How can I convert Zulu time zone string into specific datetime object?

I have a date in string format with zulu time zone. I tried to get rid of the "Z" character with regular expression, but I guess there is a more efficient way.

input:

   |index | date                | municipality  
   |------| --------------------|--------------
   | 0    | 07.02.2021 1017Z    | Algier    
   | 1    | 11.01.2019 1716Z    | Abuja     
   | 2    | 23.02.2018 1002Z    | Brüssel   
   | 3    | 19.07.2021 1459Z    | Brüssel   
   | 4    | 26.11.2019 1049Z    | Berlin    

desired outcome:

   |index | date                | municipality  
   |------| --------------------|--------------
   | 0    | 2021-02-17          | Algier    
   | 1    | 2019-01-11          | Abuja     
   | 2    | 2018-02-23          | Bruxelles     
   | 3    | 2021-07-19          | Bruxelles     
   | 4    | 2019-11-26          | Berlin    

>Solution :

Instead of getting rid of the Z character, parse it correctly. EX:

import pandas as pd
df = pd.DataFrame({'date': ['07.02.2021 1017Z', '11.01.2019 1716Z']})
df['date'] = pd.to_datetime(df['date'], format='%d.%m.%Y %H%M%z')

# df['date']
# Out[19]: 
# 0   2021-02-07 10:17:00+00:00
# 1   2019-01-11 17:16:00+00:00
# Name: date, dtype: datetime64[ns, UTC]

Note that setting the format keyword is optional, but it helps for general reliability to specify it explicitly.

You can also floor the hours if you don’t want them:

df['date'] = df['date'].dt.floor('D')

# df['date']
# Out[21]: 
# 0   2021-02-07 00:00:00+00:00
# 1   2019-01-11 00:00:00+00:00
# Name: date, dtype: datetime64[ns, UTC]

…or format to string:

df['date'].dt.strftime('%Y-%m-%d')
# 0    2021-02-07
# 1    2019-01-11
# Name: date, dtype: object