Have numpy argsort return an array of 2d indices?

Question

If we have a 1d array

arr = np.random.randint(7, size=(5))
# [3 1 4 6 2]
print np.argsort(arr)
# [1 4 0 2 3] <= The indices in the sorted order    

If we have a 2d array

arr = np.random.randint(7, size=(3, 3))
# [[5 2 4]
# [3 3 3]
# [6 1 2]]
print np.argsort(arr)
# [[1 2 0]
# [0 1 2]
# [1 2 0]] <= It sorts each row

What I need is the 2d indices that sort this matrix in its entirety. Something like this:

# [[2 1] => 1
# [0 1] => 2
# [2 2] => 2
# .
# .
# .
# [0 2] => 4
# [0 0] => 5
# [2 0]] => 6

How do I get "2d indices" for the sorting of a 2d array?

Solution 1

Apply numpy.argsort on flattened array and then unravel the indices back to (3, 3) shape:

>>> arr = np.array([[5, 2, 4],
[3, 3, 3],
[6, 1, 2]])
>>> np.dstack(np.unravel_index(np.argsort(arr.ravel()), (3, 3)))
array([[[2, 1],
        [0, 1],
        [2, 2],
        [1, 0],
        [1, 1],
        [1, 2],
        [0, 2],
        [0, 0],
        [2, 0]]])

Solution 2

From the documentation on numpy.argsort:

ind = np.unravel_index(np.argsort(x, axis=None), x.shape)

Indices of the sorted elements of a N-dimensional array.

An example:

>>> x = np.array([[0, 3], [2, 2]])
>>> x
array([[0, 3],
       [2, 2]])
>>> ind = np.unravel_index(np.argsort(x, axis=None), x.shape)
>>> ind # a tuple of arrays containing the indexes
(array([0, 1, 1, 0]), array([0, 0, 1, 1]))
>>> x[ind]  # same as np.sort(x, axis=None)
array([0, 2, 2, 3])

Comments

[MartianMartian]

jezz, what is this result...?

[Dirk Groeneveld]

I think you need a [0] at the end?

[pietz]

this looks more complex than it should be.

[AturSams]

This did not sort all the elements, it sorts rows.