使用to_datetime将字符串转换为datetime，然后减去日期时间1970-1-1并调用dt.total_seconds():

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

您可以看到，将此值转换回产生的时间是相同的:

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')

因此您可以添加新列或覆盖:

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

编辑

@Jeff建议的更好的方法:

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

您还可以看到它明显更快

You can also see that it is significantly faster