I'm trying to format floating point numbers with a maximum number of digits, but I don't want unnecessary trailing zeroes. I thought if I use g instead of f it would work (cf. this question)
def testF(d: Double) = f"$d%1.2f"
def testG(d: Double) = f"$d%1.2g"
Now this behaves rather strangely:
testF(3.1415) // --> 3.14 ok
testF(3.1) // --> 3.10 hmmm, don't want that zero
testG(3.1415) // --> 3.1 what the ?
Ok, so perhaps I need to increase the digits by one for g:
def testG(d: Double) = f"$d%1.3g"
testG(3.1415) // --> 3.14 ok
testG(3.1) // --> 3.10 grmpf
So two questions—one, why the heck is g dropping one digit and doesn't seem to care about trailing zeroes? Two, how can I have
testX(3.1415) // --> 3.14
testX(3.1) // --> 3.1
?
You could use the java DecimalFormat but it may not please the eye:
def testX(d: Double) = new java.text.DecimalFormat("#.##").format(d)
Also to answer your first question why the heck is g dropping one digit and doesn't seem to care about trailing zeroes
For the floating-point conversions 'e', 'E', and 'f' the precision is the number of digits after the decimal separator. If the conversion is 'g' or 'G', then the precision is the total number of digits in the resulting magnitude after rounding.
Formatter details
As with the C world's printf, Java's (and Scala's) Formatter includes both minimum and maximum field widths (for the %s format specifier):
// Minimum field width -- right-aligned
scala> "%23s".format(23)
res0: String = " 23"
// Minimum field width -- left-aligned
scala> "%-23s".format(23)
res1: String = "23 "
// Maximum field width with right truncation
scala> "%.3s".format(12345)
res2: String = 123
// Minimum and maximum width -- right aligned
scala> "%3.3s".format(1)
res3: String = " 1"
// Minimum and maximum width -- left aligned
scala> "%-3.3s".format(1)
res4: String = "1 "
How about this version?
The use of Formattable requires the %s format, but you're given the formatting args to interpret as you like.
A Formattable can pop up in the middle of an f-interpolation, since the normal format just uses either toString or your custom formatTo.
package object succinctly {
import java.util.{ Formattable, FormattableFlags, Formatter }
import scala.language.implicitConversions
implicit class Succinctly(val __thing: Double) extends AnyVal {
#inline def succinctly(s: String): String = s format fmtable(__thing)
#inline def succinctly: Formattable = fmtable(__thing)
}
private[this] val trailing = "(.*\\...*?)(0*)".r
private[this] def fmtable(a: Double) = new Formattable {
override def formatTo(formatter: Formatter, flags: Int, width: Int, precision: Int) = formatter.out append (
if (precision <= 0) a.toInt.toString
else s"%${width}.${precision}f" format a.asInstanceOf[java.lang.Double] match { case trailing(num, _) => num }
)
}
}
package succinctly {
import scala.language.postfixOps
import scala.math._
object Test extends App {
Console println (Pi succinctly "%1.2s")
Console println (3.1 succinctly "%1.2s")
Console println (3.0 succinctly "%1.2s")
Console println f"${3.1 succinctly}%1.2s"
}
}
It's also conceivable to write a custom interpolator that promotes doubles to trailless and changes %f to %s, though at some cost, since the f-interpolator is a macro.
Related
I have a BigDecimals that's always have 30 digits after the decimal points.
And I want to multiply them with Double and get a BigDecimal with 30 digits after decimal point.
For exemple I have:
val double = 4.0
val bd = 0.111111111111111111111111111111
def multiply(d : Double, bd : BigDecimal, scale: Int) = {
BigDecimal.valueOf(d).setScale(scale).*(bd)
}
// multiply(double,bd,30) => 0.4444444444444444000000000000000000
I expected to get 0.444444444444444444444444444444 (30 times 4 after the point)
What is wrong with my method?
Your bd isn't a BigDecimal, it's a Double:
scala> val bd = 0.111111111111111111111111111111
val bd: Double = 0.1111111111111111
It gets implicitly converted to a BigDecimal (after losing precision from Double) when you call multiply.
To create the BigDecimal, you seek, it's probably best to parse a string representation:
val bd = BigDecimal("0.111111111111111111111111111111")
// bd.scale is 30
It's also probably a good idea to explicitly set the scale again in multiply after the multiplication, as it's possible for multiplying two numbers, even with the same scale, to have greater scale than either number.
def multiply(d: Double, bd: BigDecimal, scale:Int): BigDecimal =
(BigDecimal.valueOf(d).setScale(scale) * bd).setScale(scale)
Scala's BigDecimal is similar to Java's BigDecimal.
When you multiply two BigDecimals together (using the * method on the left one), you get a new BigDecimal whose scale is the sum of the scale of its operands (inputs).
The variable bd is not a BigDecimal, so its name is misleading. It is actually a double. In order to instantiate a BigDecimal, you should use a constructor, for example: BigDecimal("0.111111111111111111111111111111").
val bd = BigDecimal("0.111111111111111111111111111111")
val double = 4.0
// There is no need to call `setScale` on `d` due to the first point noted above. Set it after multiplying:
def multiply(d : Double, bd : BigDecimal, scale: Int) =
(BigDecimal.valueOf(d) * bd).setScale(scale)
val result = multiply(double, bd, 30)
println(result)
println(result.scale)
Result:
0.444444444444444444444444444444
30
I am creating a simple function in Scala
def addOne(m: Int): Int = m + 1
Using it with integers works normally, using double it throws a type mismatch error.
addOne(2) = 3
addOne(2.1) = error: type mismatch
When I use it with a character in double quotes, it throws a type mismatch as expected.
addOne("z") = error: type mismatch
However, when using a single quotes character it returns a value for that letter.
addOne('z') = 123
What is happening here and why is it like this?
The reason you can use a Char as an argument to a function taking an Int is because Scala performs an implicit conversion from Char to Int. This specific conversion is defined in the companion object of the Char class. See here:
http://www.scala-lang.org/api/2.12.1/scala/Char$.html (It seems like SO breaks this link at the $ character. Copy-paste it instead)
The function perfoming the conversion is called char2int. It converts the Char into its corresponding Unicode value as an Int.
When the Scala compiler sees that the types Char and Int don't match, it will first check if there are any available implicit conversions. It only gives a compile error if it didn't find any. If it finds an implicit conversion, it will insert that function call into your code. Your code is therefore transformed to this:
addOne(Char.char2int('z'))
If you want to make your own implicit conversion to, for example, let your function accept String, you can define this:
// Enable implicit conversions.
import scala.language.implicitConversions
// The "implicit" modifier is the important part here, not the name of the function.
implicit def string2int(s: String) = s.toInt
Now this compiles:
// This returns 6
addOne("5")
/*
* This throws a NumberFormatException due to my implementation of string2int.
* Create your own implementation of string2int if you want it to work properly.
*/
addOne("a")
Finally: Beware that implicit conversions are very powerful and therefore can be dangerous! See TheArchetypalPaul's comment for an explanation.
It is because addOne(m: Int) [The part after colon (:) ] tells Scala you will pass Int to it, not Double, not anything else.
if you want it to work for Double, try this, but then you will always get Double as Result.
def addone (m : Double ) = m+1
addone: (m: Double)Double
scala> addone(1)
res0: Double = 2.0
scala> addone(1.1)
res1: Double = 2.1
Scala map the char type using the ASCII Table. So, 'z' is mapped to 122, which is an integer. In the method, addOne('z'), the input parameter been cast to an integer (i.e. 122).
However, the input parameter in addOne(2.1) is 2.1, which is a double and in addOne("z") is a string. They cannot be cast to an integer automatically.
def addOne(m: Int): Int = m + 1 only accept an integer for m
It also work with a single quoted character (z) because it's translated into its ASCII value. The value for 'z' is 122 and you add 1.
scala> val foo: Int = 'z'
foo: Int = 122
scala> val bar = foo + 1
bar: Int = 123
If you want to make this working with double you can specify def addOne(m: Double): Double = m + 1
What is the simplest/idiomatic way to format percentages in Scala?
I have the following solution but I'm wondering if a more concise way exists:
val value = 0.1456
val s1 = f"the float value is ${value}%.2f"
val s2= s"the percent value is ${java.text.NumberFormat.getPercentInstance.format(value)}"
value: Double = 0.1456
s1: String = the float value is 0.15
s2: String = the percent value is 15%
If you are looking for more of a concise method, the following works and goes along with your initial idea in your code. It is also easy to add on decimal placement without having to resort to the implicit functionality. Obviously these needs to be used a lot a better solution is with the implicit method.
val value = 0.1456
val s2 = val s2 = f"the percent value is ${value*100}%.0f%%"
s2: String = the percent value is 15%
just to give a couple of other runs as well (t show rounding down here):
val value2 = 0.1416
val s3 = val s2 = f"the percent value is ${value2*100}%.0f%%"
s3: String = the percent value is 14%
Example to show adding decimal places:
val s4 = f"the percent value is ${value2*100}%.1f%%"
s4: String = the percent value is 14.2%
You could use the "pimp my library" pattern to add the asPercentage method to Doubles.
implicit class DoubleAsPercentage(d: Double) {
def asPercentage = java.text.NumberFormat.getPercentInstance.format(d)
}
val s2 = s"the percent value is ${value.asPercentage}"
You can use the f string to format the percentage to the relevant decimal places you wish. In this example, it has 4 decimal places however if you only want to return two decimal places, then use .2f in the string below.
To return 3 decimal places, then use .3f etc etc. However if you want more decimal places than suggested; "here they are 4", you will end up with trailing zeros.
val percentage = 51.9938
scala> f"I scored $percentage%.8f%% in my exams"
res164: String = I scored 51.99380000% in my exams
scala> f"I scored $percentage%.2f%% in my exams"
res165: String = I scored 51.99% in my exams
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
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).
Here's another solution without BigDecimals
Truncate:
(math floor 1.23456789 * 100) / 100
Round (see rint):
(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)
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
How about :
val value = 1.4142135623730951
//3 decimal places
println((value * 1000).round / 1000.toDouble)
//4 decimal places
println((value * 10000).round / 10000.toDouble)
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.
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())
}
}
}
}
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
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))
}
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.
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("floor_val", floor(col("value"))).
withColumn("dec_val", col("value").cast(DecimalType(26,2))).
withColumn("floor2", (floor(col("value") * 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).
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.
Since the question specified rounding for doubles specifically, this seems way simpler than dealing with big integer or excessive string or numerical operations.
"%.2f".format(0.714999999999).toDouble
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
}
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:
I think previous answers are:
Plain wrong: using math.floor for example doesn't work for negative values..
Unnecessary complicated.
Here is a suggestion based on #kaito's answer (i can't comment yet):
def truncateAt(x: Double, p: Int): Double = {
val s = math.pow(10, p)
(x * s).toInt / s
}
toInt will work for positive and negative values.
I have a dynamically changing input reading from a file. The numbers are either Int or Double. Why does Scala print .0 after every Double number? Is there a way for Scala to print it the same way it reads it?
Example:
var x:Double = 1
println (x) // This prints '1.0', I want it to print '1'
x = 1.0 // This prints '1.0', which is good
I can't use Int because some of the input I get are Doubles. I can't use String or AnyVal because I perform some math operations.
Thank you,
scala> "%1.0f" format 1.0
res3: String = 1
If your input is either Int or Double, you can do it like this:
def fmt(v: Any): String = v match {
case d : Double => "%1.0f" format d
case i : Int => i.toString
case _ => throw new IllegalArgumentException
}
Usage:
scala> fmt(1.0)
res6: String = 1
scala> fmt(1)
res7: String = 1
scala> fmt(1.0f)
java.lang.IllegalArgumentException
at .fmt(<console>:7)
at .<init>(<console>:6)
at .<clinit>(<console>)
at RequestResult$.<init>(<console>:4)
at RequestResult$.<clinit>(<console>)
at RequestResult$result(<console>)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.Dele...
Otherwise, you might use BigDecimals. They are slow, but they do come with the scale, so "1", "1.0" and "1.00" are all different:
scala> var x = BigDecimal("1.0")
x: BigDecimal = 1.0
scala> x = 1
x: BigDecimal = 1
scala> x = 1.0
x: BigDecimal = 1.0
scala> x = 1.000
x: BigDecimal = 1.0
scala> x = "1.000"
x: BigDecimal = 1.000
var x:Double = 1
var y:Double = 1.0
print(x) // => 1.0
print(y) // => 1.0
If i understand you question you want scala to print x and y differently? The problem is that x and y are both a variable of the type Double and look the same.
Why do you explicitly define the type of the vars?
var x = 1
var y= 1.0
print(x) // => 1
print(y) // => 1.0
Use printf:
printf("The value is %.0f", x)
For a description of the format string, see this page from the Java SE 6 API documentation.
Note that you can ofcourse also use the Java library from Scala, so other ways to format numbers from Java can also be used from Scala. You can for example use class java.text.DecimalFormat:
val df = new java.text.DecimalFormat("#####")
println(df.format(x))
Starting with Scala 2.10 you can use the f interpolator:
scala> val x: Double = 1
x: Double = 1.0
scala> println(f"$x%.0f")
1
scala> val i = 1
i: Int = 1
scala> println(f"$i%.0f")
1
The use of a "_.0" at the end of floating point numbers is a convention. Just a way to know that the number is actually floating point and not an integer.
If you really need to "to print it the same way it reads it" you may have to rethink the way your code is structured, possibly preserving your input data. If it's just a formatting issue, the easiest way is to convert the values to integers before printing:
val x = 1.0
println(x.toInt)
If some are integers and some are not, you need a bit more code:
def fmt[T <% math.ScalaNumericConversions](n : T) =
if(n.toInt == n) n.toInt.toString else n.toString
val a : Double = 1.0
val b : Double = 1.5
val c : Int = 1
println(fmt(a))
println(fmt(b))
println(fmt(c))
The code above should print:
1
1.5
1
The signature of the fmt method accepts any type that either is a subtype of ScalaNumericConversions or can be converted to one through implicit conversions (so we can use the toInt method).
If you are working with a Double and want to format it as a String without .0 when it's a whole number and with its decimals otherwise, then you could use String::stripSuffix:
x.toString.stripSuffix(".0")
// val x: Double = 1.34 => "1.34"
// val x: Double = 1.0 => "1"
Use type inference, rather than explicit typing.
scala> val xi = 1
xi: Int = 1
scala> val xd = 1.0
xd: Double = 1.0
scala> println(xi)
1
scala> println(xd)
1.0