Given a dataframe:
>>> import pandas as pd
>>> lol = [['a', 1, 1], ['b', 1, 2], ['c', 1, 4], ['c', 2, 9], ['b', 2, 10], ['x', 2, 5], ['d', 2, 3], ['e', 3, 5], ['d', 2, 10], ['a', 3, 5]]
>>> df = pd.DataFrame(lol)
>>> df.rename(columns={0:'value', 1:'key', 2:'something'})
value key something
0 a 1 1
1 b 1 2
2 c 1 4
3 c 2 9
4 b 2 10
5 x 2 5
6 d 2 3
7 e 3 5
8 d 2 10
9 a 3 5
The goal is to keep the last N rows for the unique values of the key column.
If N=1, I could simply use the .drop_duplicates() function as such:
>>> df.drop_duplicates(subset='key', keep='last')
value key something
2 c 1 4
8 d 2 10
9 a 3 5
How do I keep the last 3 rows for each unique values of key?
I could try this for N=3:
>>> from itertools import chain
>>> unique_keys = {k:[] for k in df['key']}
>>> for idx, row in df.iterrows():
... k = row['key']
... unique_keys[k].append(list(row))
...
>>>
>>> df = pd.DataFrame(list(chain(*[v[-3:] for k,v in unique_keys.items()])))
>>> df.rename(columns={0:'value', 1:'key', 2:'something'})
value key something
0 a 1 1
1 b 1 2
2 c 1 4
3 x 2 5
4 d 2 3
5 d 2 10
6 e 3 5
7 a 3 5
But there must be a better way...
Is this what you want ?
df.groupby('key').tail(3)
Out[127]:
value key something
0 a 1 1
1 b 1 2
2 c 1 4
5 x 2 5
6 d 2 3
7 e 3 5
8 d 2 10
9 a 3 5
Does this help:
for k,v in df.groupby('key'):
print v[-2:]
value key something
1 b 1 2
2 c 1 4
value key something
6 d 2 3
8 d 2 10
value key something
7 e 3 5
9 a 3 5
来源：https://stackoverflow.com/questions/46781195/keeping-the-last-n-duplicates-in-pandas