Round down datetime to previous hour

from datetime import datetime, timedelta

n = datetime.now() - timedelta(hours=1)
new_date = datetime(year=n.year, month=n.month, day=n.day, hour=n.hour)

from datetime import datetime
import pandas as pd

currTime = datetime.now()

floored = pd.to_datetime(currTime).floor('H').to_pydatetime()

Given you want to round down to the hour, you can simply replace microsecond, second and minute with zeros:

print(datetime.now().replace(microsecond=0, second=0, minute=0))

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If you want to round down to the previous hour (as stated in the example 2017-01-11 13:26:12.0 to 2017-01-11 12:00:00.0), replace microsecond, second and minute with zeros, then subtract one hour:

from datetime import datetime, timedelta

print(datetime.now().replace(microsecond=0, second=0, minute=0) - timedelta(hours=1))

Example in the shell:

$ python3
Python 3.5.2 (default, Nov 17 2016, 17:05:23) 
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from datetime import datetime, timedelta
>>> print(datetime.now().replace(microsecond=0, second=0, minute=0) - timedelta(hours=1))
2017-01-11 16:00:00