My Scala app is being given a string that may or may not contain the token "flimFlam(*)" inside of it, where the asterisk represents any kind of text, chars, punctuation, etc. There will always only be 0 or 1 instances of "flimFlam(*)" in this string, never more.
I need to detect if the given input string contains a "flimFlam(*)" instance, and if it does, extract out whatever is inside the two parentheses. Hence, if my string contains "flimFlam(Joe)", then the result would be a string with a value of "Joe", etc.
My best attempt so far:
val inputStr : String = "blah blah flimFlam(Joe) blah blah"
// Regex must be case-sensitive for "flimFlam" (not "FLIMFLAM", "flimflam", etc.)
val flimFlamRegex = ".*flimFlam\\(.*?\\)".r
val insideTheParens = flimFlamRegex.findFirstIn(inputStr)
Can anyone spot where I'm going awry?
Use pattern matching and regex extractor
val regex = ".*flimFlam\\((.*)\\).*".r
inputStr match {
case regex(x) => println(x)
case _ => println("no match")
}
Scala REPL
scala> val inputStr : String = "blah blah flimFlam(Joe) blah blah"
inputStr: String = blah blah flimFlam(Joe) blah blah
scala> val regex = ".*flimFlam\\((.*)\\).*"
regex: String = .*flimFlam\((.*)\).*
scala> val regex = ".*flimFlam\\((.*)\\).*".r
regex: scala.util.matching.Regex = .*flimFlam\((.*)\).*
scala> inputStr match { case regex(x) => println(x); case _ => println("no match")}
Joe
You may use a capturing group around .*? and just use an unanchored regex within match block so that the pattern could stay short and "pretty" (no need for .* around the value you are looking for):
var str = "blah blah flimFlam(Joe) blah blah"
val pattern = """flimFlam\((.*?)\)""".r.unanchored
val res = str match {
case pattern(res) => println(res)
case _ => "No match"
}
See the online demo
Also, note that you do not need to double backslashes inside """-quoted string literals that helps avoid excessive backslashes.
And a hint: if the flimFlam is a whole word, add \b in front - """\bflimFlam\((.*?)\)""".
Related
Exercise: given a string with a name, then space or newline, then email, then maybe newline and some text separated by newlines capture the name and the domain of email.
So I created the following:
val regexp = "^([a-zA-Z]+)(?:\\s|\\n)\\w+#(\\w+\\.\\w+)(?:.|\\r|\\n)*".r
def fun(str: String): String = {
val result = str match {
case regexp(name, domain) => name + ' ' + domain
case _ => "invalid"
}
result
}
And started testing:
scala> val input = "oleg oleg#email.com"
scala> fun(input)
res17: String = oleg email.com
scala> val input = "oleg\noleg#email.com"
scala> fun(input)
res18: String = oleg email.com
scala> val input = """oleg
| oleg#email.com
| 7bdaf0a1be3"""
scala> fun(input)
res19: String = oleg email.com
scala> val input = """oleg
| oleg#email.com
| 7bdaf0a1be3
| """
scala> fun(input)
res20: String = invalid
Why doesn't the regexp capture the string with the newline at the end?
This part (?:\\s|\\n) can be shortened to \s as it will also match a newline, and as there is still a space before the emails where you are using multiple lines it can be \s+ to repeat it 1 or more times.
Matching any character like this (?:.|\\r|\\n)* if very inefficient due to the alternation. You can use either [\S\s]* or use an inline modifier (?s) to make the dot match a newline.
But using your pattern to just get the name and the domain of the email you don't have to match what comes after it, as you are using the 2 capturing groups in the output.
^([a-zA-Z]+)\s+\w+#(\w+\.\w+)
Regex demo
If you do want to match all that follows, you can use:
val regexp = """(?s)^([a-zA-Z]+)\s+\w+#(\w+\.\w+).*""".r
def fun(str: String): String = {
val result = str match {
case regexp(name, domain) => name + ' ' + domain
case _ => "invalid"
}
result
}
Scala demo
Note that this pattern \w+#(\w+\.\w+) is very limited for matching an email
I have a regex like this:
val myregex = "This is a (.*) text for (.*) and other thing like .*".r
If I run :
> val myregex(a,b) = "This is a test text for something and other thing like blah blah"
a: String = test
b: String = something
it is ok, and it fails is b is missing:
> val myregex(a,b) = "This is a test text for and other thing like blah blah"
scala.MatchError: This is a test text for and other thing like blah blah (of class java.lang.String)
... 33 elided
Is there a way to keep for example the value a and replace b with a fallback value (and viceversa)? Or the only solution is splitting the regex in two distincts regexs?
Your original regex requires 2 consecutive spaces between for and and.
You may change your regex to actually match the string with an optional pattern by wrapping the space and the subsequent (.*) pattern with a non-capturing group and apply the ? quantifier to it making it optional:
val myregex = "This is a (.*) text for(?: (.*))? and other thing like .*".r
val x = "This is a test text for and other thing like blah blah"
x match {
case myregex(a, b) => print(s"${a} -- ${b}");
case _ => print("none")
}
// => test -- null
See the online Scala demo. Here, there is a match, but b is just null since the second capturing group did not participate in the match (and did not get initialized).
Or the only solution is splitting the regex in two distincts regexs?
This is the only solution. Your best bet is probably to use pattern matching:
("This is a test text for something", "and other thing like blah blah") match {
case (r1(a), r2(b)) => (a, b)
case (r1(a), _) => (a, "fallback")
}
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 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")