Sort invariant for numpy.argsort with multiple dimensions

Question

numpy.argsort docs state

Returns:
index_array : ndarray, int Array of indices that sort a along the specified axis. If a is one-dimensional, <

Answer 1

The numpy issue #8708 has a sample implementation of take_along_axis that does what I need; I'm not sure if it's efficient for large arrays but it seems to work.

def take_along_axis(arr, ind, axis):
    """
    ... here means a "pack" of dimensions, possibly empty

    arr: array_like of shape (A..., M, B...)
        source array
    ind: array_like of shape (A..., K..., B...)
        indices to take along each 1d slice of `arr`
    axis: int
        index of the axis with dimension M

    out: array_like of shape (A..., K..., B...)
        out[a..., k..., b...] = arr[a..., inds[a..., k..., b...], b...]
    """
    if axis < 0:
       if axis >= -arr.ndim:
           axis += arr.ndim
       else:
           raise IndexError('axis out of range')
    ind_shape = (1,) * ind.ndim
    ins_ndim = ind.ndim - (arr.ndim - 1)   #inserted dimensions

    dest_dims = list(range(axis)) + [None] + list(range(axis+ins_ndim, ind.ndim))

    # could also call np.ix_ here with some dummy arguments, then throw those results away
    inds = []
    for dim, n in zip(dest_dims, arr.shape):
        if dim is None:
            inds.append(ind)
        else:
            ind_shape_dim = ind_shape[:dim] + (-1,) + ind_shape[dim+1:]
            inds.append(np.arange(n).reshape(ind_shape_dim))

    return arr[tuple(inds)]

which yields

>>> A = np.array([[3,2,1],[4,0,6]])
>>> B = np.array([[3,1,4],[1,5,9]])
>>> i = A.argsort(axis=-1)
>>> take_along_axis(A,i,axis=-1)
array([[1, 2, 3],
       [0, 4, 6]])
>>> take_along_axis(B,i,axis=-1)
array([[4, 1, 3],
       [5, 1, 9]])