Splitting a string with repeated characters into a list

Question

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

Answer 1

Use re.finditer():

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

Answer 2

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']

Answer 3

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']