根据多级索引Pandas DataFrame的级别0索引，自定义排序级别1索引。

Question

我有一个多层次的数据框，名为df:

arrays = [['bar', 'bar', 'baz', 'baz', 'baz', 'baz', 'foo', 'foo'],
          ['one', 'two', 'one', 'two', 'three', 'four', 'one', 'two']]

df = pd.DataFrame(np.ones([8, 4]), index=arrays)

这看起来像：

             0    1    2    3
bar one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
baz one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
    three  1.0  1.0  1.0  1.0
    four   1.0  1.0  1.0  1.0
foo one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0

我现在需要将“baz”子级别按照新顺序排序，以创建类似于“df_end”的东西：

arrays_end = [['bar', 'bar', 'baz', 'baz', 'baz', 'baz', 'foo', 'foo'],
              ['one', 'two', 'two', 'four', 'three', 'one', 'one', 'two']]

df_end = pd.DataFrame(np.ones([8, 4]), index=arrays_end)

看起来像这样：

             0    1    2    3
bar one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
baz two    1.0  1.0  1.0  1.0
    four   1.0  1.0  1.0  1.0
    three  1.0  1.0  1.0  1.0
    one    1.0  1.0  1.0  1.0
foo one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0

我想，我可能可以重新索引 baz 行：

new_index = ['two','four','three','one']

df.loc['baz'].reindex(new_index)

这将导致:

         0    1    2    3
two    1.0  1.0  1.0  1.0
four   1.0  1.0  1.0  1.0
three  1.0  1.0  1.0  1.0
one    1.0  1.0  1.0  1.0

...并将这些值插入回原始的DataFrame中：

df.loc['baz'] = df.loc['baz'].reindex(new_index)

但结果是：

             0    1    2    3
bar one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
baz one    NaN  NaN  NaN  NaN
    two    NaN  NaN  NaN  NaN
    three  NaN  NaN  NaN  NaN
    four   NaN  NaN  NaN  NaN
foo one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0

但这不是我要找的！所以我的问题是如何使用new_index来重新排序baz索引中的行。任何建议都将不胜感激。

- tomp

---

这对我来说似乎是更好的解决方案：https://dev59.com/96Dia4cB1Zd3GeqPDGj4 - Pengin

Answer 1

编辑：（以适应所需的布局）

arrays = [['bar', 'bar', 'baz', 'baz', 'baz', 'baz', 'foo', 'foo'],
          ['one', 'two', 'one', 'two', 'three', 'four', 'one', 'two']]

df = pd.DataFrame(np.arange(32).reshape([8, 4]), index=arrays)
new_baz_index = [('baz', i) for i in ['two','four','three','one']]
index = df.index.values.copy()
index[df.index.get_loc('baz')] = new_baz_index
df.reindex(index)

df.index.get_loc('baz')将获取baz部分的位置作为切片对象，我们只需在那里替换该部分。

- Tai

---

谢谢你的非常快速和有帮助的回复，Tai。这太棒了 :) - tomp

@tomp 没问题。很高兴能帮到你。 - Tai

Answer 2

更新 :-)

pd.concat([df[df.index.get_level_values(level=0)!='baz'],df.reindex(list(zip(['baz']*4,['two','four','three','one'])))])
Out[1156]: 
             0    1    2    3
bar one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
foo one    1.0  1.0  1.0  1.0
    two    1.0  1.0  1.0  1.0
baz two    1.0  1.0  1.0  1.0
    four   1.0  1.0  1.0  1.0
    three  1.0  1.0  1.0  1.0
    one    1.0  1.0  1.0  1.0

- BENY

---

嗨，Wen，再次感谢您的关注！您回答中的“df”与我的问题中的输入“df”相匹配，而不是输出。是否有缺少的步骤？ - tomp