I have a Input like

"3-August 202212.00 PM"

Expected output :

"03-08-2022,12:00:00PM"

Any suggestion recommended. I tried to convert to datetime and after strftime.But it not works due to 202212.00 pm.

Final output type will be str not datetime

Thanks in advance

CodePudding user response：

strftime works, you just need to use the correct directives in both directions:

from datetime import datetime

date = datetime.strptime('3-August 202212.00 PM',
                         '%d-%B %Y%I.%M %p')

print(date.strftime('%d-%m-%Y,%I:%M:%S%p'))

outputs

'03-08-2022,12:00:00PM'

CodePudding user response：

You can use datetime to get the result in 2 steps:

1. convert str to datetime format
2. convert datetime object to str in the new format

import datetime as dt
x = "3-August 202212.00 PM"

x = dt.datetime.strptime(x, "%d-%B %Y%I.%M %p").strftime("%d-%m-%Y,%I:%M:%S %p")
print(x)

Output:

03-08-2022,12:00:00 PM

If you have a pd.DataFrame with given strings, you can do as follows: Supposed we have a DataFrame with the following output:

import pandas as pd
y = pd.DataFrame(["3-August 202212.00 PM", "3-March 202104.00 AM", "6-July 202202.00 PM", "23-December 202009.00 AM", "27-May 201806.50 PM"], columns=['Date'])

>>> y
    Date
0   3-August 202212.00 PM
1   3-March 202104.00 AM
2   6-July 202202.00 PM
3   23-December 202009.00 AM
4   27-May 201806.50 PM

Use:

y = pd.to_datetime(y.Date, format='%d-%B %Y%I.%M %p', errors='coerce').apply(lambda x: x.strftime("%d-%m-%Y,%I:%M:%S %p"))
print(type(y))
# pd.to_datetime returns type pandas.core.series.Series so y need to be converted back to dataframe
y = pd.DataFrame(y)

will give a result:

>>> y
    Date
0   03-08-2022,12:00:00 PM
1   03-03-2021,04:00:00 AM
2   06-07-2022,02:00:00 PM
3   23-12-2020,09:00:00 AM
4   27-05-2018,06:50:00 PM

Note: Need to be careful when converting hour. in this case you will want to use %I instead of %H since:

%I : Hour (12-hour clock) as a zero-padded decimal number.

%H : Hour (24-hour clock) as a zero-padded decimal number.