I need to process phone numbers using regex and group them by (country code) (area code) (number). The input format:
country code: between 1-3 digits
, area code: between 1-3 digits
, number: between 4-10 digits
Examples:
1 877 2638277
91-011-23413627
And then I need to print out the groups like this:
CC=91,AC=011,Number=23413627
This is what I have so far:
String s = readLine
val pattern = """([0-9]{1,3})[ -]([0-9]{1,3})[ -]([0-9]{4,10})""".r
val ret = pattern.findAllIn(s)
println("CC=" + ret.group(1) + "AC=" + ret.group(2) + "Number=" + ret.group(3));
The compiler said "empty iterator." I also tried:
val (cc,ac,n) = s
and that didn't work either. How to fix this?
The problem is with your pattern. I would recommend using some tool like RegexPal to test them. Put the pattern in the first text box and your provided examples in the second one. It will highlight the matched parts.
You added spaces between your groups and [ -] separators, and it was expecting spaces there. The correct pattern is:
val pattern = """([0-9]{1,3})[ -]([0-9]{1,3})[ -]([0-9]{4,10})""".r
Also if you want to explicitly get groups then you want to get a Match returned. For an example the findFirstMatchIn function returns the first optional Match or the findAllMatchIn returns a list of matches:
val allMatches = pattern.findAllMatchIn(s)
allMatches.foreach { m =>
println("CC=" + m.group(1) + "AC=" + m.group(2) + "Number=" + m.group(3))
}
val matched = pattern.findFirstMatchIn(s)
matched match {
case Some(m) =>
println("CC=" + m.group(1) + "AC=" + m.group(2) + "Number=" + m.group(3))
case None =>
println("There wasn't a match!")
}
I see you also tried extracting the string into variables. You have to use the Regex extractor in the following way:
val Pattern = """([0-9]{1,3})[ -]([0-9]{1,3})[ -]([0-9]{4,10})""".r
val Pattern(cc, ac, n) = s
println(s"CC=${cc}AC=${ac}Number=$n")
And if you want to handle errors:
s match {
case Pattern(cc, ac, n) =>
println(s"CC=${cc}AC=${ac}Number=$n")
case _ =>
println("No match!")
}
Also you can also take a look at string interpolation to make your strings easier to understand: s"..."
Related
I have this code for extracting the repetitive : separated sections of a regex, which does not give me the right output.
val pattern = """([a-zA-Z]+)(:([a-zA-Z]+))*""".r
for (p <- pattern findAllIn "it:is:very:great just:because:it is") p match {
case pattern("it", pattern(is, pattern(very, great))) => println("it: "+ is + very+ great)
case pattern(it, _,rest) => println( it+" : "+ rest)
case pattern(it, is, very, great) => println(it +" : "+ is +" : "+ very +" : " + great)
case _ => println("match failure")
}
What am I doing wrong?
How can I write a case expression which allows me to extract each : separated part of the pattern regex?
What is the right syntax with which to solve this?
How can the match against unknown number of arguments to be extracted from a regex be done?
In this case print:
it : is : very : great
just : because : it
is
You can't use repeated capturing group like that, it only saves the last captured value as the current group value.
You can still get the matches you need with a \b[a-zA-Z]+(?::[a-zA-Z]+)*\b regex and then split each match with ::
val text = "it:is:very:great just:because:it is"
val regex = """\b[a-zA-Z]+(?::[a-zA-Z]+)*\b""".r
val results = regex.findAllIn(text).map(_ split ':').toList
results.foreach { x => println(x.mkString(", ")) }
// => it, is, very, great
// just, because, it
// is
See the Scala demo. Regex details:
\b - word boundary
[a-zA-Z]+ - one or more ASCII letters
(?::[a-zA-Z]+)* - zero or more repetitions of
: - a colon
[a-zA-Z]+ - one or more ASCII letters
\b - word boundary
I have strings in this format:
object[i].base.base_x[i] and I get lists like List(0,1).
I want to use regular expressions in scala to find the match [i] in the given string and replace the first occurance with 0 and the second with 1. Hence getting something like object[0].base.base_x[1].
I have the following code:
val stringWithoutIndex = "object[i].base.base_x[i]" // basically this string is generated dynamically
val indexReplacePattern = raw"\[i\]".r
val indexValues = List(0,1) // list generated dynamically
if(indexValues.nonEmpty){
indexValues.map(row => {
indexReplacePattern.replaceFirstIn(stringWithoutIndex , "[" + row + "]")
})
else stringWithoutIndex
Since String is immutable, I cannot update stringWithoutIndex resulting into an output like List("object[0].base.base_x[i]", "object[1].base.base_x[i]").
I tried looking into StringBuilder but I am not sure how to update it. Also, is there a better way to do this? Suggestions other than regex are also welcome.
You couldloop through the integers in indexValues using foldLeft and pass the string stringWithoutIndex as the start value.
Then use replaceFirst to replace the first match with the current value of indexValues.
If you want to use a regex, you might use a positive lookahead (?=]) and a positive lookbehind (?<=\[) to assert the i is between opening and square brackets.
(?<=\[)i(?=])
For example:
val strRegex = """(?<=\[)i(?=])"""
val res = indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
s.replaceFirst(strRegex, row.toString)
}
See the regex demo | Scala demo
How about this:
scala> val str = "object[i].base.base_x[i]"
str: String = object[i].base.base_x[i]
scala> str.replace('i', '0').replace("base_x[0]", "base_x[1]")
res0: String = object[0].base.base_x[1]
This sounds like a job for foldLeft. No need for the if (indexValues.nonEmpty) check.
indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
indexReplacePattern.replaceFirstIn(s, "[" + row + "]")
}
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)
I know Scala can split strings on regex's like this simple split on whitespace:
myString.split("\\s+").foreach(println)
What if I want to split on whitespace, accounting for the possibility that there may be a quoted string in the input (which I wish to be treated as 1 thing)?
"""This is a "very complex" test"""
In this example I want the resulting substrings to be:
This
is
a
very complex
test
While handling quoted expressions with split can be tricky, doing so with Regex matches is quite easy. We just need to match all non-whitespace character sequences with ([^\\s]+) and all quoted character sequences with \"(.*?)\" (toList added in order to avoid reiteration):
import scala.util.matching._
val text = """This is a "very complex" test"""
val regex = new Regex("\"(.*?)\"|([^\\s]+)")
val matches = regex.findAllMatchIn(text).toList
val words = matches.map { _.subgroups.flatMap(Option(_)).fold("")(_ ++ _) }
words.foreach(println)
/*
This
is
a
very complex
test
*/
Note that the solution also counts quote itself as a word boundary. If you want to inline quoted strings into surrounding expressions, you'll need to add [^\\s]* from both sides of the quoted case and adjust group boundaries correspondingly:
...
val text = """This is a ["very complex"] test"""
val regex = new Regex("([^\\s]*\".*?\"[^\\s]*)|([^\\s]+)")
...
/*
This
is
a
["very complex"]
test
*/
You can also omit quote symbols when inlining a string by splitting a regex group:
...
val text = """This is a ["very complex"] test"""
val regex = new Regex("([^\\s]*)\"(.*?)\"([^\\s]*)|([^\\s]+)")
...
/*
This
is
a
[very complex]
test
*/
In more complex scenarios, when you have to deal with CSV strings, you'd better use a CSV parser (e.g. scala-csv).
For a string like the one in question, when you do not have to deal with escaped quotation marks, nor with any "wild" quotes appearing in the middle of the fields, you may adapt a known Java solution (see Regex for splitting a string using space when not surrounded by single or double quotes):
val text = """This is a "very complex" test"""
val p = "\"([^\"]*)\"|[^\"\\s]+".r
val allMatches = p.findAllMatchIn(text).map(
m => if (m.group(1) != null) m.group(1) else m.group(0)
)
println(allMatches.mkString("\n"))
See the online Scala demo, output:
This
is
a
very complex
test
The regex is rather basic as it contains 2 alternatives, a single capturing group and a negated character class. Here are its details:
\"([^\"]*)\" - ", followed with 0+ chars other than " (captured into Group 1) and then a "
| - or
[^\"\\s]+ - 1+ chars other than " and whitespace.
You only grab .group(1) if Group 1 participated in the match, else, grab the whole match value (.group(0)).
This should work:
val xx = """This is a "very complex" test"""
var x = xx.split("\\s+")
for(i <-0 until x.length) {
if(x(i) contains "\"") {
x(i) = x(i) + " " + x(i + 1)
x(i + 1 ) = ""
}
}
val newX= x.filter(_ != "")
for(i<-newX) {
println(i.replace("\"",""))
}
Rather than using split, I used a recursive approach. Treat the input string as a List[Char], then step through, inspecting the head of the list to see if it is a quote or whitespace, and handle accordingly.
def fancySplit(s: String): List[String] = {
def recurse(s: List[Char]): List[String] = s match {
case Nil => Nil
case '"' :: tail =>
val (quoted, theRest) = tail.span(_ != '"')
quoted.mkString :: recurse(theRest drop 1)
case c :: tail if c.isWhitespace => recurse(tail)
case chars =>
val (word, theRest) = chars.span(c => !c.isWhitespace && c != '"')
word.mkString :: recurse(theRest)
}
recurse(s.toList)
}
If the list is empty, you've finished recursion
If the first character is a ", grab everything up to the next quote, and recurse with what's left (after throwing out that second quote).
If the first character is whitespace, throw it out and recurse from the next character
In any other case, grab everything up to the next split character, then recurse with what's left
Results:
scala> fancySplit("""This is a "very complex" test""") foreach println
This
is
a
very complex
test
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