Python Pandas 多索引：保持相同长度的 level=1 与所有 level=0 索引

Question

我有一个 df_ver1 具有多索引索引。我想删除所有级别 [1] 长度与 2 不同的行。下面是我的数据框。

In [13]: df_ver1
Out[13]: 
key  nm         0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
foo one  1.075770 -0.109050  1.643563 -1.469388
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061
oof two  1.340309 -1.187678 -2.211372  0.380396

我的理想输出是

In [13]: df_ver1_fixed
Out[13]: 
key  nm         0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

如您所见，我想删除只有 1 级 [1] 索引的行。换句话说，我需要在第二级有 'one' 和 'two' 索引。是否有 pythonic 方法来执行此步骤？谢谢！

Answer 1

我认为你需要：

#filter only one and two values by second level
df = df.loc[pd.IndexSlice[:, ['one','two']], :]
#filter by length
df = df[df.groupby(level=0)[df.columns[0]].transform('size') == 2]
print (df)
                0         1         2         3
key nm                                         
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

另一个解决方案是比较集合：

mask = df.reset_index()
         .groupby('key')['nm']
         .transform(lambda x: set(x) == set(['one','two']))
         .values 
df = df[mask]
print (df)
                0         1         2         3
key nm                                         
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

Answer 2

这也行。其实可以按multi-indexkey分组，过滤掉分组长度不等于2.

df.groupby(by='key').filter(lambda x: len(x) == 2) # keep groups with len 2

正如@Zero 所建议的，我们可以更具体地使用以下内容来指定满足要求的变量集，set(['one', 'two'])。

df.groupby(by='key').filter(
              lambda x: set(x.index.get_level_values('nm')) == set(['one', 'two']))

key  nm         0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

另一种方法：使用多索引选择

sz = df.groupby("key").size()
indexes = sz[sz == 2].index.tolist()  # first-level indexes that we want.
df.loc[indexes] # use loc for selection

key  nm         0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
    two -0.673690  0.113648 -1.478427  0.524988
baz one  0.404705  0.577046 -1.715002 -1.039268
    two -0.370647 -1.157892 -1.344312  0.844885
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

Python Pandas 多索引：保持相同长度的 level=1 与所有 level=0 索引

Python Pandas Multi-index: keeping same length of level=1 with all level=0 indexes

python

indexing

multi-index

multi-level

pandas