Pandas：将多索引级别作为系列
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Pandas: get multiindex level as series
提问by A User
I have a dataframe with multiple levels, eg:
我有一个具有多个级别的数据框，例如：
idx = pd.MultiIndex.from_product((['foo', 'bar'], ['one', 'five', 'three' 'four']),
names=['first', 'second'])
df = pd.DataFrame({'A': [np.nan, 12, np.nan, 11, 16, 12, 11, np.nan]}, index=idx).dropna().astype(int)
A
first second
foo five 12
four 11
bar one 16
five 12
three 11
I want to create a new column using the index level titled second, so that I get
我想使用标题为 的索引级别创建一个新列second，以便我得到
A B
first second
foo five 12 five
four 11 four
bar one 16 one
five 12 five
three 11 three
I can do this by resetting the index, copying the column, then re-applying, but that seems more round-about.
我可以通过重置索引，复制列，然后重新应用来做到这一点，但这似乎更迂回。
I tried df.index.levels[1], but that creates a sorted list, it doesn't preserve the order.
我试过df.index.levels[1]，但这会创建一个排序列表，它不会保留顺序。
If it was a single index, I would use df.indexbut in a multiindex that creates a column of tuples.
如果它是单个索引，我会df.index在创建一列元组的多索引中使用。
If this is resolved elsewhere, please share as I haven't had any luck searching the stackoverflow archives.
如果这在其他地方得到解决，请分享，因为我没有任何运气搜索 stackoverflow 档案。
回答by Alexander
df['B'] = df.index.get_level_values(level=1) # Zero based indexing.
# df['B'] = df.index.get_level_values(level='second') # This also works.
>>> df
A B
first second
foo one 12 one
two 11 two
bar one 16 one
two 12 two
three 11 three
回答by Alberto Garcia-Raboso
df['B'] = idx.to_series().str[1]
