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: 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**](http://codepad.org/7vpTqE1O)

