Permute rows and columns of a matrix

Question

Assuming that I have the following matrix/array:

array([[0, 0, 1, 1, 1],
       [0, 0, 1, 0, 1],
       [1, 1, 0, 1, 1],
       [1, 0, 1, 0, 0],
       [1, 1, 1, 0, 0]])

and I want to apply the following permutation:

1 -> 5
2 -> 4

the result should be in the end:

array([[1, 1, 1, 0, 0],
       [1, 0, 1, 0, 0],
       [1, 1, 0, 1, 1],
       [0, 0, 1, 0, 1],
       [0, 0, 1, 1, 1]])

Now, an incredibly naive (and memory costly) way of doing so might be:

a2 = deepcopy(a1)
a2[0,:] = a1[4,:]
a2[4,:] = a1[0,:]
a = deepcopy(a2)
a2[:,0] = a[:,4]
a2[:,4] = a[:,0]

a3 = deepcopy(a2)
a2[1,:] = a3[3,:]
a2[3,:] = a3[1,:]
a = deepcopy(a2)
a2[:,1] = a[:,3]
a2[:,3] = a[:,1]

But, I would like to know if there is something more efficient that does this. numpy.shuffle and numpy.permutation seem to permute only the rows of the matrix (not the columns at the same time). That doesn't work for me because the matrices are adjacency matrices (representing graphs), and I need to do the permutations which will give me a graph which is isomorphic with the original graph. Furthermore, I need to do an arbitrary number of permutations (more than one).

Thanks!

Answer 1

You can perform the swap in a one-liner using integer array indexing:

a = np.array([[0, 0, 1, 1, 1],
              [0, 0, 1, 0, 1],
              [1, 1, 0, 1, 1],
              [1, 0, 1, 0, 0],
              [1, 1, 1, 0, 0]])
b = a.copy()

# map 0 -> 4 and 1 -> 3 (N.B. Python indexing starts at 0 rather than 1)
a[[4, 3, 0, 1]] = a[[0, 1, 4, 3]]

print(repr(a))
# array([[1, 1, 1, 0, 0],
#        [1, 0, 1, 0, 0],
#        [1, 1, 0, 1, 1],
#        [0, 0, 1, 0, 1],
#        [0, 0, 1, 1, 1]])

Note that array indexing always returns a copy rather than a view - there's no way to swap arbitrary rows/columns of an array without generating a copy.

---

In this particular case you could avoid the copy by using slice indexing, which returns a view rather than a copy:

b = b[::-1] # invert the row order

print(repr(b))
# array([[1, 1, 1, 0, 0],
#        [1, 0, 1, 0, 0],
#        [1, 1, 0, 1, 1],
#        [0, 0, 1, 0, 1],
#        [0, 0, 1, 1, 1]])

---

Update:

You can use the same indexing approach to swap columns.

c = np.arange(25).reshape(5, 5)
print(repr(c))
# array([[ 0,  1,  2,  3,  4],
#        [ 5,  6,  7,  8,  9],
#        [10, 11, 12, 13, 14],
#        [15, 16, 17, 18, 19],
#        [20, 21, 22, 23, 24]])

c[[0, 4], :] = c[[4, 0], :]     # swap row 0 with row 4...
c[:, [0, 4]] = c[:, [4, 0]]     # ...and column 0 with column 4

print(repr(c))

# array([[24, 21, 22, 23, 20],
#        [ 9,  6,  7,  8,  5],
#        [14, 11, 12, 13, 10],
#        [19, 16, 17, 18, 15],
#        [ 4,  1,  2,  3,  0]])

I've used a different example array in this case - your version will yield an identical output after performing the row/column swaps which makes it difficult to understand what's going on.