Better String formatting in Scala

Question

With too many arguments, String.format easily gets too confusing. Is there a more powerful way to format a String. Like so:

"This is #{number} string".format("number" -> 1)

Or is this not possible because of type issues (format would need to take a Map[String, Any], I assume; don’t know if this would make things worse).

Or is the better way doing it like this:

val number = 1
<plain>This is { number } string</plain> text

even though it pollutes the name space?

Edit:

While a simple pimping might do in many cases, I’m also looking for something going in the same direction as Python’s format() (See: http://docs.python.org/release/3.1.2/library/string.html#formatstrings)

FWIW, I disagree that one ever needs to be in a situation where String.format has too many arguments. Use two or more strings; concat with +. But of course it would be nice if someone took the time to clone the python format for those people who prefer that style to the C style. — Rex Kerr
– Rex Kerr, Commented Oct 29, 2010 at 15:31
Sure, there are not so many applications but sometimes you just want to externalise a template string and then the advantage is evident in my opinion. Or for internationalisation stuff (though I don’t know if e.g. even the Python format is powerful enough for that). — Debilski
– Debilski, Commented Oct 29, 2010 at 15:46
@Always-Asking (via suggested edit): Those tags probably weren't in existence when this question was posted; the addition does make sense though. — Pokechu22
– Pokechu22, Commented Nov 13, 2014 at 1:12

Andrej Herich · Accepted Answer · 2012-11-20 22:34:12Z

90

In Scala 2.10 you can use string interpolation.

val height = 1.9d
val name = "James"
println(f"$name%s is $height%2.2f meters tall")  // James is 1.90 meters tall

answered Nov 20, 2012 at 22:34

Andrej Herich

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Daniel C. Sobral · Accepted Answer · 2010-10-29 23:38:00Z

Well, if your only problem is making the order of the parameters more flexible, this can be easily done:

scala> "%d %d" format (1, 2)
res0: String = 1 2

scala> "%2$d %1$d" format (1, 2)
res1: String = 2 1

And there's also regex replacement with the help of a map:

scala> val map = Map("number" -> 1)
map: scala.collection.immutable.Map[java.lang.String,Int] = Map((number,1))

scala> val getGroup = (_: scala.util.matching.Regex.Match) group 1
getGroup: (util.matching.Regex.Match) => String = <function1>

scala> val pf = getGroup andThen map.lift andThen (_ map (_.toString))
pf: (util.matching.Regex.Match) => Option[java.lang.String] = <function1>

scala> val pat = "#\\{([^}]*)\\}".r
pat: scala.util.matching.Regex = #\{([^}]*)\}

scala> pat replaceSomeIn ("This is #{number} string", pf)
res43: String = This is 1 string

Priyank Desai · Accepted Answer · 2015-10-07 22:20:45Z

19

This the answer I came here looking for:

"This is %s string".format(1)

answered Oct 7, 2015 at 22:20

Priyank Desai

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Levi Ramsey · Accepted Answer · 2021-07-22 12:50:17Z

18

You can easily implement a richer formatting yourself (with the "enhance my library" approach):

scala> implicit def RichFormatter(string: String) = new {
     |   def richFormat(replacement: Map[String, Any]) =
     |     (string /: replacement) {(res, entry) => res.replaceAll("#\\{%s\\}".format(entry._1), entry._2.toString)}
     | }
RichFormatter: (string: String)java.lang.Object{def richFormat(replacement: Map[String,Any]): String}

scala> "This is #{number} string" richFormat Map("number" -> 1)
res43: String = This is 1 string

Or on more recent Scala versions since the original answer:

implicit class RichFormatter(string: String) {
  def richFormat(replacement: Map[String, Any]): String =
    replacement.foldLeft(string) { (res, entry) =>
      res.replaceAll("#\\{%s\\}".format(entry._1), entry._2.toString)
    }
}

edited Jul 22, 2021 at 12:50

Levi Ramsey

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answered Oct 29, 2010 at 11:47

Vasil Remeniuk

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1 Comment

Debilski Over a year ago

True, I might even use % for that (as in some other languages). But still, it could be a bit more powerful… (Like python’s format maybe.)

Ruediger Keller · Accepted Answer · 2010-10-29 12:39:48Z

17

Maybe the Scala-Enhanced-Strings-Plugin can help you. Look here:

Scala-Enhanced-Strings-Plugin Documentation

answered Oct 29, 2010 at 12:39

Ruediger Keller

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3 Comments

Debilski Over a year ago

Looks pretty interesting. (Both the docu and the plugin.) But still a bit crazy (also both) because the syntax does not use any delimiters sometimes…

Alex Cruise Over a year ago

I'm using it in production and it's great! Delimiters are only required when the expression to be interpolated has much syntax beyond just a variable name.

emragins Over a year ago

For 2.10+ users, definitely check out the built in string interpolation at docs.scala-lang.org/overviews/core/string-interpolation.html. Thanks to Andrej for pointing this out.

Aldo Bucchi · Accepted Answer · 2013-01-01 22:25:21Z

5

If you're using 2.10 then go with built-in interpolation. Otherwise, if you don't care about extreme performance and are not afraid of functional one-liners, you can use a fold + several regexp scans:

val template = "Hello #{name}!"
val replacements = Map( "name" -> "Aldo" )
replacements.foldLeft(template)((s:String, x:(String,String)) => ( "#\\{" + x._1 + "\\}" ).r.replaceAllIn( s, x._2 ))

answered Jan 1, 2013 at 22:25

Aldo Bucchi

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Dominik Bucher · Accepted Answer · 2014-03-12 09:37:30Z

3

You might also consider the use of a template engine for really complex and long strings. On top of my head I have Scalate which implements amongst others the Mustache template engine.

Might be overkill and performance loss for simple strings, but you seem to be in that area where they start becoming real templates.

answered Mar 12, 2014 at 9:37

Dominik Bucher

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