pandas 数据透视表 DF 的索引计数

Question

我已经给了df：

import pandas as pd
import numpy as np
df = pd.DataFrame({"A": ["foo", "foo", "foo", "foo", "foo",
                         "bar", "bar", "bar", "bar","zz","zz"],
                  "B": ["one", "one", "one", "two", "two",
                         "one", "one", "two", "two","xy","zz"],
                   "Name":["Peter", "Amy", "Brian", "Amy", "Amy",
                         "Peter", "Brian", "Peter", "Brian","Brian","Brian"],
                  "Year": [2019, 2019, 2019, 2019,
                         2019, 2019, 2020, 2020,
                          2020,2019,2020],
                  "Values": [20, 4, 20, 5, 6, 6, 8, 9, 9,10,5]})
df_pivot = pd.pivot_table(df, values='Values', index=['Name','A', 'B'],
                    columns=['Year'], aggfunc=np.sum, fill_value=0, margins=True, 
margins_name="Totals")

print(df_pivot)

在我以我喜欢的方式旋转它之后：

Year            2019  2020  Totals
Name   A   B                      
Amy    foo one     4     0       4
           two    11     0      11
Brian  bar one     0     8       8
           two     0     9       9
       foo one    20     0      20
       zz  xy     10     0      10
           zz      0     5       5
Peter  bar one     6     0       6
           two     0     9       9
       foo one    20     0      20
Totals            71    31     102

现在我想知道从这个 DF 中提取完整的形状以及索引列。

df.shape[0] 只给我数字 3，因为这是“带数据的列”的计数，但是前三列呢？ 有没有办法计算 df 中的索引列？

我知道在这个例子中它是我定义的三个，但我希望自动计算它。

谢谢

Answer 1

如果需要计数级别，请使用MultiIndex.nlevels：

print(df_pivot.index.nlevels)
3
print(df_pivot.columns.nlevels)
1