Replace date with NaT in Pandas dataframe

In [6]:
import pandas as pd
df = pd.DataFrame({'date':[pd.datetime(1999,9,9,23,59,59), pd.datetime(2014,1,1)]* 10})
df
Out[6]:
                  date
0  1999-09-09 23:59:59
1  2014-01-01 00:00:00
2  1999-09-09 23:59:59
3  2014-01-01 00:00:00
4  1999-09-09 23:59:59
5  2014-01-01 00:00:00
6  1999-09-09 23:59:59
7  2014-01-01 00:00:00
8  1999-09-09 23:59:59
9  2014-01-01 00:00:00
10 1999-09-09 23:59:59
11 2014-01-01 00:00:00
12 1999-09-09 23:59:59
13 2014-01-01 00:00:00
14 1999-09-09 23:59:59
15 2014-01-01 00:00:00
16 1999-09-09 23:59:59
17 2014-01-01 00:00:00
18 1999-09-09 23:59:59
19 2014-01-01 00:00:00
In [9]:

import numpy as np
df.loc[df['date'] == '1999-09-09 23:59:59 ', 'date'] = pd.NaT
df
Out[9]:
         date
0         NaT
1  2014-01-01
2         NaT
3  2014-01-01
4         NaT
5  2014-01-01
6         NaT
7  2014-01-01
8         NaT
9  2014-01-01
10        NaT
11 2014-01-01
12        NaT
13 2014-01-01
14        NaT
15 2014-01-01
16        NaT
17 2014-01-01
18        NaT
19 2014-01-01

To answer your second question most pandas functions handle NaN's appropriately, you can always just drop them:

In [10]:

df.dropna()
Out[10]:
         date
1  2014-01-01
3  2014-01-01
5  2014-01-01
7  2014-01-01
9  2014-01-01
11 2014-01-01
13 2014-01-01
15 2014-01-01
17 2014-01-01
19 2014-01-01

and perform the operation just on these rows

There are some operations, especially between columns, that do not disconsider NaNs or NaTs. That is why you are getting NaTs as a result. If you want to disconsider the 1999-09-09 23:59:59 and also have a subtractable column, try to convert to NaTs and then swap the NaTs with zeros (.fillna(0)), so that, when subtracted, it will keep the value from the other column.