CODEWITHSUNDEEP
javaregex
Fabio Thomaz
Fabio Thomaz

Reputation: 98

Regex to find words with letters and numbers separated or not by symbols

I need to build a regex that match words with these patterns:

Letters and numbers:

A35, 35A, B503X, 1ABC5

Letters and numbers separated by "-", "/", "\":

AB-10, 10-AB, A10-BA, BA-A10, etc...

I wrote this regex for it:

\b[A-Za-z]+(?=[(?<!\-|\\|\/)\d]+)[(?<!\-|\\|\/)\w]+\b|\b[0-9]+(?=[(?<!\-|\\|\/)A-Za-z]+)[(?<!\-|\\|\/)\w]+\b

It works partially, but it's match only letters or only numbers separated by symbols. Example:

10-10, open-office, etc.

And I don't wanna this matches.

I guess that my regex is very repetitive and somewhat ugly. But it's what I have for now.

Could anyone help me?

I'm using java/groovy.

Thanks in advance.

Upvotes: 2

Views: 7706

Answers (5)

eyquem
eyquem

Reputation: 27585

Excuse me to write my solution in Python, I don't know enough Java to write in Java.

pat = re.compile('(?=(?:([A-Z])|[0-9])' ## This part verifies that
                 '[^ ]*'                ## there are at least one
                 '(?(1)\d|[A-Z]))'      ## letter and one digit.
                 '('   
                 '(?:(?<=[ ,])[A-Z0-9]|\A[A-Z0-9])'  # start of second group
                 '[A-Z0-9-/\\\\]*'
                 '[A-Z0-9](?= |\Z|,)'               # end of second group
                 ')',  
                 re.IGNORECASE) # this group 2 catches the string

.

My solution catches the desired string in the second group: ((?:(?<={ ,])[A-Z0-9]|\A[A-Z0-9])[A-Z0-9-/\\\\]*[A-Z0-9](?= |\Z|,))

.

The part before it verifies that one letter at least and one digit at least are present in the catched string:

(?(1)\d|[A-Z]) is a conditional regex that means "if group(1) catched something, then there must be a digit here, otherwise there must be a letter"

The group(1) is ([A-Z]) in (?=(?:([A-Z])|[0-9])

(?:([A-Z])|[0-9]) is a non-capturing group that matches a letter (catched) OR a digit, so when it matches a letter, the group(1) isn't empty

.

The flag re.IGNORECASE allows to treat strings with upper or lower cased letters.

.

In the second group, I am obliged to write (?:(?<=[ ,])[A-Z0-9]|\A[A-Z0-9]) because lookbehind assertions with non fixed length are not allowed. This part signifies one character that can't be '-' preceded by a blank or the head of the string.

At the opposite, (?= |\Z[,) means 'end of string or a comma or a blank after'

.

This regex supposes that the characters '-' , '/' , '\' can't be the first character or the last one of a captured string . Is it right ?

import re

pat = re.compile('(?=(?:([A-Z])|[0-9])' ## (from here)  This part verifies that
                 '[^ ]*'                 #              there are at least one
                 '(?(1)\d|[A-Z]))'      ## (to here)    letter and one digit.
                 '((?:(?<=[ ,])[A-Z0-9]|\A[A-Z0-9])'
                 '[A-Z0-9-/\\\\]*'
                 '[A-Z0-9](?= |\Z|,))',
                 re.IGNORECASE) # this group 2 catches the string

ch = "ALPHA13 10 ZZ 10-10 U-R open-office ,10B a10 UCS5000 -TR54 code vg4- DV-3000 SEA 300-BR  gt4/ui bn\\3K"

print [ mat.group(2) for mat in pat.finditer(ch) ]

s = "A35, 35A, B503X,1ABC5 " +\
     "AB-10, 10-AB, A10-BA, BA-A10, etc... " +\
     "10-10, open-office, etc."

print [ mat.group(2) for mat in pat.finditer(s) ]

result

['ALPHA13', '10B', 'a10', 'UCS5000', 'DV-3000', '300-BR', 'gt4/ui', 'bn\\3K']
['A35', '35A', 'B503X', '1ABC5', 'AB-10', '10-AB', 'A10-BA', 'BA-A10']

Upvotes: 1

user unknown
user unknown

Reputation: 36229

A condition (A OR NOT A) can be omited. So symbols can savely been ignored. 

for (String word : "10 10-10 open-office 10B A10 UCS5000 code DV-3000 300-BR".split (" "))
    if (word.matches ("(.*[A-Za-z].*[0-9])|(.*[0-9].*[A-Za-z].*)"))
         // do something

You didn't mention -x4, 4x-, 4-x-, -4-x or -4-x-, I expect them all to match.

My expression looks just for something-alpha-something-digits-something, where something might be alpha, digits or symbols, and the opposite: something-alpha-something-digits-something. If something else might occur, like !#$~()[]{} and so on, it would get longer.

Tested with scala:

scala> for (word <- "10 10-10 open-office 10B A10 UCS5000 code DV-3000 300-BR".split (" ")
     | if word.matches ("(.*[A-Za-z].*[0-9])|(.*[0-9].*[A-Za-z].*)")) yield word          
res89: Array[java.lang.String] = Array(10B, A10, UCS5000, DV-3000, 300-BR)

Slightly modified to filter matches: 

String s = "A35, 35A, B53X, 1AC5, AB-10, 10-AB, A10-BA, BA-A10, etc. -4x, 4x- -4-x- 10-10, oe-oe, etc";
Pattern pattern  = java.util.regex.Pattern.compile ("\\b([^ ,]*[A-Za-z][^ ,]*[0-9])[^ ,]*|([^ ,]*[0-9][^ ,]*[A-Za-z][^ ,]*)\\b");
matcher = pattern.matcher (s);
while (matcher.find ()) { System.out.print (matcher.group () + "|") }

But I still have an error, which I don't find: 

A35|35A|B53X|1AC5|AB-10|10-AB|A10-BA|BA-A10|-4x|4x|-4-x|

4x should be 4x-, and -4-x should be -4-x-.

Upvotes: 0

Java Drinker
Java Drinker

Reputation: 3167

My first pass yields

(^|\s)(?!\d+[-/\\]?\d+(\s|$))(?![A-Z]+[-/\\]?[A-Z]+(\s|$))([A-Z0-9]+[-/\\]?[A-Z0-9]+)(\s|$)

Sorry, but it's not java formatted (you'll need to edit the \ \s etc.). Also, you can't use \b b/c a word boundary is anything that is not alphanumeric and underscore, so I used \s and the start and end of the string.

This is still a bit raw

EDIT

Version 2, slightly better, but could be improved for performance by usin possessive quantifiers. It matches ABC76 AB-32 3434-F etc, but not ABC or 19\23 etc.

((?<=^)|(?<=\s))(?!\d+[-/\\]?\d+(\s|$))(?![A-Z]+[-/\\]?[A-Z]+(\s|$))([A-Z0-9]+[-/\\]?[A-Z0-9]+)((?=$)|(?=\s))

Upvotes: 0

ridgerunner
ridgerunner

Reputation: 34395

Interesting challenge. Here is a java program with a regex that picks out the types of "words" you are after:

import java.util.regex.*;
public class TEST {
    public static void main(String[] args) {
        String s = "A35, 35A, B503X, 1ABC5 " +
            "AB-10, 10-AB, A10-BA, BA-A10, etc... " +
            "10-10, open-office, etc.";
        Pattern regex = Pattern.compile(
            "# Match special word having one letter and one digit (min).\n" +
            "\\b                       # Match first word having\n" +
            "(?=[-/\\\\A-Za-z]*[0-9])  # at least one number and\n" +
            "(?=[-/\\\\0-9]*[A-Za-z])  # at least one letter.\n" +
            "[A-Za-z0-9]+              # Match first part of word.\n" +
            "(?:                       # Optional extra word parts\n" +
            "  [-/\\\\]                # separated by -, / or //\n" +
            "  [A-Za-z0-9]+            # Match extra word part.\n" +
            ")*                        # Zero or more extra word parts.\n" +
            "\\b                       # Start and end on a word boundary", 
            Pattern.COMMENTS);
        Matcher regexMatcher = regex.matcher(s);
        while (regexMatcher.find()) {
            System.out.print(regexMatcher.group() + ", ");
        } 
    }
}

Here is the correct output:

A35, 35A, B503X, 1ABC5, AB-10, 10-AB, A10-BA, BA-A10,

Note that the only complex regexes which are "ugly", are those that are not properly formatted and commented!

Upvotes: 6

hsz
hsz

Reputation: 152246

Just use this:

([a-zA-Z]+[-\/\\]?[0-9]+|[0-9]+[-\/\\]?[a-zA-Z]+)

In Java \\ and \/ should be escaped:

([a-zA-Z]+[-\\\/\\\\]?[0-9]+|[0-9]+[-\\\/\\\\]?[a-zA-Z]+)

Upvotes: 1

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