[python] How to get the last N rows of a pandas DataFrame?

I have pandas dataframe df1 and df2 (df1 is vanila dataframe, df2 is indexed by 'STK_ID' & 'RPT_Date') :

>>> df1
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

>>> df2
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20060331    3.69   5.975       NaN      5.975   2.591
       20060630    9.14  10.143       NaN     10.143   4.363
       20060930    9.49  13.854       NaN     13.854   5.901
       20061231   15.84  19.262       NaN     19.262   8.407
       20070331   17.00   6.803       NaN      6.803   2.815
       20070630   26.31  12.940       NaN     12.940   5.418
       20070930   39.12  19.977       NaN     19.977   8.452
       20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

I can get the last 3 rows of df2 by:

>>> df2.ix[-3:]
                 TClose   sales  discount  net_sales    cogs
STK_ID RPT_Date                                             
000568 20071231   45.94  29.269       NaN     29.269  12.606
       20080331   38.75  12.668       NaN     12.668   3.958
       20080630   30.09  21.102       NaN     21.102   7.431

while df1.ix[-3:] give all the rows:

>>> df1.ix[-3:]
    STK_ID  RPT_Date  TClose   sales  discount
0   000568  20060331    3.69   5.975       NaN
1   000568  20060630    9.14  10.143       NaN
2   000568  20060930    9.49  13.854       NaN
3   000568  20061231   15.84  19.262       NaN
4   000568  20070331   17.00   6.803       NaN
5   000568  20070630   26.31  12.940       NaN
6   000568  20070930   39.12  19.977       NaN
7   000568  20071231   45.94  29.269       NaN
8   000568  20080331   38.75  12.668       NaN
9   000568  20080630   30.09  21.102       NaN
10  000568  20080930   26.00  30.769       NaN

Why ? How to get the last 3 rows of df1 (dataframe without index) ? Pandas 0.10.1

This is because of using integer indices (ix selects those by label over -3 rather than position, and this is by design: see integer indexing in pandas "gotchas"*).

*In newer versions of pandas prefer loc or iloc to remove the ambiguity of ix as position or label:

df.iloc[-3:]

see the docs.

As Wes points out, in this specific case you should just use tail!

How to get the last N rows of a pandas DataFrame?

If you are slicing by position, __getitem__ (i.e., slicing with[]) works well, and is the most succinct solution I've found for this problem.

pd.__version__
# '0.24.2'

df = pd.DataFrame({'A': list('aaabbbbc'), 'B': np.arange(1, 9)})
df

   A  B
0  a  1
1  a  2
2  a  3
3  b  4
4  b  5
5  b  6
6  b  7
7  c  8

df[-3:]

   A  B
5  b  6
6  b  7
7  c  8

This is the same as calling df.iloc[-3:], for instance (iloc internally delegates to __getitem__).

As an aside, if you want to find the last N rows for each group, use groupby and GroupBy.tail:

df.groupby('A').tail(2)

   A  B
1  a  2
2  a  3
5  b  6
6  b  7
7  c  8