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.
Related
I am new to Scala and want to create a function to split Hello123 or Hello 123 into two strings as follows:
val string1 = 123
val string2 = Hello
What is the best way to do it, I have attempted to use regex matching \\d and \\D but I am not sure how to write the function fully.
Regards
You may replace with 0+ whitespaces (\s*+) that are preceded with letters and followed with digits:
var str = "Hello123"
val res = str.split("(?<=[a-zA-Z])\\s*+(?=\\d)")
println(res.deep.mkString(", ")) // => Hello, 123
See the online Scala demo
Pattern details:
(?<=[a-zA-Z]) - a positive lookbehind that only checks (but does not consume the matched text) if there is an ASCII letter before the current position in the string
\\s*+ - matches (consumes) zero or more spaces possessively, i.e.
(?=\\d) - this check is performed only once after the whitespaces - if any - were matched, and it requires a digit to appear right after the current position in the string.
Based on the given string I assume you have to match a string and a number with any number of spaces in between
here is the regex for that
([a-zA-Z]+)\\s*(\\d+)
Now create a regex object using .r
"([a-zA-Z]+)\\s*(\\d+)".r
Scala REPL
scala> val regex = "([a-zA-Z]+)\\s*(\\d+)".r
scala> val regex(a, b) = "hello 123"
a: String = "hello"
b: String = "123"
scala> val regex(a, b) = "hello123"
a: String = "hello"
b: String = "123"
Function to handle pattern matching safely
pattern match with extractors
str match {
case regex(a, b) => Some(a -> b.toInt)
case _ => None
}
Here is the function which does Regex with Pattern matching
def matchStr(str: String): Option[(String, Int)] = {
val regex = "([a-zA-Z]+)\\s*(\\d+)".r
str match {
case regex(a, b) => Some(a -> b.toInt)
case _ => None
}
}
Scala REPL
scala> def matchStr(str: String): Option[(String, Int)] = {
val regex = "([a-zA-Z]+)\\s*(\\d+)".r
str match {
case regex(a, b) => Some(a -> b.toInt)
case _ => None
}
}
defined function matchStr
scala> matchStr("Hello123")
res41: Option[(String, Int)] = Some(("Hello", 123))
scala> matchStr("Hello 123")
res42: Option[(String, Int)] = Some(("Hello", 123))
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.
I want to test if a regex including a variable that I defined before matches a string.
For instance I would like to do :
val value = "abc"
regex = "[^a-z]".r + value //something like that
if(regex matches ".abc") print("ok, it works")
So my question is : how can I add construct a regex including a variable in scala?
("[^a-z]" + value).r
is all you need
You must quote the value string to protect against special regex syntax.
scala> val value = "*"
value: String = *
scala> val oops = """[^a-z]""" + value r
oops: scala.util.matching.Regex = [^a-z]*
scala> ".*" match { case oops() => }
scala> ".....*" match { case oops() => }
They added quoting to the Scala API:
scala> import util.matching._
import util.matching._
scala> val ok = """[^a-z]""" + Regex.quote(value) r
ok: scala.util.matching.Regex = [^a-z]\Q*\E
scala> ".*" match { case ok() => }
scala> ".....*" match { case ok() => }
scala.MatchError: .....* (of class java.lang.String)
... 33 elided
You could also generalize the pattern and do additional checks:
scala> val r = """\W(\w+)""".r
r: scala.util.matching.Regex = \W(\w+)
scala> ".abc" match { case r(s) if s == "abc" => }
Parsing and building the regex itself is relatively expensive, so usually it's desirable to do it once with a general pattern.
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
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")