How to add Dates and Hours in Python

Question

Suppose I have two dates. How can I get all the intermediate dates and hours (from 00 to 23) between the two dates? Thank you very much!

from datetime import datetime,timedelta, date

date_min = (datetime.today()-timedelta(1)).strftime('%Y-%m-%d')+' 00'
date_max =  datetime.today().strftime('%Y-%m-%d')+' 00'

date_min, date_max

('2018-09-16 00', '2018-09-17 00')

how to obtain?

'2018-09-16 01', '2018-09-16 02', '2018-09-16 03', '2018-09-16 04', .... , '2018-09-16 21', '2018-09-16 22', '2018-09-16 23', '2018-09-17 00'

So you're looking for every day between date_min and date_max alongside every hour for each day? It sounds like you'd need a for loop at the least. Are you looking to do this entirely in datetime? Or would a solution using pandas like the one below work? — J0hn
– J0hn, Commented Sep 17, 2018 at 20:45

Pedro · Accepted Answer · 2018-09-17 21:25:08Z

Use pandas, it's a one-liner:

import pandas as pd
solution = pd.date_range(start=date_min , end=date_max, freq='1H')
print(solution)    
DatetimeIndex(['2018-09-16 00:00:00', '2018-09-16 01:00:00',
                   '2018-09-16 02:00:00', '2018-09-16 03:00:00',
                   '2018-09-16 04:00:00', '2018-09-16 05:00:00',
                   '2018-09-16 06:00:00', '2018-09-16 07:00:00',
                   '2018-09-16 08:00:00', '2018-09-16 09:00:00',
                   '2018-09-16 10:00:00', '2018-09-16 11:00:00',
                   '2018-09-16 12:00:00', '2018-09-16 13:00:00',
                   '2018-09-16 14:00:00', '2018-09-16 15:00:00',
                   '2018-09-16 16:00:00', '2018-09-16 17:00:00',
                   '2018-09-16 18:00:00', '2018-09-16 19:00:00',
                   '2018-09-16 20:00:00', '2018-09-16 21:00:00',
                   '2018-09-16 22:00:00', '2018-09-16 23:00:00',
                   '2018-09-17 00:00:00'],
                  dtype='datetime64[ns]', freq='H')

If you want to convert from pandas timestamp back to datetime, do:

[timestamp.to_pydatetime() for timestamp in solution]

[datetime.datetime(2018, 9, 16, 0, 0), datetime.datetime(2018, 9, 16, 1, 0), datetime.datetime(2018, 9, 16, 2, 0), datetime.datetime(2018, 9, 16, 3, 0), datetime.datetime(2018, 9, 16, 4, 0), datetime.datetime(2018, 9, 16, 5, 0), datetime.datetime(2018, 9, 16, 6, 0), datetime.datetime(2018, 9, 16, 7, 0), datetime.datetime(2018, 9, 16, 8, 0), datetime.datetime(2018, 9, 16, 9, 0), datetime.datetime(2018, 9, 16, 10, 0), datetime.datetime(2018, 9, 16, 11, 0), datetime.datetime(2018, 9, 16, 12, 0), datetime.datetime(2018, 9, 16, 13, 0), datetime.datetime(2018, 9, 16, 14, 0), datetime.datetime(2018, 9, 16, 15, 0), datetime.datetime(2018, 9, 16, 16, 0), datetime.datetime(2018, 9, 16, 17, 0), datetime.datetime(2018, 9, 16, 18, 0), datetime.datetime(2018, 9, 16, 19, 0), datetime.datetime(2018, 9, 16, 20, 0), datetime.datetime(2018, 9, 16, 21, 0), datetime.datetime(2018, 9, 16, 22, 0), datetime.datetime(2018, 9, 16, 23, 0), datetime.datetime(2018, 9, 17, 0, 0)]

To convert to the string format you mentioned:

[timestamp.to_pydatetime().strftime('%Y-%m-%d %H') for timestamp in solution]

['2018-09-16 00', '2018-09-16 01', '2018-09-16 02', '2018-09-16 03', '2018-09-16 04', '2018-09-16 05', '2018-09-16 06', '2018-09-16 07', '2018-09-16 08', '2018-09-16 09', '2018-09-16 10', '2018-09-16 11', '2018-09-16 12', '2018-09-16 13', '2018-09-16 14', '2018-09-16 15', '2018-09-16 16', '2018-09-16 17', '2018-09-16 18', '2018-09-16 19', '2018-09-16 20', '2018-09-16 21', '2018-09-16 22', '2018-09-16 23', '2018-09-17 00']

Michael Boesl · Accepted Answer · 2018-09-17 21:19:47Z

3

This will work:

date = datetime.today()-timedelta(days=1)
while date < datetime.today():
    date = date + timedelta(hours=1)
    print date.strftime('%Y-%m-%d %H')

and give you the following Output:

2018-09-17 00
2018-09-17 01
2018-09-17 02
2018-09-17 03
2018-09-17 04
2018-09-17 05
2018-09-17 06
2018-09-17 07
2018-09-17 08

You can also specify from which hour to start:

date = datetime.today().replace(hour=0, minute=0, second=0)-timedelta(days=1)

edited Sep 17, 2018 at 21:19

answered Sep 17, 2018 at 20:54

Michael Boesl

2661 silver badge9 bronze badges

2 Comments

lolo Over a year ago

And what about if I want to pass the variables date_min = (datetime.today()-timedelta(1)).strftime('%Y-%m-%d')+' 00' date_max = datetime.today().strftime('%Y-%m-%d')+' 00'

lolo Over a year ago

while date_min < date_max: date = date + timedelta(hours=1) print(date.strftime('%Y-%m-%d %H'))

Alex · Accepted Answer · 2018-09-17 20:50:02Z

2

Pure Python:

import itertools
from datetime import datetime,timedelta

date_max = datetime.today()
date_min = date_max - timedelta(hours=2)
n_hours = (date_max - date_min).total_seconds() / 3600
[date_min + timedelta(hours=x) for x in itertools.takewhile(lambda h: h <= n_hours, itertools.count())]

returns

[datetime.datetime(2018, 9, 17, 14, 49, 42, 193664),
 datetime.datetime(2018, 9, 17, 15, 49, 42, 193664),
 datetime.datetime(2018, 9, 17, 16, 49, 42, 193664)]

answered Sep 17, 2018 at 20:50

Alex

19.2k9 gold badges65 silver badges82 bronze badges

Collectives™ on Stack Overflow

How to add Dates and Hours in Python

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related