Splitting a string with repeated characters into a list

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.

Answers:

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Method 1

Use re.finditer():

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

Method 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']

Method 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']

Method 4

If you don’t want to use any libraries then here’s the code:

s = "AACBCAAB"
L = []
temp = s[0]
for i in range(1,len(s)):
    if s[i] == s[i-1]:
        temp += s[i]
    else:
        L.append(temp)
        temp = s[i]
    if i == len(s)-1:
        L.append(temp)
print(L)

Output:

['AA', 'C', 'B', 'C', 'AA', 'B']

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0