How to add 0 to 23 hours as suffix to each date in a datefield using python

Question

I am trying to add hours 0 to 23 to each each date in a datetime field, could someone please help me in achieving this.

I searched online, but was not able to find the right answer

Attached image for the input and output expected, Thanks in advance.

Answer 1

Assuming you have these dates as the index of a pandas df, you would convert to datetime and use hour as frequency. Then you can strftime to the desired output format. Ex:

import pandas as pd

df = pd.DataFrame(index=pd.to_datetime(['01/11/2020', '02/11/2020', '03/11/2020'], dayfirst=True))

new_dti = df.asfreq('H').index

new_dti.strftime('%d/%m/%Y %H:%M')

Index(['01/11/2020 00:00', '01/11/2020 01:00', '01/11/2020 02:00',
       '01/11/2020 03:00', '01/11/2020 04:00', '01/11/2020 05:00',
       '01/11/2020 06:00', '01/11/2020 07:00', '01/11/2020 08:00',
       ...
       '02/11/2020 15:00', '02/11/2020 16:00', '02/11/2020 17:00',
       '02/11/2020 18:00', '02/11/2020 19:00', '02/11/2020 20:00',
       '02/11/2020 21:00', '02/11/2020 22:00', '02/11/2020 23:00',
       '03/11/2020 00:00'],
      dtype='object')

Answer 2

This is fairly easy to achieve using a simple for loop:

from datetime import datetime, date

dates = [date(2020,11,1), date(2020,11,2)]

for dt in dates:
    for hour in range(0, 24):
        result.append(datetime(dt.year, dt.month, dt.day, hour, 0))

print(result)
>>>
[datetime.datetime(2020, 11, 1, 0, 0),
 datetime.datetime(2020, 11, 1, 1, 0),
 datetime.datetime(2020, 11, 1, 2, 0),
 datetime.datetime(2020, 11, 1, 3, 0),
...
...
...
 datetime.datetime(2020, 11, 2, 22, 0),
 datetime.datetime(2020, 11, 2, 23, 0)]