RegExp match repeated characters

Question

For example I have string:

 aacbbbqq

As the result I want to have following matches:

 (aa, c, bbb, qq)  

Answer 1

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)]

Answer 2

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

Answer 3

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.

Answer 4

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

(\w)\1*

Answer 5

You can use:

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

The output would be:

'test'

Answer 6

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