python - 在 MultiIndex 中设置级别值

Question

如何设置系列的级别值，是使用字​​典替换值，还是仅使用与系列一样长的值列表？

这是一个示例 DataFrame:

     sector from_country to_country           0
0  Textiles          FRA        AUS   47.502096
1  Textiles          FRA        USA  431.890710
2  Textiles          GBR        AUS   83.500590
3  Textiles          GBR        USA  324.836158
4      Wood          FRA        AUS   27.515607
5      Wood          FRA        USA  276.501148
6      Wood          GBR        AUS    1.406096
7      Wood          GBR        USA    8.996177

现在设置索引:

df = df.set_index(['sector', 'from_country', 'to_country']).squeeze()

例如，如果我想根据以下键/值对进行更改:

In [69]: replace_dict = {'FRA':'France', 'GBR':'UK'}
In [70]: new_vals = [replace_dict[x] for x in df.index.get_level_values('from_country')]

我希望输出看起来像:

In [68]: df.index.set_level_values(new_vals, level='from_country')
Out[68]: 
sector    from_country  to_country
Textiles  France        AUS            47.502096
                        USA           431.890710
          UK            AUS            83.500590
                        USA           324.836158
Wood      France        AUS            27.515607
                        USA           276.501148
          UK            AUS             1.406096
                        USA             8.996177

我目前正在这样做，但对我来说这似乎很愚蠢:

def set_index_values(df_or_series, new_values, level):
    """
    Replace the MultiIndex level `level` with `new_values`

    `new_values` must be the same length as `df_or_series`
    """
    levels = df_or_series.index.names
    retval = df_or_series.reset_index(level)
    retval[level] = new_values
    retval = retval.set_index(level, append=True).reorder_levels(levels).sortlevel().squeeze()
    return retval

Answer 1

有点hacky，但你可以用.index.set_levels来做到这一点:

In [11]: df1.index.levels[1]
Out[11]: Index(['FRA', 'GBR'], dtype='object', name='from_country')

In [12]: df1.index.levels[1].map(replace_dict.get)
Out[12]: array(['France', 'UK'], dtype=object)

In [13]: df1.index = df1.index.set_levels(df1.index.levels[1].map(replace_dict.get), "from_country")

In [14]: df1
Out[14]:
sector    from_country  to_country
Textiles  France        AUS            47.502096
                        USA           431.890710
          UK            AUS            83.500590
                        USA           324.836158
Wood      France        AUS            27.515607
                        USA           276.501148
          UK            AUS             1.406096
                        USA             8.996177
Name: 0, dtype: float64

注意:有一种从名称中获取级别编号的方法，但我不记得了。