python - Splitting a string with repeated characters into a list

20
votes

I am not well experienced with Regex but I have been reading a lot about it. Assume there's a string s = '111234' I want a list with the string split into L = ['111', '2', '3', '4']. My approach was to make a group checking if it's a digit or not and then check for a repetition of the group. Something like this 

L = re.findall('\d[\1+]', s)

I think that \d[\1+] will basically check for either "digit" or "digit +" the same repetitions. I think this might do what I want.

3
pythonregexstring
Do you know if the string will contain only numbers? - thefourtheye
@thefourtheye : No assume that it will contain non-digits as well - Mathews_M_J
I have impression that you were looking for r_e = "(1*)(2*)(3*)(4*)" that gives re.findall(r_e, s)[0] => ('111', '2', '3', '4'). - Grijesh Chauhan
Through list is ordered collection: If you don't need order then you can use r_e = "((?P<o>1+)|(?P<to>2+)|(?P<th>3+)|(?P<f>4+))*" then re.search(r_e, s).group('o', 'to', 'th', 'f') - Grijesh Chauhan

3 Answers

21
votes

Use re.finditer():

>>> s='111234'
>>> [m.group(0) for m in re.finditer(r"(\d)\1*", s)]
['111', '2', '3', '4']
18
votes

If you want to group all the repeated characters, then you can also use itertools.groupby, like this

from itertools import groupby
print ["".join(grp) for num, grp in groupby('111234')]
# ['111', '2', '3', '4']

If you want to make sure that you want only digits, then

print ["".join(grp) for num, grp in groupby('111aaa234') if num.isdigit()]
# ['111', '2', '3', '4']
8
votes

Try this one:

s = '111234'

l = re.findall(r'((.)\2*)', s)
## it this stage i have [('111', '1'), ('2', '2'), ('3', '3'), ('4', '4')] in l

## now I am keeping only the first value from the tuple of each list
lst = [x[0] for x in l]

print lst

output:

['111', '2', '3', '4']