Pixel neighbors in 2d array (image) using Python

By using max and min, you handle pixels at the upper and lower bounds:

im[max(i-1,0):min(i+2,i_end), max(j-1,0):min(j+2,j_end)].flatten()

EDIT: ah crap, my answer is just writing im[i-d:i+d+1, j-d:j+d+1].flatten() but written in a incomprehensible way :)

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The good old sliding window trick may help here:

import numpy as np
from numpy.lib.stride_tricks import as_strided

def sliding_window(arr, window_size):
    """ Construct a sliding window view of the array"""
    arr = np.asarray(arr)
    window_size = int(window_size)
    if arr.ndim != 2:
        raise ValueError("need 2-D input")
    if not (window_size > 0):
        raise ValueError("need a positive window size")
    shape = (arr.shape[0] - window_size + 1,
             arr.shape[1] - window_size + 1,
             window_size, window_size)
    if shape[0] <= 0:
        shape = (1, shape[1], arr.shape[0], shape[3])
    if shape[1] <= 0:
        shape = (shape[0], 1, shape[2], arr.shape[1])
    strides = (arr.shape[1]*arr.itemsize, arr.itemsize,
               arr.shape[1]*arr.itemsize, arr.itemsize)
    return as_strided(arr, shape=shape, strides=strides)

def cell_neighbors(arr, i, j, d):
    """Return d-th neighbors of cell (i, j)"""
    w = sliding_window(arr, 2*d+1)

    ix = np.clip(i - d, 0, w.shape[0]-1)
    jx = np.clip(j - d, 0, w.shape[1]-1)

    i0 = max(0, i - d - ix)
    j0 = max(0, j - d - jx)
    i1 = w.shape[2] - max(0, d - i + ix)
    j1 = w.shape[3] - max(0, d - j + jx)

    return w[ix, jx][i0:i1,j0:j1].ravel()

x = np.arange(8*8).reshape(8, 8)
print x

for d in [1, 2]:
    for p in [(0,0), (0,1), (6,6), (8,8)]:
        print "-- d=%d, %r" % (d, p)
        print cell_neighbors(x, p[0], p[1], d=d)

Didn't do any timings here, but it's possible this version has reasonable performance.

For more info, search the net with phrases "rolling window numpy" or "sliding window numpy".

I don't know about any library functions for this, but you can easily write something like this yourself using the great slicing functionality of numpy:

import numpy as np
def neighbors(im, i, j, d=1):
    n = im[i-d:i+d+1, j-d:j+d+1].flatten()
    # remove the element (i,j)
    n = np.hstack((n[:len(n)//2], n[len(n)//2+1:] ))
    return n

Of course you should do some range checks to avoid out-of-bounds access.

Have a look at scipy.ndimage.generic_filter.

As an example:

import numpy as np
import scipy.ndimage as ndimage

def test_func(values):
    print(values)
    return values.sum()


x = np.array([[1,2,3],[4,5,6],[7,8,9]])

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

results = ndimage.generic_filter(x, test_func, footprint=footprint)

By default, it will "reflect" the values at the boundaries. You can control this with the mode keyword argument.

However, if you're wanting to do something like this, there's a good chance that you can express your problem as a convolution of some sort. If so, it will be much faster to break it down into convolutional steps and use more optimized functions (e.g. most of scipy.ndimage).