How to keep first two duplicates in a pandas dataframe?

Question

I have a question in regards to finding duplicates in a dataframe, and removing duplicates in a dataframe using a specific column. Here is what I am trying to accomplish:

Is it possible to remove duplicates but keep the first 2?

Here is an example of my current dataframe called df and take a look at the bracket notes I have placed below to give you an idea.

Note: If 'Roll' = 1 then I want to look at the Date column, see if there is a second duplicate Date in that column... keep those two and delete any others.

    Date    Open    High     Low      Close  Roll  Dupes
1  19780106  236.00  237.50  234.50  235.50     0    NaN
2  19780113  235.50  239.00  235.00  238.25     0    NaN
3  19780120  238.00  239.00  234.50  237.00     0    NaN
4  19780127  237.00  238.50  235.50  236.00     1    NaN (KEEP)  
5  19780203  236.00  236.00  232.25  233.50     0    NaN (KEEP)
6  19780127  237.00  238.50  235.50  236.00     0    NaN (KEEP)
7  19780203  236.00  236.00  232.25  233.50     0    NaN (DELETE)
8  19780127  237.00  238.50  235.50  236.00     0    NaN (DELETE)
9  19780203  236.00  236.00  232.25  233.50     0    NaN (DELETE)

This is what is currently removing the dupes BUT it's removing all dupes (obviously)

df = df.drop_duplicates('Date')

EDIT: I forgot to mention something, the only duplicate I want to keep is if column 'Roll' = 1 if it does, then keep that row and the next one that matches based on column 'Date'

Answer 1

Using head with a groupby keeps the first x entries in each group, which I think accomplishes what you want.

In [52]: df.groupby('Date').head(2)
Out[52]: 
       Date   Open   High     Low   Close  Roll
1  19780106  236.0  237.5  234.50  235.50     0
2  19780113  235.5  239.0  235.00  238.25     0
3  19780120  238.0  239.0  234.50  237.00     0
4  19780127  237.0  238.5  235.50  236.00     0
5  19780203  236.0  236.0  232.25  233.50     0
6  19780127  237.0  238.5  235.50  236.00     0
7  19780203  236.0  236.0  232.25  233.50     0

Edit:

In [16]: df['dupe_count'] = df.groupby('Date')['Roll'].transform('max') + 1

In [17]: df.groupby('Date', as_index=False).apply(lambda x: x.head(x['dupe_count'].iloc[0]))
Out[17]: 
         Date   Open   High     Low   Close  Roll  Dupes  dupe_count
0 1  19780106  236.0  237.5  234.50  235.50     0    NaN           1
1 2  19780113  235.5  239.0  235.00  238.25     0    NaN           1
2 3  19780120  238.0  239.0  234.50  237.00     0    NaN           1
3 4  19780127  237.0  238.5  235.50  236.00     1    NaN           2
  6  19780127  237.0  238.5  235.50  236.00     0    NaN           2
4 5  19780203  236.0  236.0  232.25  233.50     0    NaN           1

Answer 2

Assuming Roll can only take the values 0 and 1, if you do

df.groupby(['Date', 'Roll'], as_index=False).first() 

you will get two rows for dates for which one of the rows had Roll = 1 and only one row for dates which have only Roll = 0, which I think is what you want.
If passed as_index=False so that the group keys don't end up in the index as discussed in your comment.