How to replace regex pattern hh:mm:ss with hh:mm in python

I’m sorry if this is a simple question but I have a csv file with time formats as follows: hh:mm:ss

An extract of the file looks like this:

column_name
00:00:00
01:00:00
02:00:00
03:00:00
...
23:00:00
00:00:00

I have the following regex expression to match all those times
[0-9]{2}[:][0-9]{2}[:][0-9]{2}

My question is how do I get rid of the colon and the seconds (:ss) essentially changing the format from
hh:mm:ss to hh:mm in a python script?

I managed to change all the - to / by using this line of code:

df['column_name'] = df['column_name'].str.replace('-', '/')

I tried using this line:

df['column_name'] = [re.sub(r'[0-9]{2}[:][0-9]{2}[:][0-9]{2}', r'[0-9]{2}[:][0-9]{2}', str(x)) for x in df['column_name']]

But this changed all the times to this [0-9]{2}[:][0-9]{2}

I also tried just using slicing such as [:-3] but I could not get it to work:

df['column_name'] = [re.sub(r'[0-9]{2}[:][0-9]{2}[:][0-9]{2}', [:-3], str(x)) for x in df['column_name']]

Any help would be much appreciated, Thank you

>Solution :

You can slice the string with str:

df['column_name'] = df['column_name'].str[:-3]

Or:

df['column_name'] = df['column_name'].str.rsplit(':', 1).str[0]