Remove reverse duplicates from dataframe

Question

I have a data frame with two columns, A and B. The order of A and B is unimportant in this context; for example, I would consider (0,50) and (50,0) to be duplicates. In pandas, what is an efficient way to remove these duplicates from a dataframe?

import pandas as pd

# Initial data frame.
data = pd.DataFrame({'A': [0, 10, 11, 21, 22, 35, 5, 50], 
                     'B': [50, 22, 35, 5, 10, 11, 21, 0]})
data
    A   B
0   0  50
1  10  22
2  11  35
3  21   5
4  22  10
5  35  11
6   5  21
7  50   0

# Desired output with "duplicates" removed. 
data2 = pd.DataFrame({'A': [0, 5, 10, 11], 
                      'B': [50, 21, 22, 35]})
data2
    A   B
0   0  50
1   5  21
2  10  22
3  11  35

Ideally, the output would be sorted by values of column A.

Answer 1

You can sort each row of the data frame before dropping the duplicates:

data.apply(lambda r: sorted(r), axis = 1).drop_duplicates()

#   A    B
#0  0   50
#1  10  22
#2  11  35
#3  5   21

If you prefer the result to be sorted by column A:

data.apply(lambda r: sorted(r), axis = 1).drop_duplicates().sort_values('A')

#   A    B
#0  0   50
#3  5   21
#1  10  22
#2  11  35

Answer 2

Here is bit uglier, but faster solution:

In [44]: pd.DataFrame(np.sort(data.values, axis=1), columns=data.columns).drop_duplicates()
Out[44]:
    A   B
0   0  50
1  10  22
2  11  35
3   5  21

Timing: for 8K rows DF

In [50]: big = pd.concat([data] * 10**3, ignore_index=True)

In [51]: big.shape
Out[51]: (8000, 2)

In [52]: %timeit big.apply(lambda r: sorted(r), axis = 1).drop_duplicates()
1 loop, best of 3: 3.04 s per loop

In [53]: %timeit pd.DataFrame(np.sort(big.values, axis=1), columns=big.columns).drop_duplicates()
100 loops, best of 3: 3.96 ms per loop

In [59]: %timeit big.apply(np.sort, axis = 1).drop_duplicates()
1 loop, best of 3: 2.69 s per loop