Python 如何获取pandas DataFrame的最后N行？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14663004/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to get the last N rows of a pandas DataFrame?
提问by bigbug
I have pandas dataframe df1and df2(df1 is vanila dataframe, df2 is indexed by 'STK_ID' & 'RPT_Date') :
我有Pandas 数据框df1和df2（df1 是 vanila 数据框，df2 由 'STK_ID' & 'RPT_Date' 索引）：
>>> 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:
我可以通过以下方式获取 df2 的最后 3 行：
>>> 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:
同时df1.ix[-3:]给出所有行：
>>> 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
为什么 ？如何获取df1（没有索引的数据框）的最后 3 行？熊猫 0.10.1
采纳答案by Wes McKinney
Don't forget DataFrame.tail! e.g. df1.tail(10)
不要忘记DataFrame.tail！例如df1.tail(10)
回答by Andy Hayden
This is because of using integer indices (ixselects those by labelover -3 rather than position, and this is by design: see integer indexing in pandas "gotchas"*).
这是因为使用整数索引（通过 -3 上ix的标签而不是position选择那些索引，这是设计使然：请参阅pandas“gotchas”*中的整数索引）。
*In newer versions of pandas prefer loc or iloc to remove the ambiguity of ix as position or label:
*在较新版本的熊猫中，更喜欢使用 loc 或 iloc 来消除 ix 作为位置或标签的歧义：
df.iloc[-3:]
see the docs.
请参阅文档。
As Wes points out, in this specific case you should just use tail!
正如 Wes 指出的那样，在这种特定情况下，您应该只使用 tail！
回答by cs95
How to get the last N rows of a pandas DataFrame?
如何获取pandas DataFrame的最后N行？
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.
如果您按位置切片，__getitem__（即，用 切片[]）效果很好，并且是我为这个问题找到的最简洁的解决方案。
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 (ilocinternally delegates to __getitem__).
例如，这与调用 相同df.iloc[-3:]（iloc内部委托给__getitem__）。
As an aside, if you want to find the last N rows for each group, use groupbyand GroupBy.tail:
顺便说一句，如果您想找到每个组的最后 N 行，请使用groupbyand GroupBy.tail：
df.groupby('A').tail(2)
A B
1 a 2
2 a 3
5 b 6
6 b 7
7 c 8
