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

How about :

 val value = 1.4142135623730951

//3 decimal places
println((value * 1000).round / 1000.toDouble)

//4 decimal places
println((value * 10000).round / 10000.toDouble)

Answer 2

Here's another solution without BigDecimals

Truncate:

(math floor 1.23456789 * 100) / 100

Round:

(math rint 1.23456789 * 100) / 100

Or for any double n and precision p:

def truncateAt(n: Double, p: Int): Double = { val s = math pow (10, p); (math floor n * s) / s }

Similar can be done for the rounding function, this time using currying:

def roundAt(p: Int)(n: Double): Double = { val s = math pow (10, p); (math round n * s) / s }

which is more reusable, e.g. when rounding money amounts the following could be used:

def roundAt2(n: Double) = roundAt(2)(n)

Answer 3

Since no-one mentioned the % operator yet, here comes. It only does truncation, and you cannot rely on the return value not to have floating point inaccuracies, but sometimes it's handy:

scala> 1.23456789 - (1.23456789 % 0.01)
res4: Double = 1.23

Answer 4

It's actually very easy to handle using Scala f interpolator - https://docs.scala-lang.org/overviews/core/string-interpolation.html

Suppose we want to round till 2 decimal places:

scala> val sum = 1 + 1/4D + 1/7D + 1/10D + 1/13D
sum: Double = 1.5697802197802198

scala> println(f"$sum%1.2f")
1.57

Answer 5

You can do:Math.round(<double precision value> * 100.0) / 100.0 But Math.round is fastest but it breaks down badly in corner cases with either a very high number of decimal places (e.g. round(1000.0d, 17)) or large integer part (e.g. round(90080070060.1d, 9)).

Use Bigdecimal it is bit inefficient as it converts the values to string but more relieval: BigDecimal(<value>).setScale(<places>, RoundingMode.HALF_UP).doubleValue() use your preference of Rounding mode.

If you are curious and want to know more detail why this happens you can read this:

Answer 6

For those how are interested, here are some times for the suggested solutions...

Rounding
Java Formatter: Elapsed Time: 105
Scala Formatter: Elapsed Time: 167
BigDecimal Formatter: Elapsed Time: 27

Truncation
Scala custom Formatter: Elapsed Time: 3 

Truncation is the fastest, followed by BigDecimal. Keep in mind these test were done running norma scala execution, not using any benchmarking tools.

object TestFormatters {

  val r = scala.util.Random

  def textFormatter(x: Double) = new java.text.DecimalFormat("0.##").format(x)

  def scalaFormatter(x: Double) = "$pi%1.2f".format(x)

  def bigDecimalFormatter(x: Double) = BigDecimal(x).setScale(2, BigDecimal.RoundingMode.HALF_UP).toDouble

  def scalaCustom(x: Double) = {
    val roundBy = 2
    val w = math.pow(10, roundBy)
    (x * w).toLong.toDouble / w
  }

  def timed(f: => Unit) = {
    val start = System.currentTimeMillis()
    f
    val end = System.currentTimeMillis()
    println("Elapsed Time: " + (end - start))
  }

  def main(args: Array[String]): Unit = {

    print("Java Formatter: ")
    val iters = 10000
    timed {
      (0 until iters) foreach { _ =>
        textFormatter(r.nextDouble())
      }
    }

    print("Scala Formatter: ")
    timed {
      (0 until iters) foreach { _ =>
        scalaFormatter(r.nextDouble())
      }
    }

    print("BigDecimal Formatter: ")
    timed {
      (0 until iters) foreach { _ =>
        bigDecimalFormatter(r.nextDouble())
      }
    }

    print("Scala custom Formatter (truncation): ")
    timed {
      (0 until iters) foreach { _ =>
        scalaCustom(r.nextDouble())
      }
    }
  }

}