RegExp match repeated characters
Question:
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.
Answers:
You can match that with: (w)1*
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
This will work, see a working example here: http://www.rubular.com/r/ptdPuz0qDV
(w)1*
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=xb2xb2'
>>> 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.
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)]
You can use:
re.sub(r"(w)1*", r'1', 'tessst')
The output would be:
'test'
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.
You can match that with: (w)1*
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
This will work, see a working example here: http://www.rubular.com/r/ptdPuz0qDV
(w)1*
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=xb2xb2'
>>> 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.
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)]
You can use:
re.sub(r"(w)1*", r'1', 'tessst')
The output would be:
'test'