How to truncate the time on a datetime object?

I think this is what you're looking for...

>>> import datetime
>>> dt = datetime.datetime.now()
>>> dt = dt.replace(hour=0, minute=0, second=0, microsecond=0) # Returns a copy
>>> dt
datetime.datetime(2011, 3, 29, 0, 0)

But if you really don't care about the time aspect of things, then you should really only be passing around date objects...

>>> d_truncated = datetime.date(dt.year, dt.month, dt.day)
>>> d_truncated
datetime.date(2011, 3, 29)

Use a date not a datetime if you dont care about the time.

>>> now = datetime.now()
>>> now.date()
datetime.date(2011, 3, 29)

You can update a datetime like this:

>>> now.replace(minute=0, hour=0, second=0, microsecond=0)
datetime.datetime(2011, 3, 29, 0, 0)

Four years later: another way, avoiding replace

I know the accepted answer from four years ago works, but this seems a tad lighter than using replace:

dt = datetime.date.today()
dt = datetime.datetime(dt.year, dt.month, dt.day)

Notes

• When you create a datetime object without passing time properties to the constructor, you get midnight.
• As others have noted, this assumes you want a datetime object for later use with timedeltas.
• You can, of course, substitute this for the first line: dt = datetime.datetime.now()

You cannot truncate a datetime object because it is immutable.

However, here is one way to construct a new datetime with 0 hour, minute, second, and microsecond fields, without throwing away the original date or tzinfo:

newdatetime = now.replace(hour=0, minute=0, second=0, microsecond=0)