I've searched for a solution to this problem but haven't found anything specific to this problem. My dataframe is structured like this:

   column_1    column_2     column_3
a     2           3            7
b     9           4            3
c     1           5            2
        

I want to find all permutations of the above dataframe without repeating rows or columns in each individual permutation.

The preceding isn't super clear, so here is the output I'm trying to achieve:

Out: [(2,4,2),(2,5,3),(9,3,2),(9,5,7),(1,3,3),(1,4,7)]

In other words, I expected n! results

The solution I tried was:

permutations = list(product(df['column_1'], df['column_2'], df['column_3']))
print(permutations)

This returns n^n combinations.

Any help is appreciated! THANKS

CodePudding user response：

You can use itertools.permutations on the row indices and numpy indexing:

from itertools import permutations

idx = list(permutations(range(len(df))))

df.to_numpy()[idx, np.arange(df.shape[1])].tolist()

output:

[[2, 4, 2], [2, 5, 3], [9, 3, 2], [9, 5, 7], [1, 3, 3], [1, 4, 7]]

CodePudding user response：

You can use permutations method of the itertools package. This gives you the indices you need for each column.

from itertools import permutations
indices = list(permutations('abc', 3))
print(indices)