我不确定我的请求是否可行，但我想这样做：

• 创建一个具有N个多索引和每个多索引的2列（X和Y）的 Dataframe 。然后，我想为每个多索引创建一个Z列，其中是相应多索引的X和Y列之和。
• 所有的操作都必须是矢量化的，我不希望有任何循环或硬编码（比如说，N〉1000时不会扩展）。新的列数应该是多索引的N * 3（x，y，z）。

这能做到吗？
下面是两个使用硬编码和列表理解/循环的代码示例，这不是我想要的。
列表理解：

import pandas as pd
import numpy as np

# Define N as the number of multi indexes
N = 3

# Create a list of tuples for the multi index levels
levels = [('A', 'B', 'C'), ('x', 'y')]

# Use itertools.product to generate all combinations of levels
from itertools import product
columns = list(product(*levels))

# Create a dataframe with random integers from 0 to 9 and 10 rows
df = pd.DataFrame(np.random.randint(0, 10, size=(10, len(columns))), columns=pd.MultiIndex.from_tuples(columns))

# Sum x and y columns for each multi index level using groupby and assign to z column using loop comprehension
df[[(*level, 'z') for level in levels[0]]] = df.groupby(level=0, axis=1).sum()

# Print the dataframe
print(df)

硬编码：

import pandas as pd
import numpy as np

# Define N as the number of multi indexes
N = 3

# Create a list of tuples for the multi index levels
levels = [('A', 'B', 'C'), ('x', 'y')]

# Use itertools.product to generate all combinations of levels
from itertools import product
columns = list(product(*levels))

# Create a dataframe with random integers from 0 to 9 and 10 rows
df = pd.DataFrame(np.random.randint(0, 10, size=(10, len(columns))), columns=pd.MultiIndex.from_tuples(columns))

# Sum x and y columns for each multi index level using groupby and assign to z column
df[('A', 'z')] = df.groupby(level=0, axis=1).sum()[('A')]
df[('B', 'z')] = df.groupby(level=0, axis=1).sum()[('B')]
df[('C', 'z')] = df.groupby(level=0, axis=1).sum()[('C')]

# Print the dataframe
print(df)

输出：

A     B     C      A   B   C
   x  y  x  y  x  y   z   z   z
0  8  5  9  5  9  9  13  14  18
1  7  4  6  6  0  2  11  12   2
2  4  1  5  1  5  8   5   6  13
3  5  3  5  6  2  0   8  11   2
4  4  3  5  9  3  0   7  14   3
5  9  4  8  3  3  4  13  11   7
6  0  5  7  3  6  1   5  10   7
7  2  9  2  8  0  9  11  10   9
8  5  2  7  5  1  9   7  12  10
9  7  3  9  2  5  5  10  11  10