Pandas, convert datetime column to nearest specified 12th hour

Question

I want to be able to round an entire date column to whatever two 12hr times the values are closest to.

For example, if I want the column to be rounded to either 8am or 8pm.

dates = pd.to_datetime(['2022-01-28 15:25:22.456', '2022-01-27'])

Should become this

dates2 = pd.to_datetime(['2022-01-28 20:00:00', '2022-01-27 08:00:00'])
DatetimeIndex(['2022-01-28 20:00:00', '2022-01-27 08:00:00'], dtype='datetime64[ns]', freq=None)

This was the solution I came up with thanks to other answers.

Here's a workable example.

from datetime import timedelta

step = timedelta(minutes=30)
start = datetime(2022,1,1,0,0,0)
end = datetime(2022,1,2,0,0,0)
seconds = (end - start).total_seconds()
array = []
for i in range(0, int(seconds), int(step.total_seconds())):
    array.append(start + timedelta(seconds=i))

dates = pd.Series(pd.to_datetime(array))

Everything before 8am gets rounded up to 8am, everything after is rounded up to 8pm. Anything after 8pm is rounded to 8am the next day.

dates2 = (dates-pd.Timedelta(hours=8)).dt.ceil('12H')+pd.Timedelta(hours=8)
pd.concat([dates,dates2],axis=1)

Answer 1

Use .replace to replace the nearest specific hour.

Change first and second to your preferred hour. In the code below, use 8 and 20.

dates = pd.to_datetime(['2022-01-28 15:25:22.456', '2022-01-27'])

# round time to either 8am or 8pm
def round_time(time, first, second):
    if time.hour < 12:
        return time.replace(hour=first, minute=0, second=0, microsecond=0)
    else:
        return time.replace(hour=second, minute=0, second=0, microsecond=0)

new_date = list()

for i in dates:
    # round time to either 8am or 8pm
    new_date.append(round_time(i, 8, 20))

new_dates = pd.to_datetime(new_date)
new_dates

Output:

DatetimeIndex(['2022-01-28 20:00:00', '2022-01-27 08:00:00'], dtype='datetime64[ns]', freq=None)