Question

我有一个DataFrame，在一些分组后创建了一个MultiIndex：

import numpy as np
import pandas as p
from numpy.random import randn

df = p.DataFrame({
    'A' : ['a1', 'a1', 'a2', 'a3']
  , 'B' : ['b1', 'b2', 'b3', 'b4']
  , 'Vals' : randn(4)
}).groupby(['A', 'B']).sum()

df

Output>            Vals
Output> A  B           
Output> a1 b1 -1.632460
Output>    b2  0.596027
Output> a2 b3 -0.619130
Output> a3 b4 -0.002009

如何将一个级别添加到MultiIndex，以便将其转换为：

Output>                       Vals
Output> FirstLevel A  B           
Output> Foo        a1 b1 -1.632460
Output>               b2  0.596027
Output>            a2 b3 -0.619130
Output>            a3 b4 -0.002009

Answer 1

您可以先将其添加为普通列，然后将其附加到当前索引，这样：

df['Firstlevel'] = 'Foo'
df.set_index('Firstlevel', append=True, inplace=True)

如果需要，可以通过以下方式更改订单：

df.reorder_levels(['Firstlevel', 'A', 'B'])

结果是：

                      Vals
Firstlevel A  B           
Foo        a1 b1  0.871563
              b2  0.494001
           a2 b3 -0.167811
           a3 b4 -1.353409

Answer 2

使用pandas.concat()在一行中执行此操作的好方法：

import pandas as pd

pd.concat([df], keys=['Foo'], names=['Firstlevel'])

这可以推广到许多数据框，请参阅docs。

Answer 3

我用cxrodgers answer做了一个小功能，恕我直言，IMHO是最好的解决方案，因为它完全在索引上工作，而与任何数据帧或序列无关。

我添加了一个修复程序：to_frame()方法将为没有索引级别的索引级别发明新的名称。这样，新索引将具有旧索引中不存在的名称。我添加了一些代码来还原此名称更改。

下面是代码，我已经使用了一段时间，它似乎工作正常。如果您发现任何问题或极端情况，我将不得不调整答案。

import pandas as pd

def _handle_insert_loc(loc: int, n: int) -> int:
    """
    Computes the insert index from the right if loc is negative for a given size of n.
    """
    return n + loc + 1 if loc < 0 else loc


def add_index_level(old_index: pd.Index, value: Any, name: str = None, loc: int = 0) -> pd.MultiIndex:
    """
    Expand a (multi)index by adding a level to it.

    :param old_index: The index to expand
    :param name: The name of the new index level
    :param value: Scalar or list-like, the values of the new index level
    :param loc: Where to insert the level in the index, 0 is at the front, negative values count back from the rear end
    :return: A new multi-index with the new level added
    """
    loc = _handle_insert_loc(loc, len(old_index.names))
    old_index_df = old_index.to_frame()
    old_index_df.insert(loc, name, value)
    new_index_names = list(old_index.names)  # sometimes new index level names are invented when converting to a df,
    new_index_names.insert(loc, name)        # here the original names are reconstructed
    new_index = pd.MultiIndex.from_frame(old_index_df, names=new_index_names)
    return new_index

它通过了以下单元测试代码：

import unittest

import numpy as np
import pandas as pd

class TestPandaStuff(unittest.TestCase):

    def test_add_index_level(self):
        df = pd.DataFrame(data=np.random.normal(size=(6, 3)))
        i1 = add_index_level(df.index, "foo")

        # it does not invent new index names where there are missing
        self.assertEqual([None, None], i1.names)

        # the new level values are added
        self.assertTrue(np.all(i1.get_level_values(0) == "foo"))
        self.assertTrue(np.all(i1.get_level_values(1) == df.index))

        # it does not invent new index names where there are missing
        i2 = add_index_level(i1, ["x", "y"]*3, name="xy", loc=2)
        i3 = add_index_level(i2, ["a", "b", "c"]*2, name="abc", loc=-1)
        self.assertEqual([None, None, "xy", "abc"], i3.names)

        # the new level values are added
        self.assertTrue(np.all(i3.get_level_values(0) == "foo"))
        self.assertTrue(np.all(i3.get_level_values(1) == df.index))
        self.assertTrue(np.all(i3.get_level_values(2) == ["x", "y"]*3))
        self.assertTrue(np.all(i3.get_level_values(3) == ["a", "b", "c"]*2))

        # df.index = i3
        # print()
        # print(df)

Answer 4

我认为这是一个更通用的解决方案：

# Convert index to dataframe
old_idx = df.index.to_frame()

# Insert new level at specified location
old_idx.insert(0, 'new_level_name', new_level_values)

# Convert back to MultiIndex
df.index = pandas.MultiIndex.from_frame(old_idx)

Answer 5

如何使用pandas.MultiIndex.from_tuples从头开始构建它？

df.index = p.MultiIndex.from_tuples(
    [(nl, A, B) for nl, (A, B) in
        zip(['Foo'] * len(df), df.index)],
    names=['FirstLevel', 'A', 'B'])

类似于cxrodger's solution，这是一种灵活的方法，避免了为数据帧修改基础数组。

将级别添加到pandas MultiIndex

5 个答案: