Matrix columns permutation python

Question

I try to find a solution to found all the columns permutations of a matrix. So i wrote this code but it doesn't work.

SOLVED:

#! python
import numpy

def permutation(matrix): 

    if numpy.size(matrix,1) == 1:
        return [matrix]

    #empty list
    m=[]

    # Iterate the input(matrix) and calculate the permutation 
    for i in range(numpy.size(matrix,1)):
        column = matrix[:,[i]]

        # Extract column[i] or m from the matrix. remMatrix is the remaining matrix
        remMatrix = numpy.concatenate((matrix[:,:i], matrix[:,i+1:]), axis=1)
                   
        # Generating all permutations where m is the first element
        for p in permutation(remMatrix):
            m.append(numpy.concatenate([column,p],axis=1))
    return m


#driver to test the function
matrix=numpy.matrix('1 2 3; 0 0 0')
for p in permutation(matrix): 
    print(p)

Does this answer your question? [Permute rows and columns of a matrix](https://stackoverflow.com/questions/34438098/permute-rows-and-columns-of-a-matrix) — Carlo Zanocco, Oct 02 '20 at 09:52

Martin · Accepted Answer · 2020-10-02T10:42:43.527

0

You did not handle the base case in your recursion. If you pass a matrix with a single column, then permutation returns an empty matrix. That is because line m = numpy.concatenate((column,p),axis=1) will not be reached if remMatrix is empty.

As result m is an empty array once you return it and the print statement is not called.

I also don't fully understand what you're trying to do. Wouldn't there for most matrices be multiple column permutations? Do you want all those matrices concatenated in the end?

edited Oct 02 '20 at 10:42

answered Oct 02 '20 at 10:34

Martin

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Oh thanks! in fact I'm trying to extend the code for a list, found at the following link at method number 2, for the columns of a matrix, but I'm using python since a couple of days and I can't found the complete solution https://www.geeksforgeeks.org/generate-all-the-permutation-of-a-list-in-python/ – Hike Oct 02 '20 at 11:00
Ah I see. You actually removed the base cases `if len(lst) == 0` and `if len(lst) == 1` from the example but you need them also in your case ;) – Martin Oct 02 '20 at 11:08
I tried to add the two if statement and correct some mistakes, now i had some errors, but another thing is that I 'print(p)' I obtain an array with only one row, so when I try to concatenate or append it doesn't because the dimensions along axis 0 are different. I edit the code in the question with the current – Hike Oct 02 '20 at 12:11
`m` needs to be a list that you append to. This means that your base cases need to change a bit as well: `if numpy.size(matrix, 1) == 1: return [matrix]`. It seems you actually don't need `if numpy.size(matrix, 1) == 0` case. And the append becomes: `m.append(numpy.concatenate([column, p], axis=1))` – Martin Oct 02 '20 at 12:23
now it seems work! great, thanks so much! is it better to update the code in the question with the working one? – Hike Oct 02 '20 at 12:37
You're welcome. You could have a "Solved" section with your working solution in the end of your post. Also mark my answer as solved please :) – Martin Oct 02 '20 at 12:39

Matrix columns permutation python

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