Scala Doubles, and Precision

Question

Is there a function that can truncate or round a Double? At one point in my code I would like a number like: 1.23456789 to be rounded to 1.23

Answer 1

A bit strange but nice. I use String and not BigDecimal

def round(x: Double)(p: Int): Double = {
    var A = x.toString().split('.')
    (A(0) + "." + A(1).substring(0, if (p > A(1).length()) A(1).length() else p)).toDouble
}

Answer 2

You can use scala.math.BigDecimal:

BigDecimal(1.23456789).setScale(2, BigDecimal.RoundingMode.HALF_UP).toDouble

There are a number of other rounding modes, which unfortunately aren't very well documented at present (although their Java equivalents are).

Answer 3

I wouldn't use BigDecimal if you care about performance. BigDecimal converts numbers to string and then parses it back again:

  /** Constructs a `BigDecimal` using the decimal text representation of `Double` value `d`, rounding if necessary. */
  def decimal(d: Double, mc: MathContext): BigDecimal = new BigDecimal(new BigDec(java.lang.Double.toString(d), mc), mc)

I'm going to stick to math manipulations as Kaito suggested.

Answer 4

Edit: fixed the problem that @ryryguy pointed out. (Thanks!)

If you want it to be fast, Kaito has the right idea. math.pow is slow, though. For any standard use you're better off with a recursive function:

def trunc(x: Double, n: Int) = {
  def p10(n: Int, pow: Long = 10): Long = if (n==0) pow else p10(n-1,pow*10)
  if (n < 0) {
    val m = p10(-n).toDouble
    math.round(x/m) * m
  }
  else {
    val m = p10(n).toDouble
    math.round(x*m) / m
  }
}

This is about 10x faster if you're within the range of Long (i.e 18 digits), so you can round at anywhere between 10^18 and 10^-18.

Answer 5

You may use implicit classes:

import scala.math._

object ExtNumber extends App {
  implicit class ExtendedDouble(n: Double) {
    def rounded(x: Int) = {
      val w = pow(10, x)
      (n * w).toLong.toDouble / w
    }
  }

  // usage
  val a = 1.23456789
  println(a.rounded(2))
}

Answer 6

Recently, I faced similar problem and I solved it using following approach

def round(value: Either[Double, Float], places: Int) = {
  if (places < 0) 0
  else {
    val factor = Math.pow(10, places)
    value match {
      case Left(d) => (Math.round(d * factor) / factor)
      case Right(f) => (Math.round(f * factor) / factor)
    }
  }
}

def round(value: Double): Double = round(Left(value), 0)
def round(value: Double, places: Int): Double = round(Left(value), places)
def round(value: Float): Double = round(Right(value), 0)
def round(value: Float, places: Int): Double = round(Right(value), places)

I used this SO issue. I have couple of overloaded functions for both Float\Double and implicit\explicit options. Note that, you need to explicitly mention the return type in case of overloaded functions.