Question

我有一个带有多索引索引的df_ver1。我想删除所有具有不同级别[1]长度然后是2的行。这是我的数据帧。

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

我理想的输出是

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

所以你可以看到我想删除只有1级[1]索引的行。换句话说，我需要在第二级中拥有“一个”和“两个”索引。是否有一种pythonic方法来完成这一步？谢谢！

Answer 1

这也可行。您实际上可以按多索引key进行分组，并过滤掉不等于2的组的长度。

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

正如@Zero建议的那样，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

Answer 2

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：

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

Python Pandas Multi-index：使用所有level = 0索引保持level = 1的相同长度

2 个答案: