How to round a Pandas `DatetimeIndex`?

Question

I have a pandas.DatetimeIndex, e.g.:

pd.date_range(\'2012-1-1 02:03:04.000\',periods=3,freq=\'1ms\')
>>> [2012-01-01 02:03:04, ..., 201         


        

Answer 1

For more general rounding, you can make use of the fact that Pandas Timestamp objects mostly use the standard library datetime.datetime API, including the datetime.datetime.replace() method.

So, to solve your microsecond rounding problem, you could do:

import datetime
import pandas as pd

times = pd.date_range('2012-1-1 02:03:04.499',periods=3,freq='1ms')
# Add 5e5 microseconds and truncate to simulate rounding
times_rounded = [(x + datetime.timedelta(microseconds=5e5)).replace(microsecond=0) for x in times]

from IPython.display import display
print('Before:')
display(list(times))
print('After:')
display(list(times_rounded))

Output:

Before:
[Timestamp('2012-01-01 02:03:04.499000', offset='L'),
 Timestamp('2012-01-01 02:03:04.500000', offset='L'),
 Timestamp('2012-01-01 02:03:04.501000', offset='L')]
After:
[Timestamp('2012-01-01 02:03:04', offset='L'),
 Timestamp('2012-01-01 02:03:05', offset='L'),
 Timestamp('2012-01-01 02:03:05', offset='L')]

You can use the same technique to, e.g., round to the nearest day (as long as you're not concerned about leap seconds and the like):

times = pd.date_range('2012-1-1 08:00:00', periods=3, freq='4H')
times_rounded = [(x + datetime.timedelta(hours=12)).replace(hour=0, second=0, microsecond=0) for x in times]

Inspired by this SO post: https://stackoverflow.com/a/19718411/1410871