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]]])

np.take_along_axis has an example using argsort:

>>> a = np.array([[10, 30, 20], [60, 40, 50]])

We can sort either by using sort directly, or argsort and this function

>>> np.sort(a, axis=1)
array([[10, 20, 30],
       [40, 50, 60]])
>>> ai = np.argsort(a, axis=1); ai
array([[0, 2, 1],
       [1, 2, 0]])
>>> np.take_along_axis(a, ai, axis=1)
array([[10, 20, 30],
       [40, 50, 60]])

This streamlines a process of applying ai to the array itself. We can do that directly, but it requires a bit more thought about what the index actually represents.

In this example, ai are index values along axis 1 (values like 0,1,or 2). This (2,3) has to broadcast with a (2,1) array for axis 0:

In [247]: a[np.arange(2)[:,None], ai]
Out[247]: 
array([[10, 20, 30],
       [40, 50, 60]])

Solution:

>>> a[np.arange(np.shape(a)[0])[:,np.newaxis], np.argsort(a)]
array([[1, 2, 3],
       [2, 8, 9]])

You got it right, though I wouldn't describe it as cheating the indexing.

Maybe this will help make it clearer:

In [544]: i=np.argsort(a,axis=1)

In [545]: i
Out[545]: 
array([[1, 2, 0],
       [2, 0, 1]])

i is the order that we want, for each row. That is:

In [546]: a[0, i[0,:]]
Out[546]: array([1, 2, 3])

In [547]: a[1, i[1,:]]
Out[547]: array([2, 8, 9])

To do both indexing steps at once, we have to use a 'column' index for the 1st dimension.

In [548]: a[[[0],[1]],i]
Out[548]: 
array([[1, 2, 3],
       [2, 8, 9]])

Another array that could be paired with i is:

In [560]: j=np.array([[0,0,0],[1,1,1]])

In [561]: j
Out[561]: 
array([[0, 0, 0],
       [1, 1, 1]])

In [562]: a[j,i]
Out[562]: 
array([[1, 2, 3],
       [2, 8, 9]])

If i identifies the column for each element, then j specifies the row for each element. The [[0],[1]] column array works just as well because it can be broadcasted against i.

I think of

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

as 'short hand' for j. Together they define the source row and column of each element of the new array. They work together, not sequentially.

The full mapping from a to the new array is:

[a[0,1]  a[0,2]  a[0,0]
 a[1,2]  a[1,0]  a[1,1]]

def foo(a):
    i = np.argsort(a, axis=1)
    return (np.arange(a.shape[0])[:,None], i)

In [61]: foo(a)
Out[61]: 
(array([[0],
        [1]]), array([[1, 2, 0],
        [2, 0, 1]], dtype=int32))
In [62]: a[foo(a)]
Out[62]: 
array([[1, 2, 3],
       [2, 8, 9]])

You need a little extra work to properly index the 2d array. Here's a way using advanced indexing, where np.arange is used in the first axis so that each row in sort_inds extracts values from the corresponding row in d:

d[np.arange(d.shape[0])[:,None], sort_inds]

array([[1, 1, 2, 3, 3, 4, 4, 7, 8, 9],
       [1, 3, 4, 5, 5, 5, 6, 8, 8, 9],
       [1, 2, 3, 4, 5, 6, 7, 8, 8, 8],
       [2, 2, 4, 7, 7, 8, 8, 9, 9, 9],
       [1, 1, 2, 4, 4, 7, 7, 8, 8, 8]])