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")
Related
My program is:
val pattern = "[*]prefix_([a-zA-Z]*)_[*]".r
val outputFieldMod = "TRASHprefix_target_TRASH"
var tar =
outputFieldMod match {
case pattern(target) => target
}
println(tar)
Basically, I try to get the "target" and ignore "TRASH" (I used *). But it has some error and I am not sure why..
Simple and straight forward standard library function (unanchored)
Use Unanchored
Solution one
Use unanchored on the pattern to match inside the string ignoring the trash
val pattern = "prefix_([a-zA-Z]*)_".r.unanchored
unanchored will only match the pattern ignoring all the trash (all the other words)
val result = str match {
case pattern(value) => value
case _ => ""
}
Example
Scala REPL
scala> val pattern = """foo\((.*)\)""".r.unanchored
pattern: scala.util.matching.UnanchoredRegex = foo\((.*)\)
scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah
scala> str match { case pattern(value) => value ; case _ => "no match" }
res3: String = bar
Solution two
Pad your pattern from both sides with .*. .* matches any char other than a linebreak character.
val pattern = ".*prefix_([a-zA-Z]*)_.*".r
val result = str match {
case pattern(value) => value
case _ => ""
}
Example
Scala REPL
scala> val pattern = """.*foo\((.*)\).*""".r
pattern: scala.util.matching.Regex = .*foo\((.*)\).*
scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah
scala> str match { case pattern(value) => value ; case _ => "no match" }
res4: String = bar
This will work, val pattern = ".*prefix_([a-z]+).*".r, but it distinguishes between target and trash via lower/upper-case letters. Whatever determines real target data from trash data will determine the real regex pattern.
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
Could you guys please tell me what I'm doing incorrectly trying to extract using regex pattern-matching? I have following code
val Pattern = "=".r
val Pattern(key, value) = "key=value"
And I get following exception in runtime
Exception in thread "main" scala.MatchError: key=value (of class java.lang.String)
That's more of a regular expression problem: your regex does not capture any groups, it just matches a single = character.
With
val Pattern = "([^=]*)=(.*)".r
you will get:
scala> val Pattern(key, value) = "key=value"
key: String = key
value: String = value
Edit:
Also, that won't match if the input string is empty. You can change the pattern to make it match, or (better) you can pattern match with the regex, like so:
"key=value" match {
case Pattern(k, v) => // do something
case _ => // wrong input, do nothing
}
If what you actually wanted was to split the input text with whatever the regex matches, that is also possible using Regex.split:
scala> val Pattern = "=".r
Pattern: scala.util.matching.Regex = =
scala> val Array(key, value) = Pattern.split("key=value")
key: String = key
value: String = value
I want to use this
val r = """^myprefix:(.*)""".r
val r(suffix) = line
println(suffix)
But it gives an error when the string doesn't match. How do I use a similar construct where matching is optional?
Edit: To make it clear, I need the group (.*)
You can extract match groups via pattern matching.
val r = """^myprefix:(.*)""".r
line match {
case r(group) => group
case _ => ""
}
Another way using Option:
Option(line) collect { case r(group) => group }
"""^myprefix:(.*)""".r // Regex
.findFirstMatchIn(line) // Option[Match]
.map(_ group 1) // Option[String]
This has the advantage that you can write it as a one-liner without needing to assign the regex to an intermediate value r.
In case you're wondering, group 0 is the matched string while group 1 etc are the capture groups.
try
r.findFirstIn(line)
UPD:
scala> val rgx = """^myprefix:(.*)""".r
rgx: scala.util.matching.Regex = ^myprefix:(.*)
scala> val line = "myprefix:value"
line: java.lang.String = myprefix:value
scala> for (rgx(group) <- rgx.findFirstIn(line)) yield group
res0: Option[String] = Some(value)
I would like to be able to find a match between the first letter of a word, and one of the letters in a group such as "ABC". In pseudocode, this might look something like:
case Process(word) =>
word.firstLetter match {
case([a-c][A-C]) =>
case _ =>
}
}
But how do I grab the first letter in Scala instead of Java? How do I express the regular expression properly? Is it possible to do this within a case class?
You can do this because regular expressions define extractors but you need to define the regex pattern first. I don't have access to a Scala REPL to test this but something like this should work.
val Pattern = "([a-cA-C])".r
word.firstLetter match {
case Pattern(c) => c bound to capture group here
case _ =>
}
Since version 2.10, one can use Scala's string interpolation feature:
implicit class RegexOps(sc: StringContext) {
def r = new util.matching.Regex(sc.parts.mkString, sc.parts.tail.map(_ => "x"): _*)
}
scala> "123" match { case r"\d+" => true case _ => false }
res34: Boolean = true
Even better one can bind regular expression groups:
scala> "123" match { case r"(\d+)$d" => d.toInt case _ => 0 }
res36: Int = 123
scala> "10+15" match { case r"(\d\d)${first}\+(\d\d)${second}" => first.toInt+second.toInt case _ => 0 }
res38: Int = 25
It is also possible to set more detailed binding mechanisms:
scala> object Doubler { def unapply(s: String) = Some(s.toInt*2) }
defined module Doubler
scala> "10" match { case r"(\d\d)${Doubler(d)}" => d case _ => 0 }
res40: Int = 20
scala> object isPositive { def unapply(s: String) = s.toInt >= 0 }
defined module isPositive
scala> "10" match { case r"(\d\d)${d # isPositive()}" => d.toInt case _ => 0 }
res56: Int = 10
An impressive example on what's possible with Dynamic is shown in the blog post Introduction to Type Dynamic:
object T {
class RegexpExtractor(params: List[String]) {
def unapplySeq(str: String) =
params.headOption flatMap (_.r unapplySeq str)
}
class StartsWithExtractor(params: List[String]) {
def unapply(str: String) =
params.headOption filter (str startsWith _) map (_ => str)
}
class MapExtractor(keys: List[String]) {
def unapplySeq[T](map: Map[String, T]) =
Some(keys.map(map get _))
}
import scala.language.dynamics
class ExtractorParams(params: List[String]) extends Dynamic {
val Map = new MapExtractor(params)
val StartsWith = new StartsWithExtractor(params)
val Regexp = new RegexpExtractor(params)
def selectDynamic(name: String) =
new ExtractorParams(params :+ name)
}
object p extends ExtractorParams(Nil)
Map("firstName" -> "John", "lastName" -> "Doe") match {
case p.firstName.lastName.Map(
Some(p.Jo.StartsWith(fn)),
Some(p.`.*(\\w)$`.Regexp(lastChar))) =>
println(s"Match! $fn ...$lastChar")
case _ => println("nope")
}
}
As delnan pointed out, the match keyword in Scala has nothing to do with regexes. To find out whether a string matches a regex, you can use the String.matches method. To find out whether a string starts with an a, b or c in lower or upper case, the regex would look like this:
word.matches("[a-cA-C].*")
You can read this regex as "one of the characters a, b, c, A, B or C followed by anything" (. means "any character" and * means "zero or more times", so ".*" is any string).
To expand a little on Andrew's answer: The fact that regular expressions define extractors can be used to decompose the substrings matched by the regex very nicely using Scala's pattern matching, e.g.:
val Process = """([a-cA-C])([^\s]+)""".r // define first, rest is non-space
for (p <- Process findAllIn "aha bah Cah dah") p match {
case Process("b", _) => println("first: 'a', some rest")
case Process(_, rest) => println("some first, rest: " + rest)
// etc.
}
String.matches is the way to do pattern matching in the regex sense.
But as a handy aside, word.firstLetter in real Scala code looks like:
word(0)
Scala treats Strings as a sequence of Char's, so if for some reason you wanted to explicitly get the first character of the String and match it, you could use something like this:
"Cat"(0).toString.matches("[a-cA-C]")
res10: Boolean = true
I'm not proposing this as the general way to do regex pattern matching, but it's in line with your proposed approach to first find the first character of a String and then match it against a regex.
EDIT:
To be clear, the way I would do this is, as others have said:
"Cat".matches("^[a-cA-C].*")
res14: Boolean = true
Just wanted to show an example as close as possible to your initial pseudocode. Cheers!
First we should know that regular expression can separately be used. Here is an example:
import scala.util.matching.Regex
val pattern = "Scala".r // <=> val pattern = new Regex("Scala")
val str = "Scala is very cool"
val result = pattern findFirstIn str
result match {
case Some(v) => println(v)
case _ =>
} // output: Scala
Second we should notice that combining regular expression with pattern matching would be very powerful. Here is a simple example.
val date = """(\d\d\d\d)-(\d\d)-(\d\d)""".r
"2014-11-20" match {
case date(year, month, day) => "hello"
} // output: hello
In fact, regular expression itself is already very powerful; the only thing we need to do is to make it more powerful by Scala. Here are more examples in Scala Document: http://www.scala-lang.org/files/archive/api/current/index.html#scala.util.matching.Regex
Note that the approach from #AndrewMyers's answer matches the entire string to the regular expression, with the effect of anchoring the regular expression at both ends of the string using ^ and $. Example:
scala> val MY_RE = "(foo|bar).*".r
MY_RE: scala.util.matching.Regex = (foo|bar).*
scala> val result = "foo123" match { case MY_RE(m) => m; case _ => "No match" }
result: String = foo
scala> val result = "baz123" match { case MY_RE(m) => m; case _ => "No match" }
result: String = No match
scala> val result = "abcfoo123" match { case MY_RE(m) => m; case _ => "No match" }
result: String = No match
And with no .* at the end:
scala> val MY_RE2 = "(foo|bar)".r
MY_RE2: scala.util.matching.Regex = (foo|bar)
scala> val result = "foo123" match { case MY_RE2(m) => m; case _ => "No match" }
result: String = No match