You can use datetime.replace() method -

>>> d = datetime.datetime.today().replace(microsecond=0)
>>> d
datetime.datetime(2015, 7, 18, 9, 50, 20)

If you want to format a datetime object in a specific format that is different from the standard format, it's best to explicitly specify that format:

>>> import datetime
>>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 18:21:26'

See the documentation of datetime.strftime() for an explanation of the % directives.

Starting from Python 3.6, the isoformat() method is flexible enough to also produce this format:

datetime.datetime.now().isoformat(sep=" ", timespec="seconds")

The simplest way would be to use slicing to just chop off the last three digits of the microseconds:

def format_time():
    t = datetime.datetime.now()
    s = t.strftime('%Y-%m-%d %H:%M:%S.%f')
    return s[:-3]

I strongly recommend just chopping. I once wrote some logging code that rounded the timestamps rather than chopping, and I found it actually kind of confusing when the rounding changed the last digit. There was timed code that stopped running at a certain timestamp yet there were log events with that timestamp due to the rounding. Simpler and more predictable to just chop.

If you want to actually round the number rather than just chopping, it's a little more work but not horrible:

def format_time():
    t = datetime.datetime.now()
    s = t.strftime('%Y-%m-%d %H:%M:%S.%f')
    head = s[:-7] # everything up to the '.'
    tail = s[-7:] # the '.' and the 6 digits after it
    f = float(tail)
    temp = "{:.03f}".format(f)  # for Python 2.x: temp = "%.3f" % f
    new_tail = temp[1:] # temp[0] is always '0'; get rid of it
    return head + new_tail

Obviously you can simplify the above with fewer variables; I just wanted it to be very easy to follow.

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)

dtwithoutseconds = dt.replace(second=0, microsecond=0)

http://docs.python.org/library/datetime.html#datetime.datetime.replace

To get a date string with milliseconds, use [:-3] to trim the last three digits of %f (microseconds):

>>> from datetime import datetime
>>> datetime.utcnow().strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]
'2022-09-24 10:18:32.926'

Or shorter:

>>> from datetime import datetime
>>> datetime.utcnow().strftime('%F %T.%f')[:-3]
'2022-09-24 10:18:32.926'

See the Python docs for more "%" format codes and the strftime(3) man page for the full list.

Let's say you have a mix of different formats that looks like this:

import pandas as pd

df = pd.DataFrame()
df['time'] = ['2018-06-01 06:36:40.047883+00:00', '2018-06-01 06:36:40.047883+00:00', '2018-06-04 11:30:00+00:00', '2018-06-01 06:36:40.047883']

Corresponding output:

                               time
0  2018-06-01 06:36:40.047883+00:00
1  2018-06-01 06:36:40.047883+00:00
2         2018-06-04 11:30:00+00:00
3        2018-06-01 06:36:40.047883

You wish to get to a common format by removing microseconds and anything after +. In short, you want something that is in Y-M-D H-M-S format.

Currently, let me assume that your column is in string format. So, we now convert this to a datetime format and then replace the microseconds part with 0 and get rid of it.

df['time'] = pd.to_datetime(df['time'])
df['time'] = df['time'].apply(lambda x: x.replace(microsecond = 0))

Output:

                 time
0 2018-06-01 06:36:40
1 2018-06-01 06:36:40
2 2018-06-04 11:30:00
3 2018-06-01 06:36:40