python - 将部分数据框转换为 Pandas 中的 MultiIndex

Question

我有这种 XLS 格式的数据:

+--------+---------+-------------+---------------+---------+
|   ID   |  Branch | Customer ID | Customer Name | Balance |
+--------+---------+-------------+---------------+---------+
| 111111 | Branch1 | 1           | Company A     | 10      |
+--------+---------+-------------+---------------+---------+
| 222222 | Branch2 | 2           | Company B     | 20      |
+--------+---------+-------------+---------------+---------+
| 111111 | Branch1 | 2           | Company B     | 30      |
+--------+---------+-------------+---------------+---------+
| 222222 | Branch2 | 3           | Company C     | 10      |
+--------+---------+-------------+---------------+---------+

我想用 Pandas 来处理它。 Pandas 会将其作为单张读取，但我想在这里使用 MultiIndex，例如

+--------+---------+-------------+---------------+---------+
|   ID   |  Branch | Customer ID | Customer Name | Balance |
+--------+---------+-------------+---------------+---------+
|        |         | 1           | Company A     | 10      |
+ 111111 + Branch1 +-------------+---------------+---------+
|        |         | 2           | Company B     | 30      |
+--------+---------+-------------+---------------+---------+
|        |         | 2           | Company B     | 20      |
+ 222222 + Branch2 +-------------+---------------+---------+
|        |         | 3           | Company C     | 10      |
+--------+---------+-------------+---------------+---------+

这里111111和Branch1是1级索引，1A公司是2级索引。有内置方法可以做到这一点吗？

Answer 1

如果需要的话set_index和 sort_index ，使用:

df.set_index(['ID','Branch', 'Customer ID','Customer Name'], inplace=True)
df.sort_index(inplace=True)
print (df)
                                          Balance
ID     Branch  Customer ID Customer Name         
111111 Branch1 1           Company A           10
               2           Company B           30
222222 Branch2 2           Company B           20
               3           Company C           10

但是，如果 MultiIndex 中只需要两个级别(在我的解决方案中为 a、b)，则有必要首先将第二列和第三列连接起来第四列:

df['a'] = df.ID.astype(str) + '_' + df.Branch
df['b'] = df['Customer ID'].astype(str) + '_' + df['Customer Name']
#delete original columns
df.drop(['ID','Branch', 'Customer ID','Customer Name'], axis=1, inplace=True)

df.set_index(['a','b'], inplace=True)
df.sort_index(inplace=True)
print (df)
                            Balance
a              b                   
111111_Branch1 1_Company A       10
               2_Company B       30
222222_Branch2 2_Company B       20
               3_Company C       10

如果需要通过前一列聚合最后一列，请使用 groupby与 GroupBy.mean :

df = df.groupby(['ID','Branch', 'Customer ID','Customer Name'])['Balance'].mean().to_frame()
print (df)
                                          Balance
ID     Branch  Customer ID Customer Name         
111111 Branch1 1           Company A           10
               2           Company B           30
222222 Branch2 2           Company B           20
               3           Company C           10

<小时/>

如果在列中使用MultiIndex，则需要元组 set_index :

df.columns = pd.MultiIndex.from_arrays([['a'] * 2 + ['b']* 2 + ['c'], df.columns])
print (df)
        a                    b                     c
       ID   Branch Customer ID Customer Name Balance
0  111111  Branch1           1     Company A      10
1  222222  Branch2           2     Company B      20
2  111111  Branch1           2     Company B      30
3  222222  Branch2           3     Company C      10

df.set_index([('a','ID'), ('a','Branch'), 
              ('b','Customer ID'), ('b','Customer Name')], inplace=True)
df.sort_index(inplace=True)
print (df)
                                                              c
                                                        Balance
(a, ID) (a, Branch) (b, Customer ID) (b, Customer Name)        
111111  Branch1     1                Company A               10
                    2                Company B               30
222222  Branch2     2                Company B               20
                    3                Company C               10