python - Rounding datetime based on time of day

Question

I have a pandas dataframe with timestamps shown below:

6/30/2019 3:45:00 PM

I would like to round the date based on time. Anything before 6AM will be counted as the day before.

6/30/2019  5:45:00 AM -> 6/29/2019
6/30/2019  6:30:00 AM -> 6/30/2019

What I have considered doing is splitting date and time into 2 different columns then using an if statement to shift the date (if time >= 06:00 etc). Just wondering there is a built in function in pandas to do this. Ive seen posts of people rounding up and down based on the closest hour but never a specific time threshold (6AM).

Thank you for the help!

Answer 1

there could be a better way to do this.. But this is one way of doing it.

import pandas as pd    

def checkDates(d):
    if d.time().hour < 6:
        return d - pd.Timedelta(days=1)
    else:
        return d

ls = ["12/31/2019  3:45:00 AM", "6/30/2019  9:45:00 PM", "6/30/2019  10:45:00 PM", "1/1/2019  4:45:00 AM"]
df = pd.DataFrame(ls, columns=["dates"])
df["dates"] = df["dates"].apply(lambda d: checkDates(pd.to_datetime(d)))
print (df)
                dates
0 2019-12-30 03:45:00
1 2019-06-30 21:45:00
2 2019-06-30 22:45:00
3 2018-12-31 04:45:00

Also note i am not taking into consideration of the time. when giving back the result.. if you just want the date at the end of it you can just get that out of the datetime object doing something like this

print ((pd.to_datetime("12/31/2019 3:45:00 AM")).date()) >>> 2019-12-31

if understand python well and dont want anyone else(in the future) to understand what your are doing one liner to the above is.

df["dates"] = df["dates"].apply(lambda d: pd.to_datetime(d) - pd.Timedelta(days=1) if pd.to_datetime(d).time().hour < 6 else pd.to_datetime(d))