Scala array pattern matching using regular expressions

Question

How to pattern-match for instance the first string element in an array using regular expressions?

Consider for example

Array("col",1) match {
  case Array("""col | row""", n, _*) => n
  case _ => 0
}

which delivers 0, although the desired result would be 1.

Many Thanks.

score 7 · Accepted Answer · edited May 23 '17 at 12:28

7

A Regex instance provides extractors automatically, so you can use one directly in a pattern match expression:

val regex = "col|row".r

Array("col",1) match {
  case Array(regex(), n, _*) => n
  case _ => 0
}

Also: in a more general QA about regexps in Scala, sschaef has provided a very nice string interpolation for pattern matching usage (e.g. r"col|row" in this example). A potential caveat: the interpolation creates a fresh Regex instance on every call - so, if you use the same regex a large number of times, it may more efficient to store it in a val instead (as in this answer).

edited May 23 '17 at 12:28

Community

1
1

answered Mar 03 '14 at 13:05

mikołak

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That is definitely better than my solution – serejja Mar 03 '14 at 13:24
@serejja : well, potentially more efficient, and arguably more readable - maybe ;). Thanks in any case. – mikołak Mar 03 '14 at 18:38
1

Many Thanks for alĺ the comments. – elm Mar 03 '14 at 18:42

score 2 · Answer 2 · answered Mar 03 '14 at 12:55

2

Don't know if it is the best solution, but the working one:

Array("col", 1) match {
  case Array(str: String, n, _*) if str.matches("col|row") => n //note that spaces are removed in the pattern
  case _ => 0
}

answered Mar 03 '14 at 12:55

serejja

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1

Note that it's an `Array[Any]` and your result is also still `Any`. – som-snytt Mar 03 '14 at 17:18

Scala array pattern matching using regular expressions

2 Answers2