Round up double to 2 decimal places - swift

Question

How do I round up currentRatio to two decimal places?
let currentRatio = Double (rxCurrentTextField.text!)! / Double (txCurrentTextField.text!)!
railRatioLabelField.text! = "\(currentRatio)"

Answer 1

Use a format string to round up to two decimal places and convert the double to a String:
let currentRatio = Double (rxCurrentTextField.text!)! / Double (txCurrentTextField.text!)!
railRatioLabelField.text! = String(format: "%.2f", currentRatio)
Example:
let myDouble = 3.141
let doubleStr = String(format: "%.2f", myDouble) // "3.14"
If you want to round up your last decimal place, you could do something like this (thanks Phoen1xUK):
let myDouble = 3.141
let doubleStr = String(format: "%.2f", ceil(myDouble*100)/100) // "3.15"

Answer 2

(Swift 4.2 Xcode 11)
Simple to use Extension:-
extension Double {
func round(to places: Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
Use:-
if let distanceDb = Double(strDistance) {
cell.lblDistance.text = "\(distanceDb.round(to:2)) km"
}

Answer 3

Updated to SWIFT 4 and the proper answer for the question
If you want to round up to 2 decimal places you should multiply with 100 then round it off and then divide by 100
var x = 1.5657676754
var y = (x*100).rounded()/100
print(y) // 1.57

Answer 4

Consider using NumberFormatter for this purpose, it provides more flexibility if you want to print the percentage sign of the ratio or if you have things like currency and large numbers.
let amount = 10.000001
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.maximumFractionDigits = 2
let formattedAmount = formatter.string(from: amount as NSNumber)!
print(formattedAmount) // 10

Answer 5

Adding to above answer if we want to format Double multiple times, we can use protocol extension of Double like below:
extension Double {
var dollarString:String {
return String(format: "$%.2f", self)
}
}
let a = 45.666
print(a.dollarString) //will print "$45.67"

Answer 6

Just a quick follow-up answer for noobs like me:
You can make the other answers super easily implementable by using a function with an output. E.g.
func twoDecimals(number: Float) -> String{
return String(format: "%.2f", number)
}
This way, whenever you want to grab a value to 2 decimal places you just type
twoDecimals('Your number here')
...
Simples!
P.s. You could also make it return a Float value, or anything you want, by then converting it again after the String conversion as follows:
func twoDecimals(number: Float) -> Float{
let stringValue = String(format: "%.2f", number)
return Float(stringValue)!
}
Hope that helps.

Answer 7

The code for specific digits after decimals is:
var roundedString = String(format: "%.2f", currentRatio)
Here the %.2f tells the swift to make this number rounded to 2 decimal places.

Answer 8

#Rounded, A swift 5.1 property wrapper
Example :
struct GameResult {
#Rounded(rule: NSDecimalNumber.RoundingMode.up,scale: 4)
var score: Decimal
}
var result = GameResult()
result.score = 3.14159265358979
print(result.score) // 3.1416

Answer 9

Maybe also:
// Specify the decimal place to round to using an enum
public enum RoundingPrecision {
case ones
case tenths
case hundredths
case thousands
}
extension Double {
// Round to the specific decimal place
func customRound(_ rule: FloatingPointRoundingRule, precision: RoundingPrecision = .ones) -> Double {
switch precision {
case .ones: return (self * Double(1)).rounded(rule) / 1
case .tenths: return (self * Double(10)).rounded(rule) / 10
case .hundredths: return (self * Double(100)).rounded(rule) / 100
case .thousands: return (self * Double(1000)).rounded(rule) / 1000
}
}
}
let value: Double = 98.163846
print(value.customRound(.toNearestOrEven, precision: .ones)) //98.0
print(value.customRound(.toNearestOrEven, precision: .tenths)) //98.2
print(value.customRound(.toNearestOrEven, precision: .hundredths)) //98.16
print(value.customRound(.toNearestOrEven, precision: .thousands)) //98.164
Keeps decimals, does not truncate but rounds
See for more details even specified rounding rules

Answer 10

String(format: "%.2f", Double(round(1000*34.578)/1000))
Output: 34.58

Answer 11

Try this , you will get a better result instead of 0.0
extension Double {
func rounded(toPlaces places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
func toRoundedString(toPlaces places:Int) -> String {
let amount = self.rounded(toPlaces: places)
let str_mount = String(amount)
let sub_amountStrings = str_mount.split(separator: ".")
if sub_amountStrings.count == 1
{
var re_str = "\(sub_amountStrings[0])."
for _ in 0..<places
{
re_str += "0"
}
return re_str
}
else if sub_amountStrings.count > 1, "\(sub_amountStrings[1])".count < places
{
var re_str = "\(sub_amountStrings[0]).\(sub_amountStrings[1])"
let tem_places = (places - "\(sub_amountStrings[1])".count)
for _ in 0..<tem_places
{
re_str += "0"
}
return re_str
}
return str_mount
}
}

Answer 12

if you give it 234.545332233 it will give you 234.54
let textData = Double(myTextField.text!)!
let text = String(format: "%.2f", arguments: [textData])
mylabel.text = text

Answer 13

Just single line of code:
let obj = self.arrayResult[indexPath.row]
let str = String(format: "%.2f", arguments: [Double((obj.mainWeight)!)!])