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

Answer 1

Here's a pretty intuitive way, it may be a little barbaric looking but it works in one line.

Basically, the logic is get your current date, then subtract all the hours minutes, second and microseconds you have, and then add one day ( that's if you want to round up, otherwise just leave it if you're rounding down.)

   from datetime import date,timedelta

    Rounded_date =  ( the_date - timedelta(hours=date.hour, minutes=date.minute,
    seconds=date.second, microseconds=date.microsecond) ) + timedelta(days=1)