如何使用Pandas groupby对某些列进行分组，而不对其他列进行分组？
table_D = pd.DataFrame({
'Geo_ID': [1, 1, 1, 1, 2, 3, 4, 4, 5],
'A_Code': [12, 12, 12, 65, 65, 65, 65, 98, 98],
'A_Cost': [2, 9, 1, 10, 6, 7, 7, 6, 2],
}, columns=['Geo_ID', 'A_Code', 'A_Cost'])
table_D_dummies = pd.get_dummies(data = table_D, columns = ["A_Code"])
table_D_dummies_grouped = table_D_dummies.groupby(by = ["Geo_ID"]).sum()
如下所示，这可以按Geo_ID正确计算费用。不幸的是，它也通过A_Code求和。
A_Code_12，A_Code_65和A_Code_98应该分别组合 。此外，在实际数据集中，有超过100个A_Code。
table_D
+--------+--------+--------+
| Geo_ID | A_Code | A_Cost |
+--------+--------+--------+
| 1 | 12 | 2 |
| 1 | 12 | 9 |
| 1 | 12 | 1 |
| 1 | 65 | 10 |
| 2 | 65 | 6 |
| 3 | 65 | 7 |
| 4 | 65 | 7 |
| 4 | 98 | 6 |
| 5 | 98 | 2 |
+--------+--------+--------+
table_D_dummies
+---+--------+--------+-----------+-----------+-----------+
| | Geo_ID | A_Cost | A_Code_12 | A_Code_65 | A_Code_98 |
+---+--------+--------+-----------+-----------+-----------+
| 0 | 1 | 2 | 1 | 0 | 0 |
| 1 | 1 | 9 | 1 | 0 | 0 |
| 2 | 1 | 1 | 1 | 0 | 0 |
| 3 | 1 | 10 | 0 | 1 | 0 |
| 4 | 2 | 6 | 0 | 1 | 0 |
| 5 | 3 | 7 | 0 | 1 | 0 |
| 6 | 4 | 7 | 0 | 1 | 0 |
| 7 | 4 | 6 | 0 | 0 | 1 |
| 8 | 5 | 2 | 0 | 0 | 1 |
+---+--------+--------+-----------+-----------+-----------+
table_D_dummies_grouped
+--------+--------+-----------+-----------+-----------+
| Geo_ID | A_Cost | A_Code_12 | A_Code_65 | A_Code_98 |
+--------+--------+-----------+-----------+-----------+
| 1 | 22 | 3 | 1 | 0 |
| 2 | 6 | 0 | 1 | 0 |
| 3 | 7 | 0 | 1 | 0 |
| 4 | 13 | 0 | 1 | 1 |
| 5 | 2 | 0 | 0 | 1 |
+--------+--------+-----------+-----------+-----------+
答案 0 :(得分：2)
您没有使用虚拟表，而是对原始数据框进行了分组：
table_D_dummies = pd.get_dummies(data = table_D, columns = ["A_Code"])
table_D_dummies_grouped = table_D.groupby(by = ["Geo_ID"]).sum()
您要在此处将table_D_dummies分组：
>>> table_D_dummies
Geo_ID A_Cost A_Code_12 A_Code_65 A_Code_98
0 1 2 1 0 0
1 1 9 1 0 0
2 1 1 1 0 0
3 1 10 0 1 0
4 2 6 0 1 0
5 3 7 0 1 0
6 4 7 0 1 0
7 4 6 0 0 1
8 5 2 0 0 1
>>> table_D_dummies.groupby(by = ["Geo_ID"]).sum()
A_Cost A_Code_12 A_Code_65 A_Code_98
Geo_ID
1 22 3 1 0
2 6 0 1 0
3 7 0 1 0
4 13 0 1 1
5 2 0 0 1
如果您需要汇总每个假人的成本 ，请将这些成本添加到分组列中：
>>> table_D_dummies.groupby(by = [
... "Geo_ID",
... *(c for c in table_D_dummies.columns if c.startswith('A_Code_'))
... ]).sum()
A_Cost
Geo_ID A_Code_12 A_Code_65 A_Code_98
1 0 1 0 10
1 0 0 12
2 0 1 0 6
3 0 1 0 7
4 0 0 1 6
1 0 7
5 0 0 1 2