Round Double to 1 decimal place kotlin: from 0.044999 to 0.1

Question

I have a Double variable that is 0.0449999 and I would like to round it to 1 decimal place 0.1 .

I am using Kotlin but the Java solution is also helpful.

val number:Double = 0.0449999

I tried getting 1 decimal place with these two solutions:

1. val solution = Math.round(number * 10.0) / 10.0
2. val solution = String.format("%.1f", number)

The problem is that I get 0.0 in both cases because it rounds the number from 0.04 to 0.0. It doesn't take all decimals and round it.

I would like to obtain 0.1: 0.045 -> 0.05 -> 0.1

Answer 1

Finally I did what Andy Turner suggested, rounded to 3 decimals, then to 2 and then to 1:

Answer 1:

val number:Double = 0.0449999
val number3digits:Double = String.format("%.3f", number).toDouble()
val number2digits:Double = String.format("%.2f", number3digits).toDouble()
val solution:Double = String.format("%.1f", number2digits).toDouble()

Answer 2:

val number:Double = 0.0449999
val number3digits:Double = Math.round(number * 1000.0) / 1000.0
val number2digits:Double = Math.round(number3digits * 100.0) / 100.0
val solution:Double = Math.round(number2digits * 10.0) / 10.0

Result:

0.045 ? 0.05 ? 0.1

Note: I know it is not how it should work but sometimes you need to round up taking into account all decimals for some special cases so maybe someone finds this useful.

Answer 2

I know some of the above solutions work perfectly but I want to add another solution that uses ceil and floor concept, which I think is optimized for all the cases.

If you want the highest value of the 2 digits after decimal use below code.

import java.math.BigDecimal 
import java.math.RoundingMode
import java.text.DecimalFormat

here, 1.45678 = 1.46

fun roundOffDecimal(number: Double): Double? {
    val df = DecimalFormat("#.##")
    df.roundingMode = RoundingMode.CEILING
    return df.format(number).toDouble()
}

If you want the lowest value of the 2 digits after decimal use below code.

here, 1.45678 = 1.45

fun roundOffDecimal(number: Double): Double? {
    val df = DecimalFormat("#.##")
    df.roundingMode = RoundingMode.FLOOR
    return df.format(number).toDouble()
}

Here a list of all available flags: CEILING, DOWN, FLOOR, HALF_DOWN, HALF_EVEN, HALF_UP, UNNECESSARY, UP

The detailed information is given in docs

Answer 3

The BigDecimal rounding features several RoundingModes, including those rounding up (away from zero) or towards positive infinity. If that's what you need, you can perform rounding by calling setScale as follows:

val number = 0.0449999
val rounded = number.toBigDecimal().setScale(1, RoundingMode.UP).toDouble()
println(rounded) // 0.1

Note, however, that it works in a way that will also round anything between 0.0 and 0.1 to 0.1 (e.g. 0.00001 ? 0.1).

^{The .toBigDecimal() extension is available since Kotlin 1.2.}

Answer 4

1. Method (using Noelia's idea):

You can pass the number of desired decimal places in a string template and make the precision variable this way:

fun Number.roundTo(
  numFractionDigits: Int
) = "%.${numFractionDigits}f".format(this, Locale.ENGLISH).toDouble()

2. Method (numeric, no string conversion)

fun Double.roundTo(numFractionDigits: Int): Double {
  val factor = 10.0.pow(numFractionDigits.toDouble())
  return (this * factor).roundToInt() / factor
}

_{One could create an overload for Float as well.}

Answer 5

An example of extension functions for Float and Double, round to n decimal positions.

fun Float.roundTo(n : Int) : Float {
    return "%.${n}f".format(this).toFloat()
}

fun Double.roundTo(n : Int) : Double {
    return "%.${n}f".format(this).toDouble()
}

Answer 6

Always beware of Locale!

With unspecified locale you can get occasional issue (e.g. with Portugies locale) such as

Fatal Exception: java.lang.NumberFormatException
For input string: "0,1"

1. Solution using DecimalFormat approach

fun Float.roundToOneDecimalPlace(): Float {
    val df = DecimalFormat("#.#", DecimalFormatSymbols(Locale.ENGLISH)).apply {
        roundingMode = RoundingMode.HALF_UP
    }
    return df.format(this).toFloat()
}

2. Solution using string format approach

fun Float.roundTo(decimalPlaces: Int): Float {
    return "%.${decimalPlaces}f".format(Locale.ENGLISH,this).toFloat()
}

Answer 7

Try this, its work for me

     val number = 0.045
     var filterUserPrice: String? = "%.2f".format(number)
     Log.v("afterRoundoff"," : $filterUserPrice")// its print filterUserPrice is 0.05

Answer 8

Try this way for two decimal value return as string

private fun getValue(doubleValue: Double): String {
    return String.format(Locale.US, "%.2f", doubleValue)
}

Answer 9

For new comers

Using String.format for decimal precision can lead to problems for different languages.

Use the following code to convert Double to as many decimal places as you want.

val price = 6.675668

//to convert to 2 decimal places
totalTime = Math.round(totalTime * 100.0) / 100.00
// totalTime = 6.68

//to convert to 4 decimal places
totalTime = Math.round(totalTime * 10000.0) / 10000.00
// totalTime = 6.6757

Answer 10

In Kotlin I just use this function:

fun roundTheNumber(numInDouble: Double): String {

        return "%.2f".format(numInDouble)

    }

Answer 11

Use this extension function:

toBigDecimal(MathContext(3, RoundingMode.HALF_EVEN)).toPlainString()