Regex Pattern Within Case Class Pattern Using Scala - regex

This has gotta be something stupid, but I'm wondering if someone can help me out here. The following regex pattern match within a case class match is not working as I would expect. Can someone provide some insight? Thanks.
object Confused {
case class MyCaseClass(s: String)
val WS = """\s*""".r
def matcher(myCaseClass: MyCaseClass) = myCaseClass match {
case MyCaseClass(WS(_)) => println("Found WS")
case MyCaseClass(s) => println(s"Found >>$s<<")
}
def main(args: Array[String]): Unit = {
val ws = " "
matcher(MyCaseClass(ws))
}
}
I would expect the the first case in the pattern match to be the one that matches, but it is not.
This prints
Found >> <<

It should be:
val WS = """(\s*)""".r
For your question, you want to match a pattern of spaces, In Scala,
A regular expression is used to determine whether a string matches a
pattern and, if it does, to extract or transform the parts that match.
for extracting match parts we need to use group to pattern a string. It means that we need to use parentheses to around our pattern string.
Example:
val date = """(\d\d\d\d)-(\d\d)-(\d\d)""".r
"2004-01-20" match {
case date(year, month, day) => s"$year was a good year for PLs."
}

Related

How to use match with regular expressions in Scala

I am starting to learn Scala and want to use regular expressions to match a character from a string so I can populate a mutable map of characters and their value (String values, numbers etc) and then print the result.
I have looked at several answers on SO and gone over the Scala Docs but can't seem to get this right. I have a short Lexer class that currently looks like this:
class Lexer {
private val tokens: mutable.Map[String, Any] = collection.mutable.Map()
private def checkCharacter(char: Character): Unit = {
val Operator = "[-+*/^%=()]".r
val Digit = "[\\d]".r
val Other = "[^\\d][^-+*/^%=()]".r
char.toString match {
case Operator(c) => tokens(c) = "Operator"
case Digit(c) => tokens(c) = Integer.parseInt(c)
case Other(c) => tokens(c) = "Other" // Temp value, write function for this
}
}
def lex(input: String): Unit = {
val inputArray = input.toArray
for (s <- inputArray)
checkCharacter(s)
for((key, value) <- tokens)
println(key + ": " + value)
}
}
I'm pretty confused by the sort of strange method syntax, Operator(c), that I have seen being used to handle the value to match and am also unsure if this is the correct way to use regex in Scala. I think what I want this code to do is clear, I'd really appreciate some help understanding this. If more info is needed I will supply what I can
This official doc has lot's of examples: https://www.scala-lang.org/api/2.12.1/scala/util/matching/Regex.html. What might be confusing is the type of the regular expression and its use in pattern matching...
You can construct a regex from any string by using .r:
scala> val regex = "(something)".r
regex: scala.util.matching.Regex = (something)
Your regex becomes an object that has a few useful methods to be able to find matching groups like findAllIn.
In Scala it's idiomatic to use pattern matching for safe extraction of values, thus Regex class also has unapplySeq method to support pattern matching. This makes it an extractor object. You can use it directly (not common):
scala> regex.unapplySeq("something")
res1: Option[List[String]] = Some(List(something))
or you can let Scala compiler call it for you when you do pattern matching:
scala> "something" match {
| case regex(x) => x
| case _ => ???
| }
res2: String = something
You might ask why exactly this return type on unapply/unapplySeq. The doc explains it very well:
The return type of an unapply should be chosen as follows:
If it is just a test, return a Boolean. For instance case even().
If it returns a single sub-value of type T, return an Option[T].
If you want to return several sub-values T1,...,Tn, group them in an optional tuple Option[(T1,...,Tn)].
Sometimes, the number of values to extract isn’t fixed and we would
like to return an arbitrary number of values, depending on the input.
For this use case, you can define extractors with an unapplySeq method
which returns an Option[Seq[T]]. Common examples of these patterns
include deconstructing a List using case List(x, y, z) => and
decomposing a String using a regular expression Regex, such as case
r(name, remainingFields # _*) =>
In short your regex might match one or more groups, thus you need to return a list/seq. It has to be wrapped in an Option to comply with extractor contract.
The way you are using regex is correct, I would just map your function over the input array to avoid creating mutable maps. Perhaps something like this:
class Lexer {
private def getCharacterType(char: Character): Any = {
val Operator = "([-+*/^%=()])".r
val Digit = "([\\d])".r
//val Other = "[^\\d][^-+*/^%=()]".r
char.toString match {
case Operator(c) => "Operator"
case Digit(c) => Integer.parseInt(c)
case _ => "Other" // Temp value, write function for this
}
}
def lex(input: String): Unit = {
val inputArray = input.toArray
val tokens = inputArray.map(x => x -> getCharacterType(x))
for((key, value) <- tokens)
println(key + ": " + value)
}
}
scala> val l = new Lexer()
l: Lexer = Lexer#60f662bd
scala> l.lex("a-1")
a: Other
-: Operator
1: 1

Scala regex find matches in middle of string [duplicate]

This question already has an answer here:
Working regex fails when using Scala pattern matching
(1 answer)
Closed 5 years ago.
I have written the following code in scala:
val regex_str = "([a-z]+)(\\d+)".r
"_abc123" match {
case regex_str(a, n) => "found"
case _ => "other"
}
which returns "other", but if I take off the leading underscore:
val regex_str = "([a-z]+)(\\d+)".r
"abc123" match {
case regex_str(a, n) => "found"
case _ => "other"
}
I get "found". How can I find any ([a-z]+)(\\d+) instead of just at the beginning? I am used to other regex languages where you use a ^ to specify beginning of the string, and the absence of that just gets all matches.
Scala regex patterns default as "anchored", i.e. bound to beginning and end of target string.
You'll get the expected match with this.
val regex_str = "([a-z]+)(\\d+)".r.unanchored
Hi May be you need something like this,
val regex_str = "[^>]([a-z]+)(\\d+)".r
"_abc123" match {
case regex_str(a, n) => println(s"found $a $n")
case _ => println("other")
}
This will avoid the first character from your string.
Hope this helps!
The unapplySeq of the Regex tries to capture the whole input by default (treats the pattern as if it was between ^ and $).
There are two ways to capture inside the input:
use .* before and after the captures: val regex_str = ".*([a-z]+)(\\d+).*".r
do the same with .unanchored: val regex_str = "([a-z]+)(\\d+)".r.unanchored
Otherwise scala treats regular expression anchors the same way as in other languages; this one is an exception made for semantic reasons.
The regex extractor in scala pattern-matching attempts to match the entire string. If you want to skip some junk-characters in the beginning and in the end, prepend a . with a reluctant quantifier to the regex:
val regex_str = ".*?([a-z]+)(\\d+).*".r
val result = "_!+<>__abc123_%$" match {
case regex_str(a, n) => s"found a = '$a', n = '$n'"
case _ => "no match"
}
println(result)
This outputs:
found a = 'abc', n = '123'
Otherwise, don't use the pattern match with the extractor, use "...".r.findAllIn to find all matches.

Scala regex get parameters in path

regex noob here.
example path:
home://Joseph/age=20/race=human/height=170/etc
Using regex, how do I grab everything after the "=" between the /Joseph/ path and /etc? I'm trying to create a list like
[20, human, 170]
So far I have
val pattern = ("""(?<=Joseph/)[^/]*""").r
val matches = pattern.findAllIn(path)
The pattern lets me just get "age=20" but I thought findAllIn would let me find all of the "parameter=" matches. And after that, I'm not sure how I would use regex to just obtain the "20" in "age=20", etc.
Code
See regex in use here
(?:(?<=/Joseph/)|\G(?!\A)/)[^=]+=([^=/]+)
Usage
See code in use here
object Main extends App {
val path = "home://Joseph/age=20/race=human/height=170/etc"
val pattern = ("""(?:(?<=/Joseph/)|\G(?!\A)/)[^=]+=([^=/]+)""").r
pattern.findAllIn(path).matchData foreach {
m => println(m.group(1))
}
}
Results
Input
home://Joseph/age=20/race=human/height=170/etc
Output
20
human
170
Explanation
(?:(?<=/Joseph/)|\G(?!\A)/) Match the following
(?<=/Joseph/) Positive lookbehind ensuring what precedes matches /Joseph/ literally
\G(?!\A)/ Assert position at the end of the previous match and match / literally
[^=]+ Match one or more of any character except =
= Match this literally
([^=/]+) Capture one or more of any character except = and / into capture group 1
Your pattern looks for the pattern directly after Joseph/, which is why only age=20 matched, maybe just look after =?
val s = "home://Joseph/age=20/race=human/height=170/etc"
// s: String = home://Joseph/age=20/race=human/height=170/etc
val pattern = "(?<==)[^/]*".r
// pattern: scala.util.matching.Regex = (?<==)[^/]*
pattern.findAllIn(s).toList
// res3: List[String] = List(20, human, 170)

Cannot retrive a group from Scala Regex match

I am struggling with regexps in Scala (2.11.5), I have a followin string to parse (example):
val string = "http://sth.com/sth/56,57597,14058913,Article_title,,5.html"
I want to extract third numeric value in the string above (it needs to be third after a slash because there can be other groups following), in order to do that I have the following regex pattern:
val pattern = """\/\d+,\d+,(\d+)""".r
I have been trying to retrieve the group for the third sequence of digits, but nothing seems to work for me.
val matchList = pattern.findAllMatchIn(string).foreach(println)
val matchListb = pattern.findAllIn(string).foreach(println)
I also tried using matching pattern.
string match {
case pattern(a) => println(a)
case _ => "What's going on?"
}
and got the same results. Either whole regexp is returned or nothing.
Is there an easy way to retrieve a group form regexp pattern in Scala?
You can use group method of scala.util.matching.Regex.Match to get the result.
val string = "http://sth.com/sth/56,57597,14058913,Article_title,,5.html"
val pattern = """\/\d+,\d+,(\d+)""".r
val result = pattern.findAllMatchIn(string) // returns iterator of Match
.toArray
.headOption // returns None if match fails
.map(_.group(1)) // select first regex group
// or simply
val result = pattern.findFirstMatchIn(string).map(_.group(1))
// result = Some(14058913)
// result will be None if the string does not match the pattern.
// if you have more than one groups, for instance:
// val pattern = """\/(\d+),\d+,(\d+)""".r
// result will be Some(56)
Pattern matching is usually the easiest way to do it, but it requires a match on the full string, so you'll have to prefix and suffix your regex pattern with .*:
val string = "http://sth.com/sth/56,57597,14058913,Article_title,,5.html"
val pattern = """.*\/\d+,\d+,(\d+).*""".r
val pattern(x) = string
// x: String = 14058913

Scala capture group using regex

Let's say I have this code:
val string = "one493two483three"
val pattern = """two(\d+)three""".r
pattern.findAllIn(string).foreach(println)
I expected findAllIn to only return 483, but instead, it returned two483three. I know I could use unapply to extract only that part, but I'd have to have a pattern for the entire string, something like:
val pattern = """one.*two(\d+)three""".r
val pattern(aMatch) = string
println(aMatch) // prints 483
Is there another way of achieving this, without using the classes from java.util directly, and without using unapply?
Here's an example of how you can access group(1) of each match:
val string = "one493two483three"
val pattern = """two(\d+)three""".r
pattern.findAllIn(string).matchData foreach {
m => println(m.group(1))
}
This prints "483" (as seen on ideone.com).
The lookaround option
Depending on the complexity of the pattern, you can also use lookarounds to only match the portion you want. It'll look something like this:
val string = "one493two483three"
val pattern = """(?<=two)\d+(?=three)""".r
pattern.findAllIn(string).foreach(println)
The above also prints "483" (as seen on ideone.com).
References
regular-expressions.info/Lookarounds
val string = "one493two483three"
val pattern = """.*two(\d+)three.*""".r
string match {
case pattern(a483) => println(a483) //matched group(1) assigned to variable a483
case _ => // no match
}
Starting Scala 2.13, as an alternative to regex solutions, it's also possible to pattern match a String by unapplying a string interpolator:
"one493two483three" match { case s"${x}two${y}three" => y }
// String = "483"
Or even:
val s"${x}two${y}three" = "one493two483three"
// x: String = one493
// y: String = 483
If you expect non matching input, you can add a default pattern guard:
"one493deux483three" match {
case s"${x}two${y}three" => y
case _ => "no match"
}
// String = "no match"
You want to look at group(1), you're currently looking at group(0), which is "the entire matched string".
See this regex tutorial.
def extractFileNameFromHttpFilePathExpression(expr: String) = {
//define regex
val regex = "http4.*\\/(\\w+.(xlsx|xls|zip))$".r
// findFirstMatchIn/findAllMatchIn returns Option[Match] and Match has methods to access capture groups.
regex.findFirstMatchIn(expr) match {
case Some(i) => i.group(1)
case None => "regex_error"
}
}
extractFileNameFromHttpFilePathExpression(
"http4://testing.bbmkl.com/document/sth1234.zip")