大熊猫可互换双重索引？

Question

I have a DataFrame and I build a dual index. 'start' values don't exist in 'end' index values and versa.

c_weights.rename(columns={0:'start',1:'end',2:'metric',3:'angular',4:'special',5:'cos_pi'}, inplace=True)
c_weights.set_index(['start','end'],inplace=True)
c_weights.head()

Id like to be able to call something like: c_weights.loc[1,638] or c_weights.loc[638,1] and get the same line of data. To make it clear, the two index combinations are always unique. How this can be bone?

Answer 1

Anyways, for the first case, you can just index using ix and passing a tuple on the row index

c_weights.ix[(1,638)]

For the second case, I guess it'll depend whether you know off hand or not if you're trying to pass the end first, in which case I'd just construct a tuple in a right way or reverse it ( (638,1)[::-1] = (1, 638) )

To get to your point: since you say you have mutually exclusive start and end, you can also use the following list comprehension

l = (start, end) # l = (end, start) returns the same
c_weights.ix[[x for x in c_weights.index if (x ==  l) or (x == l[::-1])]]

If you also have unique index, you can simplify this to:

c_weights.ix[[x for x in c_weights.index if (x[0] ==  l[0]) or (x[1] == l[1])]]

Answer 2

A dataframe is a wrapper around an numpy ndarray in which a row and column index are assigned. We can define a second dataframe with different row or column indices and access the same ndarray. For example, let's first define df1 , then define df2 with the same data, but swap the levels in a MultiIndex row index. Leave the columns the same.

import pandas as pd
import numpy as np

np.random.seed([3,1415])

df1 = pd.DataFrame(np.random.rand(4, 2),
                   pd.MultiIndex.from_product([('a', 'b'), (1, 2)]),
                   ['col1', 'col2'])
df2 = pd.DataFrame(df1.values, df1.index.swaplevel(0, 1), df1.columns)

print df1

         col1      col2
a 1  0.444939  0.407554
  2  0.460148  0.465239
b 1  0.462691  0.016545
  2  0.850445  0.817744

print df2

         col1      col2
1 a  0.444939  0.407554
2 a  0.460148  0.465239
1 b  0.462691  0.016545
2 b  0.850445  0.817744

We can see the data is the same, the indices are swapped. Accessing data from df1 is the same data as from df1 to the point of co-mutability. Let's change something in df1 and look at df2

df1.loc[('a', 1), 'col1'] = 1.
print df2

         col1      col2
1 a  1.000000  0.407554
2 a  0.460148  0.465239
1 b  0.462691  0.016545
2 b  0.850445  0.817744

Now that we're convinced, let's observe that we now have 2 dataframes from which we can access the same data. Let's define a function to do what the OP asked for.

ambigui_t = lambda t: df1.loc[t] if t in df.index else df2.loc[t]

print ambigui_t(('a', 1))

col1    1.000000
col2    0.407554
Name: (a, 1), dtype: float64

print ambigui_t((1, 'a'))

col1    1.000000
col2    0.407554
Name: (1, a), dtype: float64