Scala Escape Character Regex - regex

Question

How can I write an expression to filter inputs so that it would be in the format of
(AAA) where A is a number from 0-9.
EX: (123), (592), (999)

Answer 1

Usually you want to do more than filter.
scala> val r = raw"\(\d{3}\)".r
r: scala.util.matching.Regex = \(\d{3}\)
scala> List("(123)", "xyz", "(456)").filter { case r() => true case _ => false }
res0: List[String] = List((123), (456))
scala> import PartialFunction.{cond => when}
import PartialFunction.{cond=>when}
scala> List("(123)", "xyz", "(456)").filter(when(_) { case r() => true })
res1: List[String] = List((123), (456))
Keeping all matches from each input:
scala> List("a(123)b", "xyz", "c(456)d").flatMap(s =>
| r.findAllMatchIn(s).map(_.matched).toList)
res2: List[String] = List((123), (456))
scala> List("a(123)b", "xyz", "c(456)d(789)e").flatMap(s =>
| r.findAllMatchIn(s).map(_.matched).toList)
res3: List[String] = List((123), (456), (789))
Keeping just the first:
scala> val r = raw"(\(\d{3}\))".r.unanchored
r: scala.util.matching.UnanchoredRegex = (\(\d{3}\))
scala> List("a(123)b", "xyz", "c(456)d(789)e").flatMap(r.unapplySeq(_: String)).flatten
res4: List[String] = List((123), (456))
scala> List("a(123)b", "xyz", "c(456)d(789)e").collect { case r(x) => x }
res5: List[String] = List((123), (456))
Keeping entire lines that match:
scala> List("a(123)b", "xyz", "c(456)d(789)e").collect { case s # r(_*) => s }
res6: List[String] = List(a(123)b, c(456)d(789)e)
Java API:
scala> import java.util.regex._
import java.util.regex._
scala> val p = Pattern.compile(raw"(\(\d{3}\))")
p: java.util.regex.Pattern = (\(\d{3}\))
scala> val q = p.asPredicate
q: java.util.function.Predicate[String] = java.util.regex.Pattern$$Lambda$1107/824691524#3234474
scala> List("(123)", "xyz", "(456)").filter(q.test)
res0: List[String] = List((123), (456))

Answer 2

Typically you create regexes by using the .r method available on strings, such as "[0-9]".r. However, as you have noticed, that means you can't interpolate escape characters, as the parser thinks you want to insert escape characters into the string, not the regex.
For this, you can use Scala's triple-quoted strings, which create strings of the exact character sequence, including backslashes and newlines.
To create a regex like you describe, you could write """\(\d\d\d\)""".r. Here's an example of it in use:
val regex = """\(\d\d\d\)""".r.pattern
Seq("(123)", "(---)", "456").filter(str => regex.matcher(str).matches)