Show first 10 rows of multi-index pandas dataframe

Here is an answer. Maybe there is a better way to do that (with indexing ?), but I thing it works. The principle seems complex but is quite simple:

• Index the DataFrame by year and username.
• Group the DataFrame by year which is the first level (=0) of the index
• Apply two operations on the sub DataFrame obtained by the groupby (one for each year)
  • sort the index by count in ascending order sort_index(by='count')-> the row with more counts will be at the tail of the DataFrame
  • Only keep the last top rows (2 in this case) by using the negative slicing notation ([-top:]). The tail method could also be used (tail(top)) to improve readability.
• Dropping the unnecessary level created for year droplevel(0)

---

# Test data    
df = pd.DataFrame({'year': [2010, 2010, 2010, 2011,2011,2011, 2012, 2012, 2013, 2013, 2014, 2014],
                  'username': ['b','a','a','c','c','d','e','f','g','i','h','j'],
                  'count': [400, 505, 678, 677, 505, 505, 677, 505, 677, 505, 677, 505]})
df = df.set_index(['year','username'])

top = 2
df = df.groupby(level=0).apply(lambda df: df.sort_index(by='count')[-top:])
df.index = df.index.droplevel(0)
df

               count
year username       
2010 a           505
     a           678
2011 d           505
     c           677
2012 f           505
     e           677
2013 i           505
     g           677
2014 j           505
     h           677