Question

问题：

我希望在每个配对的uid和零售商中只选择最新的价格记录。

数据：

import pandas as pd
import numpy as np
data = {"uid":{"0":"123","1":"345","2":"678","3":"123","4":"345","5":"123","6":"678","7":"369","8":"890","9":"678"},"retailer":{"0":"GUY","1":"GUY","2":"GUY","3":"GUY","4":"GUY","5":"GAL","6":"GUY","7":"GAL","8":"GAL","9":"GUY"},"upload date":{"0":"11/17/17","1":"11/17/17","2":"11/16/17","3":"11/16/17","4":"11/16/17","5":"11/17/17","6":"11/17/17","7":"11/17/17","8":"11/17/17","9":"11/15/17"},"price":{"0":12.00,"1":1.23, "2":34.00, "3":69.69, "4":13.72, "5":49.98, "6":98.02, "7":1.02,"8":98.23,"9":12.69}}
df = pd.DataFrame(data=data)
df = df[['uid','retailer','upload date','price']]
df['upload date']=pd.to_datetime(df['upload date'])

解决方案：

idx = df.groupby(['uid','retailer'])['upload date'].max().rename('upload date')
idx.reset_index(inplace=True)
solution = idx.merge(df, how='left', on=['uid','retailer','upload date'])

问题：

我希望能够利用指数来实现我的解决方案。或者，我希望能够使用连接，或者使用保留原始数据帧的索引的函数找到每个配对的最大日期。

加入错误：

idx.set_index(['uid','retailer','upload date']).join(df, on=['uid','retailer','upload date'])

返回：

ValueError: len(left_on) must equal the number of levels in the index of "right"

Answer 1

IIUC，idxmax

df.loc[df.groupby(['uid','retailer'])['upload date'].idxmax()]
Out[168]: 
   uid retailer upload date  price
5  123      GAL  2017-11-17  49.98
0  123      GUY  2017-11-17  12.00
1  345      GUY  2017-11-17   1.23
7  369      GAL  2017-11-17   1.02
6  678      GUY  2017-11-17  98.02
8  890      GAL  2017-11-17  98.23

或reindex

df.reindex(df.groupby(['uid','retailer'])['upload date'].idxmax().values)

如果您希望文档<{1}} ：在关键列的索引或上加入与其他DataFrame相关的列

join

要获得预期的输出，您需要在最后添加idx.set_index(['uid','retailer','upload date']).join(df.set_index(['uid','retailer','upload date'])) Out[175]: price uid retailer upload date 123 GAL 2017-11-17 49.98 GUY 2017-11-17 12.00 345 GUY 2017-11-17 1.23 369 GAL 2017-11-17 1.02 678 GUY 2017-11-17 98.02 890 GAL 2017-11-17 98.23

或做类似

的事情

.reset_index()

PANDAS选择包含最近配对观察的行

1 个答案: