python – 在pandas multiindex的第二级中选择数据帧的子集

这是我的数据框：

 iterables = [['bar', 'baz', 'foo', 'qux'], ['one', 'two', 'three', 'four']]  
 mindex = pd.MultiIndex.from_product(iterables, names=['first', 'second'])   
 df = pd.DataFrame(np.random.randn(16, 3), index=mindex)

它看起来像这样：

                     0         1         2
first second                              
bar   one    -0.445212 -2.208192 -1.297759
      two     1.521942  0.592622 -1.677931
      three   0.709292  0.348715 -0.766430
      four   -1.812516 -0.982077 -1.155860
baz   one    -0.375230 -0.267912  2.621249
      two    -1.041991 -0.752277 -0.494512
      three  -1.029389 -0.331234  0.950335
      four   -1.357269  0.653581  1.289331
foo   one     0.980196  0.865067 -0.780575
      two    -1.641748  0.220253  2.141745
      three   0.272158 -0.320238  0.787176
      four   -0.265425 -0.767928  0.695651
qux   one    -0.117099  1.089503 -0.692016
      two    -0.203240 -0.314236  0.010321
      three   1.425749  0.268420 -0.886384
      four    0.181717 -0.268686  1.186988

我想为第一个索引中的每个元素选择数据帧的子集,以便仅使用来自多索引的第二级的一个和三个索引值.

我已经检查了in the advanced indexing section的文档,但没有太大的成功.可以从第二个索引级别中选择一个特定的索引值：

df.loc['bar','one']
Out[74]: 
0   -0.445212
1   -2.208192
2   -1.297759
Name: (bar, one), dtype: float64

但不是一个价值元组,因为这：

df.loc[('bar',('one','three'))]

导致错误：

KeyError: “None of [(‘one’, ‘three’)] are in the [columns]”

我希望.loc基本上通过这个命令传递bar,然后是第二级索引值为1和3的行.

如何基于多索引级别子集执行此类子选择？

@R_404_5620@:

添加：用于选择所有列：

a = df.loc[('bar',('one','three')), :]
print (a)
                     0         1         2
first second                              
bar   one    -0.902444  2.115037 -0.065644
      three   2.095998  0.768128  0.413566

与IndexSlice类似的解决方案：

idx = pd.IndexSlice
a = df.loc[idx['bar', ('one','three')], :]
print (a)
                     0         1         2
first second                              
bar   one    -0.515183 -0.858751  0.854838
      three   2.315598  0.402738 -0.184113

正如@Brad所罗门所提到的,如果想要所有第一级的价值：

df1 = df.loc[(slice(None), ['one', 'three']), :]

idx = pd.IndexSlice
df1 = df.loc[idx[:, ('one','three')], :]

print (df1)
                     0         1         2
first second                              
bar   one    -0.266926  1.105319  1.768572
      three  -0.632492 -1.642508 -0.779770
baz   one    -0.380545 -1.632120  0.435597
      three   0.018085  2.114032  0.888008
foo   one     0.539179  0.164681  1.598194
      three   0.051494  0.872987 -1.882287
qux   one    -1.361244 -1.520816  2.678428
      three   0.323771 -1.691334 -1.826938