切片后如何重置 MultiIndex [重复]

Question

我经常将数据框特定级别的值作为我应该做什么的指南。在这种情况下，我使用pd.IndexSlice 对数据框进行切片并引用生成的数据框的索引。问题是生成的数据帧的索引与原始索引相同。我需要它是原始索引的子集，它尊重我制作的切片。

设置

import pandas as pd

def produce_df(rows, columns, row_names=None, column_names=None):
    """rows is a list of lists that will be used to build a MultiIndex
    columns is a list of lists that will be used to build a MultiIndex"""
    row_index = pd.MultiIndex.from_product(rows, names=row_names)
    col_index = pd.MultiIndex.from_product(columns, names=column_names)
    return pd.DataFrame(index=row_index, columns=col_index)

df = produce_df([['a', 'b'], ['c', 'd']], [['1', '2'], ['3', '4']],
                row_names=['alpha1', 'alpha2'], column_names=['number1', 'number2'])

print df

number1          1         2     
number2          3    4    3    4
alpha1 alpha2                    
a      c       NaN  NaN  NaN  NaN
       d       NaN  NaN  NaN  NaN
b      c       NaN  NaN  NaN  NaN
       d       NaN  NaN  NaN  NaN

索引如下：

print df.index

MultiIndex(levels=[[u'a', u'b'], [u'c', u'd']],
           labels=[[0, 0, 1, 1], [0, 1, 0, 1]],
           names=[u'alpha1', u'alpha2'])

然后我切片：

islc = pd.IndexSlice[['a'], :]
df2 = df.loc[islc, :]
print df2

number1          1         2     
number2          3    4    3    4
alpha1 alpha2                    
a      c       NaN  NaN  NaN  NaN
       d       NaN  NaN  NaN  NaN

这是预期的切片。索引是什么样的：

MultiIndex(levels=[[u'a', u'b'], [u'c', u'd']],
           labels=[[0, 0], [0, 1]],
           names=[u'alpha1', u'alpha2'])

df.index.levels[0] 里面还有'b'。

问题：如何在切片后重置MultiIndex。

Answer 1

这可行，但很笨拙。我觉得这应该是我不在寻找的地方的一个选择。

df2.index = pd.MultiIndex.from_tuples(df2.index.to_series().values, names=df.index.names)

print df2

number1          1         2     
number2          3    4    3    4
alpha1 alpha2                    
a      c       NaN  NaN  NaN  NaN
       d       NaN  NaN  NaN  NaN

print df2.index

MultiIndex(levels=[[u'a'], [u'c', u'd']],
           labels=[[0, 0], [0, 1]],
           names=[u'alpha1', u'alpha2'])

'b' 已从 df2.index.levels[0] 消失