Pandas get topmost n records within each group

Question

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 1

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

Answer 2

Since 0.14.1, you can now do nlargest and nsmallest on a groupby object:

In [23]: df.groupby('id')['value'].nlargest(2)
Out[23]: 
id   
1   2    3
    1    2
2   6    4
    5    3
3   7    1
4   8    1
dtype: int64

There's a slight weirdness that you get the original index in there as well, but this might be really useful depending on what your original index was.

If you're not interested in it, you can do .reset_index(level=1, drop=True) to get rid of it altogether.

(Note: From 0.17.1 you'll be able to do this on a DataFrameGroupBy too but for now it only works with Series and SeriesGroupBy.)

Answer 3

Sometimes sorting the whole data ahead is very time consuming. We can groupby first and doing topk for each group:

g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)

Answer 4

df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))

• Here sort values ascending false gives similar to nlargest and True gives similar to nsmallest.
• The value inside the head is the same as the value we give inside nlargest to get the number of values to display for each group.
• reset_index is optional and not necessary.

Answer 5

This works for duplicated values

If you have duplicated values in top-n values, and want only unique values, you can do like this:

import pandas as pd

ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])

    id first_name last_name department  salary
24  12   Shandler      Bing      Audit  110000
25  14      Jason       Tom      Audit  100000
26  16     Celine    Anston      Audit  100000
27  15    Michale   Jackson      Audit   70000

If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:

(df.groupby('department')['salary']
 .apply(lambda ser: ser.drop_duplicates().nlargest(3))
 .droplevel(level=1)
 .sort_index()
 .reset_index()
)

This gives

department  salary
0   Audit   110000
1   Audit   100000
2   Audit   70000
3   Management  250000
4   Management  200000
5   Management  150000
6   Sales   220000
7   Sales   200000
8   Sales   150000