How to convert datetime object to milliseconds

You can try pd.to_datetime(df['actualDateTime'], unit='ms')

http://pandas.pydata.org/pandas-docs/stable/generated/pandas.to_datetime.html

says this will denote in epoch, with variations 's','ms', 'ns' ...

Update

If you want in epoch timestamp of the form 14567899..

import pandas as pd
import time
t = pd.Timestamp('2015-10-19 07:22:00')
time.mktime(t.timetuple())

>> 1445219520.0

Latest update

df = pd.DataFrame(s1)
df1 = pd.to_datetime(df['Date'])
pd.DatetimeIndex(df1)
>>>DatetimeIndex(['2015-10-20 07:21:00', '2015-10-19 07:18:00',
           '2015-10-19 07:15:00'],
          dtype='datetime64[ns]', freq=None)
df1.astype(np.int64) 
>>>0    1445325660000000000
1    1445239080000000000
2    1445238900000000000
df1.astype(np.int64) // 10**9
>>>0    1445325660
1    1445239080
2    1445238900
Name: Date, dtype: int64

This will return milliseconds from epoch

timestamp_object.timestamp() * 1000

Timestamps in pandas are always in nanoseconds.

This gives you milliseconds since the epoch (1970-01-01):

df['actualDateTime'] = df['actualDateTime'].astype(np.int64) / int(1e6)

pandas.to_datetime is to convert string or few other datatype to pandas datetime[ns]

In your instance initial 'actualDateTime' is not having milliseconds.So, if you are parsing a column which has milliseconds you will get data.

for example,

df
Out[60]: 
                         a  b
0  2015-11-02 18:04:32.926  0
1  2015-11-02 18:04:32.928  1
2  2015-11-02 18:04:32.927  2

df.a
Out[61]: 
0    2015-11-02 18:04:32.926
1    2015-11-02 18:04:32.928
2    2015-11-02 18:04:32.927
Name: a, dtype: object

df.a = pd.to_datetime(df.a)

df.a
Out[63]: 
0   2015-11-02 18:04:32.926
1   2015-11-02 18:04:32.928
2   2015-11-02 18:04:32.927
Name: a, dtype: datetime64[ns]

df.a.dt.nanosecond
Out[64]: 
0    0
1    0
2    0
dtype: int64

df.a.dt.microsecond
Out[65]: 
0    926000
1    928000
2    927000
dtype: int64