CODEWITHSUNDEEP
regexscala
Laxmikanth Samudrala
Laxmikanth Samudrala

Reputation: 2243

scala regular expression for string patterns

val fields = Set("abc", "abcd", "abc per", "uio abcd", "yrabc", "entry", "peer")

def matchFields(param: String): Set[String] = ????

if I am providing input "abc" to matchFields method above; I am expecting the response as Set("abc", "abcd", "abc per", "uio abcd", "yrabc").

Can I know the implementation suggestions in scala with the help of regular expressions ?

Upvotes: 0

Views: 66

Answers (1)

dk14
dk14

Reputation: 22374

No need for regular expressions here:

def matchFields(param: String): Set[String] = fields.filter(_.contains(param))

scala> matchFields("abc")
res0: Set[String] = Set(uio abcd, abc, abc per, yrabc, abcd)

contains checks that one string is a substring of another, filter filters out elements, that not match given predicate.

If you really really want regexps:

import scala.util.matching._

scala> def matchFields(R: Regex): Set[String] = fields.collect{case str@R() => str}
matchFields: (R: scala.util.matching.Regex)Set[String]

scala> matchFields(".*abc.*".r)
res5: Set[String] = Set(uio abcd, abc, abc per, yrabc, abcd)

Or:

scala> def matchFields(R: Regex): Set[String] = fields.flatMap(R.findFirstIn)
matchFields: (R: scala.util.matching.Regex)Set[String]

scala> matchFields(".*abc.*".r)
res7: Set[String] = Set(uio abcd, abc, abc per, yrabc, abcd)

.* means 0 or more of any symbol. .r creates Regex from String

Upvotes: 1

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