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         


        

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.