Round timestamp to nearest day in Python

Question

In Python 2.7.2 I am getting the seconds since epoch using:

sec_since_epoch = (date_obj - datetime(1970, 1, 1, 0, 0)).total_seconds()

Now I want to round these seconds to the nearest day e.g. if:

datetime.fromtimestamp(sec_since_epoch)

corresponds to datetime(2013, 12, 14, 5, 0, 0)

I want the new timestamp to correspond to datetime(2013, 12, 14, 0, 0, 0)

I know the ugly way of doing it, but is there an elegant way ?

Answer 1

You can use datetime.timetuple() to manipulate with the date. E.g. in this way:

from datetime import datetime


dt = datetime(2013, 12, 14, 5, 0, 0)
dt = datetime(*dt.timetuple()[:3]) # 2013-12-14 00:00:00
print dt.strftime('%s') # 1386997200

DEMO