Scala Doubles, and Precision
How about :
val value = 1.4142135623730951
//3 decimal places
println((value * 1000).round / 1000.toDouble)
//4 decimal places
println((value * 10000).round / 10000.toDouble)
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)
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).
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