Convert a column of datetimes to epoch in Python

I know this is old but I believe the correct (and cleanest) way is the single line below:

calls['DATE'].apply(lambda x: x.timestamp())

This assumes calls['DATE'] is a datetime64[ns] type. If not, convert it with:

pd.to_datetime(calls['DATE'], format="%Y-%m-%d %H:%m:00.000")

Explanation

To get the epoch value (in seconds) of a pd.Timestamp, use:

pd.Timestamp('20200101').timestamp()

This should give you 1577836800.0. You can cast to an int if you want. The reason it is a float is because any subsecond time will be in the decimal part.

For completeness, you can also get the raw epoch value (in nanoseconds) using this:

pd.Timestamp('20200101').value

Gives 1577836800000000000 which is the epoch of the date above. The .value attribute is the number of nanoseconds since epoch so we divide by 1e6 to get to milliseconds. Divide by 1e9 if you want epoch in seconds as the first call.

convert the string to a datetime using to_datetime and then subtract datetime 1970-1-1 and call dt.total_seconds():

In [2]:
import pandas as pd
import datetime as dt
df = pd.DataFrame({'date':['2011-04-24 01:30:00.000']})
df

Out[2]:
                      date
0  2011-04-24 01:30:00.000

In [3]:
df['date'] = pd.to_datetime(df['date'])
df

Out[3]:
                 date
0 2011-04-24 01:30:00

In [6]:    
(df['date'] - dt.datetime(1970,1,1)).dt.total_seconds()

Out[6]:
0    1303608600
Name: date, dtype: float64

You can see that converting this value back yields the same time:

In [8]:
pd.to_datetime(1303608600, unit='s')

Out[8]:
Timestamp('2011-04-24 01:30:00')

So you can either add a new column or overwrite:

In [9]:
df['epoch'] = (df['date'] - dt.datetime(1970,1,1)).dt.total_seconds()
df

Out[9]:
                 date       epoch
0 2011-04-24 01:30:00  1303608600

EDIT

better method as suggested by @Jeff:

In [3]:
df['date'].astype('int64')//1e9

Out[3]:
0    1303608600
Name: date, dtype: float64

In [4]:
%timeit (df['date'] - dt.datetime(1970,1,1)).dt.total_seconds()
%timeit df['date'].astype('int64')//1e9

100 loops, best of 3: 1.72 ms per loop
1000 loops, best of 3: 275 µs per loop

You can also see that it is significantly faster