[FIXED] Sort n-dimensional numpy array by sorting highest dimension and afterwards reordering entire segments

Issue

I've got a list of ndarrays (all arrays are of the same dimensions but there can be n dimensions) which I need to compare with each other to eliminate duplicates. For that reason I am trying to sort all arrays equally.

A quick eample for 2d and 3d arrays:

[[1, 0]  == [[0, 1]   ==  [[1, 0]  ==  [[0, 1]
 [0, 1]]     [0, 1]]       [1, 0]]      [1, 0]]

[[[1,1],[0,2]],[[0,0],[0,0]]] == [[[0, 0], [0, 0]],[[1, 1],[0, 2]]] == [[[0, 0], [0, 0]],[[0, 2], [[1, 1]]] ...

I don't care how it is ordered, i just want to find, handle and afterwards eliminate duplicates.

I tried using numpy.sort(array) which can sadly miss duplicates (e.g. [[0,0],[0,1]] and [[0,1],[0,0]]). If I sort by every dimension starting with the highest it destroys my structure (eg. [[1, 1], [0, 2]] becomes [[0, 1], [1, 2]]).

From a nested list point of view, I think, the best idea is to first sort all lists at depth n. Afterwards sort the nested lists at depth n-1 ... 0 by first ordering by sum and afterwards if they are equal by comparing the elements.

Solution

If I understand correctly, you could use a recursive sorter. I'm providing a python solution here since I believe there is no advantage of using numpy here:

def sorter(lst):
    if isinstance(lst, list):
        return sorted(map(sorter, lst))
    else:
        return lst

Or, maybe more generic using a try/except (this would also work on numpy arrays):

def sorter(lst):
    try:
        return sorted(map(sorter, lst))
    except TypeError:
        return lst

Examples:

example1 = ([[1, 0], [0, 1]],
            [[0, 1], [0, 1]],
            [[1, 0], [1, 0]],
            [[0, 1], [1, 0]],
            [[0, 0], [1, 1]], # different
           )

example2 = ([[[1,1],[0,2]],[[0,0],[0,0]]],
            [[[0,0],[0,0]],[[1,1],[0,2]]],
            [[[0,0],[0,0]],[[0,2],[1,1]]],
            [[[0,0],[0,0]],[[1,1],[2,0]]],
            [[[0,0],[0,0]],[[1,2],[1,0]]], # different
           )

[sorter(x) for x in example1]

[[[0, 1], [0, 1]],
 [[0, 1], [0, 1]],
 [[0, 1], [0, 1]],
 [[0, 1], [0, 1]],
 [[0, 0], [1, 1]]] # different

[sorter(x) for x in example2]

[[[[0, 0], [0, 0]], [[0, 2], [1, 1]]],
 [[[0, 0], [0, 0]], [[0, 2], [1, 1]]],
 [[[0, 0], [0, 0]], [[0, 2], [1, 1]]],
 [[[0, 0], [0, 0]], [[0, 2], [1, 1]]],
 [[[0, 0], [0, 0]], [[0, 1], [1, 2]]]] # different

Answered By - mozway