Rounding time off to the nearest second - Python

Question

I have a large dataset with more than 500 000 date & time stamps that look like this:

date        time
2017-06-25 00:31:53.993
2017-06-25 00:32:31.224
2017-06-25 00:33:11.223
2017-06-25 00:33:53.876
2017-06-25 00:34:31.219
2017-06-25 00:35:12.634 

How do I round these timestamps off to the nearest second?

My code looks like this:

readcsv = pd.read_csv(filename)
log_date = readcsv.date
log_time = readcsv.time

readcsv['date'] = pd.to_datetime(readcsv['date']).dt.date
readcsv['time'] = pd.to_datetime(readcsv['time']).dt.time
timestamp = [datetime.datetime.combine(log_date[i],log_time[i]) for i in range(len(log_date))]

So now I have combined the dates and times into a list of datetime.datetime objects that looks like this:

datetime.datetime(2017,6,25,00,31,53,993000)
datetime.datetime(2017,6,25,00,32,31,224000)
datetime.datetime(2017,6,25,00,33,11,223000)
datetime.datetime(2017,6,25,00,33,53,876000)
datetime.datetime(2017,6,25,00,34,31,219000)
datetime.datetime(2017,6,25,00,35,12,634000)

Where do I go from here? The df.timestamp.dt.round('1s') function doesn't seem to be working? Also when using .split() I was having issues when the seconds and minutes exceeded 59

Many thanks

Answer 1

Without any extra packages, a datetime object can be rounded to the nearest second with the following simple function:

import datetime as dt

def round_seconds(obj: dt.datetime) -> dt.datetime:
    if obj.microsecond >= 500_000:
        obj += dt.timedelta(seconds=1)
    return obj.replace(microsecond=0)