如何访问熊猫数据框中的多级索引？

Question

I would like to call those row with same index.

so this is the example data frame,

arrays = [np.array(['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux']),
np.array(['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two'])]

df = pd.DataFrame(np.random.randn(8, 4), index=arrays)

In [16]: df
Out[16]: 
                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
    two  0.357021 -0.674600 -1.776904 -0.968914
qux one -1.294524  0.413738  0.276662 -0.472035
    two -0.013960 -0.362543 -0.006154 -0.923061

I would like to select

                0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
baz one  0.404705  0.577046 -1.715002 -1.039268
foo one  1.075770 -0.109050  1.643563 -1.469388
qux one -1.294524  0.413738  0.276662 -0.472035

or even as this format

            0         1         2         3
one -0.424972  0.567020  0.276232 -1.087401
one  0.404705  0.577046 -1.715002 -1.039268
one  1.075770 -0.109050  1.643563 -1.469388
one -1.294524  0.413738  0.276662 -0.472035

I have tried df['bar','one] and it's not working. I am now sure how should I access the multi-level index.

Answer 1

You can use MultiIndex slicing (use slice(None) instead of colon):

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

Result:

                0         1         2         3
bar one -0.424972  0.567020  0.276232 -1.087401
baz one  0.404705  0.577046 -1.715002 -1.039268
foo one  1.075770 -0.109050  1.643563 -1.469388
qux one -1.294524  0.413738  0.276662 -0.472035

Finally you can drop the first index column:

df.index = df.index.droplevel(0)

Result:

            0         1         2         3
one -0.424972  0.567020  0.276232 -1.087401
one  0.404705  0.577046 -1.715002 -1.039268
one  1.075770 -0.109050  1.643563 -1.469388
one -1.294524  0.413738  0.276662 -0.472035

Answer 2

Use DataFrame.xs and if need both levels add drop_level=False :

df1 = df.xs('one', level=1, drop_level=False)
print (df1)
bar one -0.424972  0.567020  0.276232 -1.087401
baz one  0.404705  0.577046 -1.715002 -1.039268
foo one  1.075770 -0.109050  1.643563 -1.469388
qux one -1.294524  0.413738  0.276662 -0.472035

For second remove first level by DataFrame.reset_index with drop=True , so possible select by label with DataFrame.loc :

df2 = df.reset_index(level=0, drop=True).loc['one']
#alternative
#df2 = df.xs('one', level=1, drop_level=False).reset_index(level=0, drop=True)
print (df2)
            0         1         2         3
one -0.424972  0.567020  0.276232 -1.087401
one  0.404705  0.577046 -1.715002 -1.039268
one  1.075770 -0.109050  1.643563 -1.469388
one -1.294524  0.413738  0.276662 -0.472035

More common is used xs without duplicated levels - so after select one is removed this level:

df3 = df.xs('one', level=1)
print (df3)
            0         1         2         3
bar -0.424972  0.567020  0.276232 -1.087401
baz  0.404705  0.577046 -1.715002 -1.039268
foo  1.075770 -0.109050  1.643563 -1.469388
qux -1.294524  0.413738  0.276662 -0.472035

Answer 3

Since the question involves multi-indexing and the sequence of the index is 'bar' and then 'one' which can be verified by using df.index command:

MultiIndex([('bar', 'one'),
            ('bar', 'two'),
            ('baz', 'one'),
            ('baz', 'two'),
            ('foo', 'one'),
            ('foo', 'two'),
            ('qux', 'one'),
            ('qux', 'two')],
           )

The output that you are looking for can be accessed using df.loc[('bar','one')]

The output it produces is

0    0.162693
1    0.420518
2   -0.152041
3   -1.039439
Name: (bar, one), dtype: float64