CODEWITHSUNDEEP
pythonregex
Hugo
Hugo

Reputation: 109

Looking for the right RE

Just started with regular expressions... Im looking for a regular expression
something like \b\d\d\b but the digits may not be the same.(e.g 23 should match
but 22 should not) I,ve tried a lot ( involving backreferences ) but they all failed.
I've tried RE's with the code below ( python 2.7.3) but nothing matched so far 

import re
# accept a raw string(e) as input
# and return a function with an argument
# 'string' which returns a re.Match object
# on succes. Else it returns None
def myMatch(e):
    RegexObj= re.compile(e)
    return RegexObj.match

menu= raw_input
expr= "expression\n:>"
Quit= 'q'
NewExpression= 'r'
str2match= "string to match\n:>"
validate= myMatch(menu(expr))
# exits when the user # hits 'q'
while True:                     
    # set the string to match or hit 'q' or 'r'
    option = menu(str2match)
    if option== Quit: break 
    #invokes when the user hits 'r'
    #setting the new expression
    elif option== NewExpression:
        validate= myMatch(menu(expr))
        continue
    reMatchObject= validate(option) 
    # we have a match ! 
    if reMatchObject:           
        print "Pattern: ",reMatchObject.re.pattern
        print "group(0): ",reMatchObject.group()
        print "groups: ",reMatchObject.groups()
    else:
        print "No match found "

Upvotes: 1

Views: 105

Answers (1)

JDB
JDB

Reputation: 25810

You can use backreferencing and a negative lookahead.

\b(\d)(?!\1)\d\b

The backreference is replaced with whatever was matched in the first group: (\d)

A negative lookahead prevents the match from succeeding if the following characters match the expression.

So this basically says match a number (we'll call it "N"). If the next character is N, fail the match. If not, match one more number.

Upvotes: 5

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