datediff in seconds in pandas
t1 = pd.to_datetime('1/1/2015 01:00')
t2 = pd.to_datetime('1/1/2015 03:30')
print pd.Timedelta(t2 - t1).seconds / 3600.0datediff in seconds in pandas
t1 = pd.to_datetime('1/1/2015 01:00')
t2 = pd.to_datetime('1/1/2015 03:30')
print pd.Timedelta(t2 - t1).seconds / 3600.0between date pandas
df[df.some_date.between(start_date, end_date)]Copyright © 2021 Codeinu
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