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Noelia
Noelia

Reputation: 4146

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

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

Upvotes: 126

Views: 216211

Answers (15)

Areeb Momin
Areeb Momin

Reputation: 447

If you want a pure kotlin solution, that is usually required in KMP projects.

This extension method worked for me:

fun Double.round(decimals: Int): Double {
    val factor = 10.0.pow(decimals)
    return (this * factor).roundToInt() / factor
}

Upvotes: 2

Homayoon Ahmadi
Homayoon Ahmadi

Reputation: 2833

yet another extension function:

fun Number.roundDecimal(digits: Int): Double {
    require(digits >= 0) { "Number of digits must be non-negative." }

    val multiplier = 10.0.pow(digits.toDouble())
    return round(this.toDouble().times(multiplier)) / multiplier
}

Upvotes: 0

Arun Aditya
Arun Aditya

Reputation: 1104

If You want pure Kotlin solution then you can use roundToInt() method of Kolin

val result = ((number * 1000).roundToInt())/1000.0

Upvotes: 0

Saiful Sazib
Saiful Sazib

Reputation: 501

Use Kotlin Extension Function and then call where you need.

fun Double.roundTo(): Double = this.times(100.0).roundToInt() / 100.0
fun Long.roundTo(): Double = this.times(100.0).roundToInt() / 100.0
fun Float.roundTo(): Double = this.times(100.0).roundToInt() / 100.0

Example:

val amount = 156.26588
val roundAmount = amount.roundTo()
Log.d("Amount:" "$roundAmount") // Amount:156.26

Upvotes: 2

jafar_aml
jafar_aml

Reputation: 369

In Kotlin I just use this function:

fun roundTheNumber(numInDouble: Double): String {

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

    }

Upvotes: 19

Mehranjp73
Mehranjp73

Reputation: 511

Use this extension function:

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

Upvotes: -1

Usman Zafer
Usman Zafer

Reputation: 1361

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

Upvotes: 2

Toufiqul Haque Mamun
Toufiqul Haque Mamun

Reputation: 121

Try this way for two decimal value return as string

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

Upvotes: 0

Shirsh Shukla
Shirsh Shukla

Reputation: 5973

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

Upvotes: 3

Willi Mentzel
Willi Mentzel

Reputation: 29844

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.

Upvotes: 33

Gaurang Goda
Gaurang Goda

Reputation: 3854

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

Upvotes: 82

Ignacio Garcia
Ignacio Garcia

Reputation: 1153

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()
}

Upvotes: 30

Noelia
Noelia

Reputation: 4146

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.

Upvotes: 170

kotoMJ
kotoMJ

Reputation: 803

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()
}

Upvotes: 13

hotkey
hotkey

Reputation: 147901

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.

Upvotes: 65

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