List of pd.IndexSlice

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Input :

midx = pd.MultiIndex.from_product([['A0','A1'], ['B0','B1','B2','B3']])
columns = ['foo', 'bar']
t = pd.DataFrame(np.arange(16).reshape((len(midx), len(columns))),
                    index=midx, columns=columns)

Output:

In [6]: t
Out[6]:
       foo  bar
A0 B0    0    1
   B1    2    3
   B2    4    5
   B3    6    7
A1 B0    8    9
   B1   10   11
   B2   12   13
   B3   14   15

Question:

Let v = pd.IndexSlice['A0','B0':'B1']. I can access t.loc[v,:]:

In [13]: t.loc[v,:]
Out[13]:
       foo  bar
A0 B0    0    1
   B1    2    3

From [v,v], how can I get the following dataframe?

In [13]: f([v,v])
Out[13]:
       foo  bar
A0 B0    0    1
   B1    2    3
   B0    0    1
   B1    2    3

I can't specify a list of those multi-index slicers: t.loc[[v,v],:] will raise an Exception: Unashable type: 'slice'. Why is it so? When using scalars, I can easily list the desired locs: t.loc[[('A0','B0'),('A0','B1')],:].

pandas indexing python
2 Answers

Try:

t.loc[tuple(zip(v, v)),:]

It should be :

u = (['A0', 'A1'], [slice('B0', 'B1', None), slice('B0', 'B1', None)])

try to have it in the above form.

t.loc[u,:]

        foo bar
A0  B0  0   1
    B1  2   3
A1  B0  8   9
    B1  10  11
user_66603730

Use a little known nor used parameter axis=0 of loc:

You can also specify the axis argument to .loc to interpret the passed slicers on a single axis.

Advanced indexing with hierarchical index

t.loc(axis=0)[v]

Output:

       foo  bar
A0 B0    0    1
   B1    2    3

Work around per updated comments and question clarification:

pd.concat([t.loc[v,:]]*2)

Output:

       foo  bar
A0 B0    0    1
   B1    2    3
   B0    0    1
   B1    2    3
user_63615310
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