Rounding datetime to the nearest hour - python-3.x

Question

I have a question very similar to this one and this one but I'm stuck on some rounding issue.
I have a time series from a netCDF file and I'm trying to convert them to a datetime format. The format of the time series is in 'days since 1990-01-01 00:00:00'. Eventually I want output in the format .strftime('%Y%m%d.%H%M'). So for example I read my netCDF file as follows
import netCDF4
nc = netCDF4.Dataset(file_name)
time = np.array(nc['time'][:])
I then have
In [180]: time[0]
Out[180]: 365
In [181]: time[1]
Out[181]: 365.04166666651145
I then did
In [182]: start = datetime.datetime(1990,1,1)
In [183]: delta = datetime.timedelta(time[1])
In [184]: new_time = start + delta
In [185]: print(new_time.strftime('%Y%m%d.%H%M'))
19910101.0059
Is there a a way to "round" to the nearest hour so I get 19910101.0100?

Answer 1

You can round down with datetime.replace(), and round up by adding an hour to the rounded down value using datetime.timedelta(hours=1).
import datetime
def round_to_hour(dt):
dt_start_of_hour = dt.replace(minute=0, second=0, microsecond=0)
dt_half_hour = dt.replace(minute=30, second=0, microsecond=0)
if dt >= dt_half_hour:
# round up
dt = dt_start_of_hour + datetime.timedelta(hours=1)
else:
# round down
dt = dt_start_of_hour
return dt
Note that since we're using replace the values we're not replacing (like the timezone - tzinfo) will be preserved.

Answer 2

I don't think datetime provides a way to round times, you'll have to provide the code to do that yourself. Something like this should work:
def round_to_hour(dt):
round_delta = 60 * 30
round_timestamp = datetime.datetime.fromtimestamp(dt.timestamp() + round_delta)
round_dt = datetime.datetime.fromtimestamp(round_timestamp)
return round_dt.replace(microsecond=0, second=0, minute=0)