How to avoid using a var when doing multiple string replacements in a string

Question

This is what I tried:

def doStringReplcements(originalStr: String, replacementsMap: Map[String,String]): String = {
  var newStr = originalStr
  replacementsMap.foreach { pair => 
     newStr = newStr.replaceAllLiterally(pair._1, pair._2)
  }
  newStr
}

But functional programming style recommends avoiding vars so how do I do this with just vals?

Your code doesn't work, functional or not? replaceAllLiterally returns the new string, it doesn't change newStr. newStr = newStr.replaceAllLiterally is needed... — The Archetypal Paul
– The Archetypal Paul, Commented Sep 30, 2015 at 11:38

rahilb · Accepted Answer · 2015-09-30 11:08:54Z

4

Consider using the foldLeft function on Map.

e.g.

replacementsMap.foldLeft(originalStr){ case (accumulator, (target, replacement)) => 
    accumulator.replaceAllLiterally(target, replacement)      
}

answered Sep 30, 2015 at 11:08

rahilb

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1 Comment

som-snytt Over a year ago

This has two disadvantages, making multiple passes over the text, and also performing replacements on previous replacements: "ab" -> "Ab" then "Ab" is eligible for match, even though not in the original text.

som-snytt · Accepted Answer · 2015-09-30 14:47:28Z

2

Consider a regex:

scala> import util.matching.Regex
import util.matching.Regex

scala> val text = "one in hand, two in bush"
text: String = one in hand, two in bush

scala> val reps = Map("one"->"lots","two"->"more")
reps: scala.collection.immutable.Map[String,String] = Map(one -> lots, two -> more)

scala> val r = reps.keys.mkString("|").r
r: scala.util.matching.Regex = one|two

scala> r.replaceAllIn(text, m=>reps(m.matched))
res0: String = lots in hand, more in bush

answered Sep 30, 2015 at 14:47

som-snytt

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