val AlphabetPattern = "^([a-z]+)".r
def stringMatch(s: String) = s match {
case AlphabetPattern() => println("found")
case _ => println("not found")
}
If I try,
stringMatch("hello")
I get "not found", but I expected to get "found".
My understanding of the regex,
[a-z] = in the range of 'a' to 'z'
+ = one more of the previous pattern
^ = starts with
So regex AlphabetPattern is "all strings that start with one or more alphabets in the range a-z"
Surely I am missing something, want to know what.
Replace case AlphabetPattern() with case AlphabetPattern(_) and it works. The extractor pattern takes a variable to which it binds the result. Here we discard it but you could use x or whatever.
edit: Further to Randall's comment below, if you check the docs for Regex you'll see that it has an unapplySeq rather than an unapply method, which means it takes multiple variables. If you have the wrong number, it won't match, rather like
list match { case List(a,b,c) => a + b + c }
won't match if list doesn't have exactly 3 elements.
There are some issues with the match statement. s match is matching on the value of s which is checked against AlphabetPattern and _ which always evaluates to _ since s is never equal to "^([a-z]+)".r. Use one of the find methods in Scala.Util.Regex to look for a match with the given `Regex.
For example, using findFirstIn to find the first match of a string in AlphabetPattern.
scala> AlphabetPattern.findFirstIn("hello")
res0: Option[String] = Some(hello)
The stringMatch method using findFirstIn and a case statement:
scala> def stringMatch(s: String) = AlphabetPattern findFirstIn s match {
| case Some(s) => println("Found: " + s)
| case None => println("Not found")
| }
stringMatch: (s:String)Unit
scala> stringMatch("hello")
Found: hello
Related
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.
I want to extract part of a String that match one of the tow regex patterns i defined:
//should match R0010, R0100,R0300 etc
val rPat="[R]{1}[0-9]{4}".r
// should match P.25.01.21 , P.27.03.25 etc
val pPat="[P]{1}[.]{1}[0-9]{2}[.]{1}[0-9]{2}[.]{1}[0-9]{2}".r
When I now define my method to extract the elements as:
val matcher= (s:String) => s match {case pPat(el)=> println(el) // print the P.25.01.25
case rPat(el)=>println(el) // print R0100
case _ => println("no match")}
And test it eg with:
val pSt=" P.25.01.21 - Hello whats going on?"
matcher(pSt)//prints "no match" but should print P.25.01.21
val rSt= "R0010 test test 3,870"
matcher(rSt) //prints also "no match" but should print R0010
//check if regex is wrong
val pHead="P.25.01.21"
pHead.matches(pPat.toString)//returns true
val rHead="R0010"
rHead.matches(rPat.toString)//return true
I'm not sure if the regex expression are wrong but the matches method works on the elements. So what is wrong with the approach?
When you use pattern matching with strings, you need to bear in mind that:
The .r pattern you pass will need to match the whole string, else, no match will be returned (the solution is to make the pattern .r.unanchored)
Once you make it unanchored, watch out for unwanted matches: R[0-9]{4} will match R1234 in CSR123456 (solutions are different depending on what your real requirements are, usually word boundaries \b are enough, or negative lookarounds can be used)
Inside a match block, the regex matching function requires a capturing group to be present if you want to get some value back (you defined it as el in pPat(el) and rPat(el).
So, I suggest the following solution:
val rPat="""\b(R\d{4})\b""".r.unanchored
val pPat="""\b(P\.\d{2}\.\d{2}\.\d{2})\b""".r.unanchored
val matcher= (s:String) => s match {case pPat(el)=> println(el) // print the P.25.01.25
case rPat(el)=>println(el) // print R0100
case _ => println("no match")
}
Then,
val pSt=" P.25.01.21 - Hello whats going on?"
matcher(pSt) // => P.25.01.21
val pSt2_bad=" CP.2334565.01124.212 - Hello whats going on?"
matcher(pSt2_bad) // => no match
val rSt= "R0010 test test 3,870"
matcher(rSt) // => R0010
val rSt2_bad = "CSR00105 test test 3,870"
matcher(rSt2_bad) // => no match
Some notes on the patterns:
\b - a leading word boundary
(R\d{4}) - a capturing group matching exactly 4 digits
\b - a trailing word boundary
Due to the triple quotes used to define the string literal, there is no need to escape the backslashes.
Introduce groups in your patterns:
val rPat=".*([R]{1}[0-9]{4}).*".r
val pPat=".*([P]{1}[.]{1}[0-9]{2}[.]{1}[0-9]{2}[.]{1}[0-9]{2}).*".r
...
scala> matcher(pSt)
P.25.01.21
scala> matcher(rSt)
R0010
If code is written in the following way, the desired outcome will be generated. Reference API documentation followed is http://www.scala-lang.org/api/2.12.1/scala/util/matching/Regex.html
//should match R0010, R0100,R0300 etc
val rPat="[R]{1}[0-9]{4}".r
// should match P.25.01.21 , P.27.03.25 etc
val pPat="[P]{1}[.]{1}[0-9]{2}[.]{1}[0-9]{2}[.]{1}[0-9]{2}".r
def main(args: Array[String]) {
val pSt=" P.25.01.21 - Hello whats going on?"
val pPatMatches = pPat.findAllIn(pSt);
pPatMatches.foreach(println)
val rSt= "R0010 test test 3,870"
val rPatMatches = rPat.findAllIn(rSt);
rPatMatches.foreach(println)
}
Please, let me know if that works for you.
Struggling with my first (ever) Scala regex here. I need to see if a given String matches the regex: "animal<[a-zA-Z0-9]+,[a-zA-Z0-9]+>".
So, some examples:
animal<0,sega> => valid
animal<fizz,buzz> => valid
animAl<fizz,buzz> => illegal; animAl contains upper-case (and this is case-sensitive)
animal<fizz,3d> => valid
animal<,3d> => illegal; there needs to be something [a-zA-Z0-9]+ between '<' and ','
animal<fizz,> => illegal; there needs to be something [a-zA-Z0-9]+ between ',' and '>'
animal<fizz,%> => illegal; '%' doesn't match [a-zA-Z0-9]+
etc.
My best attempt so far:
val animalRegex = "animal<[a-zA-Z0-9]+,[a-zA-Z0-9]+>".r
animalRegex.findFirstIn("animal<fizz,buzz")
Unfortunately that's where I'm hitting a brick wall. findFirstIn and all the other obvious methods available of animalRegex all return Option[String] types. I was hoping to find something that returns a boolean, so something like:
val animalRegex = "animal<[a-zA-Z0-9]+,[a-zA-Z0-9]+>".r
if(animalRegex.matches("animal<fizz,buzz>")) {
val leftOperand : String = getLeftOperandSomehow(...)
val rightOperand : String = getRightOperandSomehow(...)
}
So I need the equivalent of Java's matches method, and then need a way to access the "left operand" (that is, the value of the first [a-zA-Z0-9]+ group, which in the current case is "fizz"), and then ditto for the right/second operand ("buzz"). Any ideas where I'm going awry?
To be able to extract the matched parts from your string, you'll need to add capture groups to your regex expression, like so (note the parentheses):
val animalRegex = "animal<([a-zA-Z0-9]+),([a-zA-Z0-9]+)>".r
Then, you can use Scala's pattern matching to check for a match and extract the operands from the string:
val str = "animal<fizz,3d>"
val result = str match {
case animalRegex(op1,op2) => s"$op1, $op2"
case _ => "Did not match"
}
In this example, result will contain "fizz, 3d"
I have this verbose code that does shortcircuit Regex extraction / matching in Scala. This attempts to match a string with the first Regex, if that doesn't match, it attempts to match the string with the second Regex.
val regex1 : scala.util.matching.Regex = "^a(b)".r
val regex2 : scala.util.matching.Regex = "^c(d)".r
val s = ?
val extractedGroup1 : Option[String] = s match { case regex1(v) => Some(v) case _ => None }
val extractedGroup2 : Option[String] = s match { case regex2(v) => Some(v) case _ => None}
val extractedValue = extractedGroup1.orElse(extractedGroup2)
Here are the results:
s == "ab" then extractedValue == "b"
s == "cd" then extractedValue == "c"
s == "gg" then extractedValue == None.
My question is how can we combine the two regex into a single regex with the regex or operator, and still use Scala extractors. I tried this, but it's always the None case.
val regex : scala.util.matching.Regex = "^a(b)$ | ^c(d)$".r
val extractedValue: s match { case regex(v) => Some(v) case _ => None }
Don't use quality of life spaces within the regex, although they feel very scala-esque, they might be taken literary and your program expects that there should be a whitespace after the endOfString or space before the startOfString, which is obviously never the case. Try ^(?:a(b)|c(d))$, which is the same thing you did without repeating ^ and $.
Your own ^a(b)$|^c(d)$ can work too (if you remove the whitespaces).
Also, do you really get c out of cd? Judging by your regex, you should be getting d, if we're talking about capture groups.
Also, note that you're extracting capture groups. If you combine the regexes, an extracted d will be $2, while b would be $1.
Assume I have a Regex pattern I want to match many Strings to.
val Digit = """\d""".r
I just want to check whether a given String fully matches the Regex. What is a good and idiomatic way to do this in Scala?
I know that I can pattern match on Regexes, but this is syntactically not very pleasing in this case, because I have no groups to extract:
scala> "5" match { case Digit() => true case _ => false }
res4: Boolean = true
Or I could fall back to the underlying Java pattern:
scala> Digit.pattern.matcher("5").matches
res6: Boolean = true
which is not elegant, either.
Is there a better solution?
Answering my own question I'll use the "pimp my library pattern"
object RegexUtils {
implicit class RichRegex(val underlying: Regex) extends AnyVal {
def matches(s: String) = underlying.pattern.matcher(s).matches
}
}
and use it like this
import RegexUtils._
val Digit = """\d""".r
if (Digit matches "5") println("match")
else println("no match")
unless someone comes up with a better (standard) solution.
Notes
I didn't pimp String to limit the scope of potential side effects.
unapplySeq does not read very well in that context.
I don't know Scala all that well, but it looks like you can just do:
"5".matches("\\d")
References
http://langref.org/scala/pattern-matching/matching
For the full match you may use unapplySeq. This method tries to match target (whole match) and returns the matches.
scala> val Digit = """\d""".r
Digit: scala.util.matching.Regex = \d
scala> Digit unapplySeq "1"
res9: Option[List[String]] = Some(List())
scala> Digit unapplySeq "123"
res10: Option[List[String]] = None
scala> Digit unapplySeq "string"
res11: Option[List[String]] = None
"""\d""".r.unapplySeq("5").isDefined //> res1: Boolean = true
"""\d""".r.unapplySeq("a").isDefined //> res2: Boolean = false
Using Standard Scala library and a pre-compiled regex pattern and pattern matching (which is scala state of the art):
val digit = """(\d)""".r
"2" match {
case digit( a) => println(a + " is Digit")
case _ => println("it is something else")
}
more to read: http://www.scala-lang.org/api/2.12.1/scala/util/matching/index.html
The answer is in the regex:
val Digit = """^\d$""".r
Then use the one of the existing methods.