Pandas替代SQL Case when语句

Question

我是熊猫的新手，并且从sql迁移而来。 我有一个问题，我试图用熊猫替换sql-case语句。从高层次上讲，我有一个输入数据框和一个引用表。 我基于引用创建计算列。 表格示例输入数据------------ + ----------- + ---- + ------------ + --- -+ ------ + |

 STUDENT_ID | UG_MAJOR  | C1 |     C2     | C3  |  C4  |
+------------+-----------+----+------------+-----+------+
|        123 | MATH      | A  | 8000-10000 | 12% | 9000 |
|        234 | ALL_OTHER | B  | 1500-2000  | 10% | 1500 |
|        345 | ALL_OTHER | A  | 2800-3000  | 8%  | 2300 |
|        456 | ALL_OTHER | A  | 8000-10000 | 12% | 3200 |
|        980 | ALL_OTHER | C  | 1000-2500  | 15% | 2700 |
+------------+-----------+----+------------+-----+------+

参考数据

---------+---------+---------+
| REF_COL | REF_VAL | REF_SCR |
+---------+---------+---------+
| C1      | A       |      10 |
| C1      | B       |      20 |
| C1      | C       |      30 |
| C1      | NULL    |       0 |
| C1      | MISSING |       0 |
| C1      | A       |      20 |
| C1      | B       |      30 |
| C1      | C       |      40 |
| C1      | NULL    |      10 |
| C1      | MISSING |      10 |
| C2      | <1000   |       0 |
| C2      | >1000   |      20 |
| C2      | >7000   |      30 |
| C2      | >9500   |      40 |
| C2      | MISSING |       0 |
| C2      | NULL    |       0 |
| C3      | <3%     |       5 |
| C3      | >3%     |      10 |
| C3      | >5%     |     100 |
| C3      | >7%     |     200 |
| C3      | >10%    |     300 |
| C3      | NULL    |       0 |
| C3      | MISSING |       0 |
| C4      | <5000   |      10 |
| C4      | >5000   |      20 |
| C4      | >10000  |      30 |
| C4      | >15000  |      40 |
+---------+---------+---------+

预期产量

----------+-----------+----+------------+-----+------+--------+--------+--------+---------+
| Req.Output |           |    |            |     |      |        |        |        |         |
+------------+-----------+----+------------+-----+------+--------+--------+--------+---------+
| STUDENT_ID | UG_MAJOR  | C1 | C2         | C3  | C4   | C1_SCR | C2_SCR | C3_SCR | TOT_SCR |
| 123        | MATH      | A  | 8000-10000 | 12% | 9000 |        |        |        |         |
| 234        | ALL_OTHER | B  | 1500-2000  | 10% | 1500 |        |        |        |         |
| 345        | ALL_OTHER | A  | 2800-3000  | 8%  | 2300 |        |        |        |         |
| 456        | ALL_OTHER | A  | 8000-10000 | 12% | 3200 |        |        |        |         |
| 980        | ALL_OTHER | C  | 1000-2500  | 15% | 2700 |        |        |        |         |
+------------+-----------+----+------------+-----+------+--------+--------+--------+---------+

传统的SQL方式是

select student_id, 
UG_MAJOR, 
C1,
case 
when UG_MAJOR ='MATH' AND when C1 IS NULL THEN 0
when UG_MAJOR ='MATH' AND when C1 ='MISSING' THEN 0
when UG_MAJOR ='MATH' AND when C1 ='A' THEN 10
when UG_MAJOR ='MATH' AND when C1 ='B' THEN 20
when UG_MAJOR ='MATH' AND when C1 ='C' THEN 30

when UG_MAJOR ='ALL_OTHER' AND when C1 IS NULL THEN 0
when UG_MAJOR ='ALL_OTHER' AND when C1 ='MISSING' THEN 0
when UG_MAJOR ='ALL_OTHER' AND when C1 ='A' THEN 20
when UG_MAJOR ='ALL_OTHER' AND when C1 ='B' THEN 30
when UG_MAJOR ='ALL_OTHER' AND when C1 ='C' THEN 40

ELSE 'TBD' END AS C1_SCR,

C2,
CASE 
WHEN C2 IS NULL THEN 0
WHEN C2 ='Missing' OR C2 = . THEN 0
WHEN C2<=1000 THEN 0
WHEN C2 >1000 AND C2<=7000 THEN 20
WHEN C2 >7000 AND C2<=9500 THEN 30
WHEN C2 >9500 THEN 40
ELSE 'TBD' 
END AS C2_SCR

FROM REF_INPUT
GROUP BY 1,2,3,4,5,6

我想知道是否有一种优雅的方式来处理大熊猫？ 感谢Par

Answer 1

正如上面的一些评论中所提到的，由于没有提供一些东西，因此我在创建此解决方案时做了一些假设，但这是我尝试返回所请求的df的尝试，即使它不是很优雅...

dfc = df.copy()
dfc['c1_scr'] = 'TBD'
dfc = dfc.loc[((dfc.ug_major=='MATH')&(dfc.c1.isnull()))
              |((dfc.ug_major=='MATH')&(dfc.c1=='Missing'))
              |((dfc.ug_major=='ALL_OTHER')&(dfc.c1=='Missing'))
              |((dfc.ug_major=='MATH')&(dfc.c1.isnull())),
              'c1_scr'] = 0
dfc = dfc.loc[((dfc.ug_major=='MATH')&(dfc.c1=='A')),'c1_scr'] = 10
dfc = dfc.loc[((dfc.ug_major=='MATH')&(dfc.c1=='B'))
              |((dfc.ug_major=='ALL_OTHER')&(dfc.c1=='A'))
              ,'c1_scr'] = 20
dfc = dfc.loc[((dfc.ug_major=='MATH')&(dfc.c1=='C'))
              |((dfc.ug_major=='ALL_OTHER')&(dfc.c1=='A'))
              ,'c1_scr'] = 30
dfc = dfc.loc[((dfc.ug_major=='ALL_OTHER')&(dfc.c1=='C')),'c1_scr'] = 40
dfc['c2_scr'] = 'TBD'
dfc = dfc.loc[(dfc.c2.isnull())
              |(dfc.c2=='MISSING')
              |(dfc.c2=='.')
              |(dfc.c2<=1000)
              ,'c2_scr'] = 0
dfc = dfc.loc[(dfc.c2>1000)
              &(dfc.c2<=7000)
              ,'c2_scr'] = 20
dfc = dfc.loc[(dfc.c2>7000)
              &(dfc.c2<=9500)
              ,'c2_scr'] = 30
dfc = dfc.loc[(dfc.c2>9500),'c2_scr'] = 40
dfc = dfc[['student_id','ug_major','c1','c1_scr'
           ,'c2','c2_scr']
         ].groupby(['student_id','ug_major','c1','c1_scr','c2','c2_scr'])
print(dfc.head())