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

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Those are great answers in this thread  In order to better show the difference  here is just an example  The reason I put it here b c during my work the numbers are required to be NOT half-up       import org apache spark sql types       val values   List 1 2345 2 9998 3 4567 4 0099 5 1231      val df   values toDF     df show        ------        value       ------       1 2345       2 9998       3 4567       4 0099       5 1231       ------       val df2   df withColumn  quot floor val quot   floor col  quot value quot          withColumn  quot dec val quot   col  quot value quot   cast DecimalType 26 2         withColumn  quot floor2 quot    floor col  quot value quot     100 0  100 0  cast DecimalType 26 2         df2 show    ------ --------- ------- ------    value floor val dec val floor2   ------ --------- ------- ------   1 2345         1    1 23   1 23   2 9998         2    3 00   2 99   3 4567         3    3 46   3 45   4 0099         4    4 01   4 00   5 1231         5    5 12   5 12   ------ --------- ------- ------   floor function floors to the largest interger less than current value  DecimalType by default will enable HALF UP mode  not just cut to precision you want  If you want to cut to a certain precision without using HALF UP mode  you can use above solution instead   or use scala math BigDecimal  where you have to explicitly define rounding modes

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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

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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 gt  1 23456789 -  1 23456789   0 01  res4  Double   1 23

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How about      val value   1 4142135623730951    3 decimal places println  value   1000  round   1000 toDouble     4 decimal places println  value   10000  round   10000 toDouble

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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

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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

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You can do Math round  lt double precision value gt    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  lt value gt   setScale  lt places gt   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

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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    gt  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       gt          textFormatter r nextDouble                       print  Scala Formatter         timed          0 until iters  foreach       gt          scalaFormatter r nextDouble                       print  BigDecimal Formatter         timed          0 until iters  foreach       gt          bigDecimalFormatter r nextDouble                       print  Scala custom Formatter  truncation          timed          0 until iters  foreach       gt          scalaCustom r nextDouble

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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  lt  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

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Since the question specified rounding for doubles  specifically  this seems way simpler than dealing with big integer or excessive string or numerical operations   quot   2f quot  format 0 714999999999  toDouble

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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 gt  val sum   1   1 4D   1 7D   1 10D   1 13D sum  Double   1 5697802197802198  scala gt  println f  sum 1 2f   1 57

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Recently  I faced similar problem and I solved it using following approach  def round value  Either Double  Float   places  Int        if  places  lt  0  0   else       val factor   Math pow 10  places      value match         case Left d    gt   Math round d   factor    factor        case Right f    gt   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

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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  gt  A 1  length    A 1  length   else p   toDouble

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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

[scala] Scala Doubles, and Precision

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