.string(from: number).unwrap
return formattedNumber
}
}
The first one converts 71238.12
with maxFraction of 8 to: 71238.12000000
but the second one with maxFraction of 8 converts it to: 71238.12
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