Permutation of symmetric array in Python

Question

I want to have the permutation for this symmetric matrix, if we move the the second column to the third the second row also should go to the third row.

array([[ 0.        ,  0.06377803,  0.1157737 ,  0.19542195],
       [ 0.06377803,  0.        ,  0.14754803,  0.23185761],
       [ 0.1157737 ,  0.14754803,  0.        ,  0.0843134 ],
       [ 0.19542195,  0.23185761,  0.0843134 ,  0.        ]])

This is code for permutation on a list:

import numpy as np
x=[]
def perm(a, k=0):

    if k == len(a):
        x.extend(a)
#        print (a )


    else:
      for i in range(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([0,1,2,3])
a=np.asarray(x).reshape((24,4))
print(a)

Output:

But I want to have the permutation for the above array which is 4*4. For simplicity if we have a 3*3 array, we want something like below which is K!=6 but when k=4 then we have to get k! which is 24 permutations

Are you saying you want to permute the matrix in such a way that it status symmetrical? — Mad Physicist
– Mad Physicist, Commented Sep 3, 2018 at 20:28
Fix your indentation and use a return value instead of a global. — Mad Physicist
– Mad Physicist, Commented Sep 3, 2018 at 20:29
Thank you for you comment, if you look at the image you will see there is no problem for diagonal and off-diagonal elements because the diagonal still is zero, — user8523104
– user8523104, Commented Sep 3, 2018 at 20:54
I think the image is really clear what I want to get, permutation for the above array vertically and horizontally please look the image to see what I wan to have — user8523104
– user8523104, Commented Sep 3, 2018 at 21:19

xdze2 · Accepted Answer · 2018-09-03 21:27:34Z

3

import numpy as np
from itertools import permutations

n = 3
a = np.arange(n**2).reshape(n, n)

for perm in permutations(range(a.shape[0])):
    b = np.zeros_like(a)
    b[:, :] = a[perm, :]
    b[:, :] = b[:, perm]
    print(b)

gives the 6 following matrices:

[[0 1 2]
 [3 4 5]
 [6 7 8]]
[[0 2 1]
 [6 8 7]
 [3 5 4]]
[[4 3 5]
 [1 0 2]
 [7 6 8]]
[[4 5 3]
 [7 8 6]
 [1 2 0]]
[[8 6 7]
 [2 0 1]
 [5 3 4]]
[[8 7 6]
 [5 4 3]
 [2 1 0]]

Is this the question?

answered Sep 3, 2018 at 21:27

xdze2

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