熊貓具有多個聚合函數的多個索引

Question

使用此樣本數據幀數據：

+------+--------+------+-------+------+--------+
| NAME |  JOB   | YEAR | MONTH | DAYS | SALARY |
+------+--------+------+-------+------+--------+
| Bob  | Worker | 2013 |    12 |    3 |     17 |
| Mary | Employ | 2013 |    12 |    5 |     23 |
| Bob  | Worker | 2014 |     1 |   10 |    100 |
| Bob  | Worker | 2014 |     1 |   11 |    110 |
| Mary | Employ | 2014 |     1 |   15 |    200 |
| Bob  | Worker | 2014 |     2 |    8 |     80 |
| Mary | Employ | 2014 |     2 |    5 |    190 |
+------+--------+------+-------+------+--------+

是否有一種簡單的方法無需手動創建所有樞軸零件即可獲得這樣的輸出？

index=JOB,MAX(YEAR),NAME,SUM(DAYS)  
columns=MONTH  
values=SUM(SALARY)

                                +-----------+-------------+-------------+
                                |     MONTH |           1 |           2 |
    +--------+-----------+------+-----------+-------------+-------------+
    |  JOB   | MAX(YEAR) | NAME | SUM(DAYS) | SUM(SALARY) | SUM(SALARY) |
    +--------+-----------+------+-----------+-------------+-------------+
    | Employ |      2014 | Mary |        29 |         210 |         190 |
    | Worker |      2014 | Bob  |        20 |         200 |          80 |
    +--------+-----------+------+-----------+-------------+-------------+

Answer 1

從...開始：

In [179]: df
Out[179]: 
   NAME     JOB  YEAR  MONTH  DAYS  SALARY
0   Bob  Worker  2013     12     3      17
1  Mary  Employ  2013     12     5      23
2   Bob  Worker  2014      1    10     100
3   Bob  Worker  2014      1    11     110
4  Mary  Employ  2014      1    15     200
5   Bob  Worker  2014      2     8      80
6  Mary  Employ  2014      2     5     190

我們可以獲得想要使用的大多數數據

result = df.groupby(['JOB', 'NAME', 'MONTH', 'YEAR']).sum().reset_index(['MONTH'])

#                   MONTH  DAYS  SALARY
# JOB    NAME YEAR                     
# Employ Mary 2014      1    15     200
#             2014      2     5     190
#             2013     12     5      23
# Worker Bob  2014      1    21     210
#             2014      2     8      80
#             2013     12     3      17

為此，我們加上天數的總和：

total_days = df.groupby(['JOB', 'NAME', 'YEAR'])[['DAYS']].sum()
total_days.columns = ['SUM(DAYS)']

#                   SUM(DAYS)
# JOB    NAME YEAR           
# Employ Mary 2013          5
#             2014         20
# Worker Bob  2013          3
#             2014         29

result = result.join(total_days)
del result['DAYS']
#                   MONTH  SALARY  SUM(DAYS)
# JOB    NAME YEAR                          
# Employ Mary 2013     12      23          5
#             2014      1     200         20
#             2014      2     190         20
# Worker Bob  2013     12      17          3
#             2014      1     210         29
#             2014      2      80         29

要選擇與max(YEAR)關聯的行，我們計算

max_year = df.groupby(['JOB', 'NAME'])[['YEAR']].max()
max_year = max_year.set_index('YEAR', drop=False, append=True)

#                   YEAR
# JOB    NAME YEAR      
# Employ Mary 2014  2014
# Worker Bob  2014  2014

因此選擇可以表示為左連接：

result = max_year.join(result)
del result['YEAR']

#                   MONTH  SALARY  SUM(DAYS)
# JOB    NAME YEAR                          
# Employ Mary 2014      1     200         20
#             2014      2     190         20
# Worker Bob  2014      1     210         29
#             2014      2      80         29

現在我們可以將MONTH移到這樣的層次列級別：

result = result.set_index(['SUM(DAYS)', 'MONTH'], append=True)
result = result.unstack('MONTH')
result = result.reset_index(['SUM(DAYS)'])

產生

                  SUM(DAYS)  SALARY     
MONTH                             1    2
JOB    NAME YEAR                        
Employ Mary 2014         20     200  190
Worker Bob  2014         29     210   80