蟒蛇大熊猫计算由

Question

列平均列在大熊猫0.18.1，蟒蛇2.7.6：

假设我们有如下表：

ID,FROM_YEAR,FROM_MONTH,YEARMONTH,AREA,AREA2 
1,2015,1,201501,200,100 
1,2015,2,201502,200,100 
1,2015,3,201503,200,100 
1,2015,4,201504,200,100 
1,2015,5,201505,200,100 
1,2015,6,201506,200,100 
1,2015,7,201507,200,100 
1,2015,8,201508,200,100 
1,2015,9,201509,200,100 
1,2015,10,201510,200,100 
1,2015,11,201511,200,100 
1,2015,12,201512,200,100 
1,2016,1,201601,100,200 
1,2016,2,201602,100,200 
1,2016,3,201603,100,200 
1,2016,4,201604,100,200 
1,2016,5,201605,100,200 
1,2016,6,201606,100,200 
1,2016,7,201607,100,200 
1,2016,8,201608,100,200 
1,2016,9,201609,100,200 
1,2016,10,201610,100,200 
1,2016,11,201611,100,200 
1,2016,12,201612,100,200 

有没有什么办法，我们可以做同样的事情作为在python熊猫中的以下MySQL查询（合并功能可能可以工作，但有什么办法可以避免昂贵的合并/连接在Python熊猫）？

SELECT 
ID, 
FROM_YEAR, 
'A' AS TYPE, 
AVG(AREA) AS AREA, 
AVG(AREA2) AS AREA2 
FROM table GROUP BY ID,FROM_YEAR 

UNION ALL 

SELECT 
ID, 
FROM_YEAR, 
'B' AS TYPE, 
AVG(AREA) AS AREA, 
AVG(AREA2) AS AREA2 
FROM table GROUP BY ID,FROM_YEAR; 

这里的目标是获得在以下格式的历年平均面积和AREA2列：

ID,FROM_YEAR,TYPE,AREA,AREA2 
1,2015,A,200,100 
1,2016,A,100,200 
1,2015,B,200,100 
1,2016,B,100,200 

可以在任何大师指教？

=================================一个扩展问题========== =======

感谢您的回答！我只是遇到一个连续12个案例的另一个问题：

所需的输出：

ID,FROM_YEAR,FROM_MONTH,YEARMONTH,AREA,AREA2 
1,2015,1,201501,NULL,NULL 
1,2015,2,201502,NULL,NULL 
1,2015,3,201503,NULL,NULL 
1,2015,4,201504,NULL,NULL 
1,2015,5,201505,NULL,NULL 
1,2015,6,201506,NULL,NULL 
1,2015,7,201507,NULL,NULL 
1,2015,8,201508,NULL,NULL 
1,2015,9,201509,NULL,NULL 
1,2015,10,201510,NULL,NULL 
1,2015,11,201511,NULL,NULL 
1,2015,12,201512,200,100 

下面的代码

agg=df.groupby(['ID','FROM_YEAR'])[['AREA','AREA2']].rolling(window=12).mean() 

才会产生这样的结果，其中FROM_MONTH和YEARMONTH失踪。

ID,FROM_YEAR,AREA,AREA2 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,NULL,NULL 
1,2015,200,100 

任何人都可以启发？谢谢！

Answer 1

您可以使用pandas.concat这里只涉及一个聚集和不调用merge/join过程：用`assign`和`concat`

agg = df.groupby(['ID', 'FROM_YEAR'], as_index=False)[["AREA", "AREA2"]].mean() 

pd.concat([agg.assign(TYPE = t) for t in ["A", "B"]], ignore_index=True)