Scala Doubles, and Precision

问题

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

回答1:

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).

回答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(p: Int) = roundAt(2)(p)

回答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

回答4:

How about :

 val value = 1.4142135623730951

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

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

回答5:

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.

回答6:

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))
}

回答7:

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())
      }
    }
  }

}

回答8:

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.

回答9:

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.

回答10:

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
}

回答11:

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:

来源：https://stackoverflow.com/questions/11106886/scala-doubles-and-precision

标签

scala

double

truncate

rounding