Pandas get topmost n records within each group

Suppose I have pandas DataFrame like this:

df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4], 'value':[1,2,3,1,2,3,4,1,1]})

which looks like:

   id  value
0   1      1
1   1      2
2   1      3
3   2      1
4   2      2
5   2      3
6   2      4
7   3      1
8   4      1

I want to get a new DataFrame with top 2 records for each id, like this:

   id  value
0   1      1
1   1      2
3   2      1
4   2      2
7   3      1
8   4      1

I can do it with numbering records within group after groupby:

dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index()

which looks like:

   id  level_1  index  value
0   1        0      0      1
1   1        1      1      2
2   1        2      2      3
3   2        0      3      1
4   2        1      4      2
5   2        2      5      3
6   2        3      6      4
7   3        0      7      1
8   4        0      8      1

then for the desired output:

dfN[dfN['level_1'] <= 1][['id', 'value']]

Output:

   id  value
0   1      1
1   1      2
3   2      1
4   2      2
7   3      1
8   4      1

But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).

Answer

Did you try

df.groupby('id').head(2)

Output generated:

       id  value
id             
1  0   1      1
   1   1      2 
2  3   2      1
   4   2      2
3  7   3      1
4  8   4      1

(Keep in mind that you might need to order/sort before, depending on your data)

EDIT: As mentioned by the questioner, use

df.groupby('id').head(2).reset_index(drop=True)

to remove the MultiIndex and flatten the results:

    id  value
0   1      1
1   1      2
2   2      1
3   2      2
4   3      1
5   4      1