I have a dataframe like the below and want to remove trailing zeros in pairs of 2.
col1
99990000
11100000
22220000
data = {'col1': ['99990000', '11100000', '22220000']}
df = pd.DataFrame(data=data)
desired result
col1
9999
1110
2222
The below removes all trailing zeros not and not keeping 1110?
df['col1'].str.replace('00+$','')
df['col1'].str.rstrip('00')
Your 00+ applies the repeater + only to the last 0, you need to use a group:
df['col1'].str.replace('(00)+$','')
Output:
0 9999
1 1110
2 2222
Name: col1, dtype: object
Just to conceptualize the regex, here is an example. The solution from @mozway is much better but this focuses on the substitution only, ignoring pandas:
import re
col1 = [re.sub(r'(00)+$', '', num) for num in
['99990000', '11100000', '22220000']]
# ['9999', '1110', '2222']
