CODEWITHSUNDEEP
pythonregex
Andrew
Andrew

Reputation: 8470

RegExp match repeated characters

For example I have string: 

 aacbbbqq

As the result I want to have following matches: 

 (aa, c, bbb, qq)

I know that I can write something like this: 

 ([a]+)|([b]+)|([c]+)|...

But I think i's ugly and looking for better solution. I'm looking for regular expression solution, not self-written finite-state machines.

Upvotes: 38

Views: 40743

Answers (8)

Aditya Rama Narayana Vuyyuru
Aditya Rama Narayana Vuyyuru

Reputation: 1

This raw solution may be usefull..

    string = "helllllo worlddd hhiii "
    i = 0
    j = 1
    b = ''
    l = []
    for a in range(len(string)-1):
        if string[i] !=  string[j]:
            j = j+1
            i = j-1
            if b:
                l.append(b)
                b = ''
        elif string[i] == string[j]:
            if j-i == 1:
                b += string[i:j+1]
            else:
                b += string[i]
            j = j+1
    print(l)

output:

['lllll', 'ddd', 'hh', 'iii']

Upvotes: 0

Kate Ishchenko
Kate Ishchenko

Reputation: 21

You can try something like this:

import re

string = 'aacbbbqq'
result = re.findall(r'((\w)\2*?)', string)
output = [x[0] for x in result]

print(output)

Output will be :

['aa', 'c', 'bbb', 'qq']

Upvotes: 2

Wesam Nabki
Wesam Nabki

Reputation: 2614

You can use:

re.sub(r"(\w)\1*", r'\1', 'tessst')

The output would be:

'test'

Upvotes: 4

SwiftsNamesake
SwiftsNamesake

Reputation: 1578

The findall method will work if you capture the back-reference like so:

result = [match[1] + match[0] for match in re.findall(r"(.)(\1*)", string)]

Upvotes: 5

tzot
tzot

Reputation: 95901

Generally

The trick is to match a single char of the range you want, and then make sure you match all repetitions of the same character:

>>> matcher= re.compile(r'(.)\1*')

This matches any single character (.) and then its repetitions (\1*) if any.

For your input string, you can get the desired output as:

>>> [match.group() for match in matcher.finditer('aacbbbqq')]
['aa', 'c', 'bbb', 'qq']

NB: because of the match group, re.findall won't work correctly.

Other ranges

In case you don't want to match any character, change accordingly the . in the regular expression:

>>> matcher= re.compile(r'([a-z])\1*') # only lower case ASCII letters
>>> matcher= re.compile(r'(?i)([a-z])\1*') # only ASCII letters
>>> matcher= re.compile(r'(\w)\1*') # ASCII letters or digits or underscores
>>> matcher= re.compile(r'(?u)(\w)\1*') # against unicode values, any letter or digit known to Unicode, or underscore

Check the latter against u'hello²²' (Python 2.x) or 'hello²²' (Python 3.x):

>>> text= u'hello=\xb2\xb2'
>>> print('\n'.join(match.group() for match in matcher.finditer(text)))
h
e
ll
o
²²

\w against non-Unicode strings / bytearrays might be modified if you first have issued a locale.setlocale call.

Upvotes: 26

Rakesh Sankar
Rakesh Sankar

Reputation: 9415

This will work, see a working example here: http://www.rubular.com/r/ptdPuz0qDV

(\w)\1*

Upvotes: 8

DrTyrsa
DrTyrsa

Reputation: 31951

itertools.groupby is not a RexExp, but it's not self-written either. :-) A quote from python docs:

# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D

Upvotes: 26

Qtax
Qtax

Reputation: 33908

You can match that with: (\w)\1*

Upvotes: 47

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