根据0级索引自定义排序多索引Pandas DataFrame的1级索引

Question

我有一个多索引DataFrame， 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)

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 '子级别排序为新的顺序，以创建看起来像df_end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：

             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

我以为我可以重新索引baz行：

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

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

这使：

         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

...并将这些值插回到原始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

这不是我想要的！ 所以我的问题是如何使用new_index重新排序baz索引中的行。 任何建议将不胜感激。

Answer 1

编辑:(以适合所需的布局）

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零件的位置作为切片对象，我们只替换那里的零件。

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