I'm looking for a way to calculate the average absolute difference between neighboring elements in a NumPy array. Namely, given an array like

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

The value for the middle square will be 2.5 (aka (4 3 2 1 1 2 3 4)/8). I know with SciPy's correlate2d you can compute the average difference, but, as far as I know, not the average absolute difference (i.e. for the example above, correlate2d would give 0 - (-4 -3 -2 -1 1 2 3 4)/8 - not 2.5).

Is there a fast way to do this in Python? I don't want to iterate over the elements since this will be running for very large arrays many times.

CodePudding user response：

Why not write it by hand and numba.jit the result?

arr = (np.arange(10**6) 1).reshape(10**3,10**3)

@nb.njit
def get_avg_abs_diff(arr):
    n,m = arr.shape
    out = np.empty((n,m))
    for i in range(n):
        for j in range(m):
            a = max(0,i-1)
            b = i 2
            c = max(0,j-1)
            d = j 2
            neighbours = arr[a:b,c:d]
            out[i,j] = np.sum(np.abs(neighbours-arr[i,j]))/(neighbours.shape[0]*neighbours.shape[1]-1)
    return out

get_avg_abs_diff(arr)

This takes about 160ms for the 1000 by 1000 array (down from non jited version which takes 10.6s).

For your array I get

array([[2.66666667, 2.2       , 2.        ],
       [2.6       , 2.5       , 2.6       ],
       [2.        , 2.2       , 2.66666667]])