Python set datetime hour to be a specific time

Question

I am trying to get the date to be yesterday at 11.30 PM.

Here is my code:

    import datetime
    yesterday = datetime.date.today () - datetime.timedelta (days=1) 
    PERIOD=yesterday.strftime ('%Y-%m-%d') 
    new_period=PERIOD.replace(hour=23, minute=30)
    print new_period

however i am getting this error:

TypeError: replace() takes no keyword arguments

any help would be appreciated.

Answer 1

First, change datetime.date.today() to datetime.datetime.today() so that you can manipulate the time of the day.

Then call replace before turning the time into a string.

So instead of:

PERIOD=yesterday.strftime ('%Y-%m-%d') 
new_period=PERIOD.replace(hour=23, minute=30)

Do this:

new_period=yesterday.replace(hour=23, minute=30).strftime('%Y-%m-%d')
print new_period

Also keep in mind that the string you're converting it to displays no information about the hour or minute. If you're interested in that, add %H for hour and %M for the minute information to your format string.

Answer 2

You can use datetime.combine(date, time, tzinfo=self.tzinfo)

import datetime
yesterday = datetime.date.today () - datetime.timedelta (days=1)
t = datetime.time(hour=23, minute=30)
print(datetime.datetime.combine(yesterday, t))

Answer 3

Is this what you want?

from datetime import datetime
yesterday = datetime(2014, 5, 12, 23, 30)
print yesterday

Edited

from datetime import datetime
import calendar

diff = 60 * 60 * 24

yesterday = datetime(*datetime.fromtimestamp(calendar.timegm(datetime.today().utctimetuple()) - diff).utctimetuple()[:3], hour=23, minute=30)

print yesterday

Answer 4

In one line, here you go

import datetime

new_period=(datetime.datetime.today()- datetime.timedelta (days=1)).replace(hour=23, minute=30)