Question

我正在使用XLS格式的这种数据形式：

+--------+---------+-------------+---------------+---------+
|   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。熊猫会把它读成单张，但我想在这里使用MultiIndex，比如

+--------+---------+-------------+---------------+---------+
|   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      |
+--------+---------+-------------+---------------+---------+

此处111111和Branch1是1级索引，1 Company A是2级索引。有内置的方法吗？

Answer 1

如果只需要set_index和sort_index，请使用：

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

但如果MultiIndex（我的解决方案中a，b只需要两个级别，则必须先连接第二列，第三列连接第四列：

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

如果需要按前一列汇总最后一列，请将groupby与GroupBy.mean一起使用：

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

如果在set_index中使用MultiIndex列需要tuples：

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

将部分数据帧转换为Pandas中的MultiIndex

1 个答案: