Python Dataframe find difference between datetime rows and convert to seconds

Question

I have a data frame with the DateTime index. I want to find the difference between row datetimes and convert it into seconds. My code:

import datetime
df = 
index                   Col_A   datetime
2019-10-01 08:10:53  400.5334 2019-10-01 08:10:53
2019-10-01 08:10:57  401.2720 2019-10-01 08:10:57
2019-10-01 08:10:59  401.7874 2019-10-01 08:10:59
2019-10-01 08:11:02  402.1579 2019-10-01 08:11:02
2019-10-01 08:11:06  402.6045 2019-10-01 08:11:06

df['timedif'] = df['datetime'].diff()
df['timedif1'] = df['timedif'].total_seconds()

Present output:

print(df['timedif'])
2019-10-01 08:10:53               NaT
2019-10-01 08:10:57   0 days 00:00:04
2019-10-01 08:10:59   0 days 00:00:02
2019-10-01 08:11:02   0 days 00:00:03
2019-10-01 08:11:06   0 days 00:00:04
Name: datetime, Length: 1604993, dtype: timedelta64[ns]

AttributeError: 'Series' object has no attribute 'total_seconds'

How do I convert the timedif column into total seconds?

Z Li · Accepted Answer · 2021-01-15 21:37:53Z

1

Simply do:

df['timedif1'] = df['timedif'] / np.timedelta64(1, 's')

Example:

df = pd.DataFrame({'datetime':[datetime.datetime(2019,10,1,8,10,53), datetime.datetime(2019,10,1,8,10,57),
                              datetime.datetime(2019,10,2,8,10,57)]})
df['timedif'] = df['datetime'].diff()
df['timedif1'] = df['timedif'] / np.timedelta64(1, 's')

    datetime            timedif         timedif1
0   2019-10-01 08:10:53 NaT             NaN
1   2019-10-01 08:10:57 0 days 00:00:04 4.0
2   2019-10-02 08:10:57 1 days 00:00:00 86400.0

answered Jan 15, 2021 at 21:37

Z Li

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