Python - Convert datetime column into seconds [duplicate]

Answer 1

Let's assume your DataFrame's name is df.

If you want to create a new column with the seconds you should do the following:

df['newColumn'] = df['Time'].dt.total_seconds()

Answer 2

I think you need total_seconds:

print (df['col'].dt.total_seconds())

Sample:

df = pd.DataFrame({'date1':pd.date_range('2015-01-01', periods=3),
                   'date2':pd.date_range('2015-01-01 02:00:00', periods=3, freq='23H')})

print (df)
       date1               date2
0 2015-01-01 2015-01-01 02:00:00
1 2015-01-02 2015-01-02 01:00:00
2 2015-01-03 2015-01-03 00:00:00

df['diff'] = df['date2'] - df['date1']
df['seconds'] = df['diff'].dt.total_seconds()

print (df)
       date1               date2     diff  seconds
0 2015-01-01 2015-01-01 02:00:00 02:00:00   7200.0
1 2015-01-02 2015-01-02 01:00:00 01:00:00   3600.0
2 2015-01-03 2015-01-03 00:00:00 00:00:00      0.0

---

df['diff'] = df['date2'] - df['date1']
df['diff'] = df['diff'].dt.total_seconds()

print (df)
       date1               date2    diff
0 2015-01-01 2015-01-01 02:00:00  7200.0
1 2015-01-02 2015-01-02 01:00:00  3600.0
2 2015-01-03 2015-01-03 00:00:00     0.0

If need cast to int:

df['diff'] = df['date2'] - df['date1']
df['diff'] = df['diff'].dt.total_seconds().astype(int)

print (df)
       date1               date2  diff
0 2015-01-01 2015-01-01 02:00:00  7200
1 2015-01-02 2015-01-02 01:00:00  3600
2 2015-01-03 2015-01-03 00:00:00     0