Pixel neighbors in 2d array (image) using Python

Question

I have a numpy array like this:

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

I need to create a function let's call it "neighbors" with the following input parameter:

• x: a numpy 2d array
• (i,j): the index of an element in a 2d array
• d: neighborhood radius

As output I want to get the neighbors of the cell i,j with a given distance d. So if I run

neighbors(im, i, j, d=1) with i = 1 and j = 1 (element value = 5) 

I should get the indices of the following values: [1,2,3,4,6,7,8,9]. I hope I make it clear. Is there any library like scipy which deal with this?

I've done something working but it's a rough solution.

def pixel_neighbours(self, p):

    rows, cols = self.im.shape

    i, j = p[0], p[1]

    rmin = i - 1 if i - 1 >= 0 else 0
    rmax = i + 1 if i + 1 < rows else i

    cmin = j - 1 if j - 1 >= 0 else 0
    cmax = j + 1 if j + 1 < cols else j

    neighbours = []

    for x in xrange(rmin, rmax + 1):
        for y in xrange(cmin, cmax + 1):
            neighbours.append([x, y])
    neighbours.remove([p[0], p[1]])

    return neighbours

How can I improve this?

Answer 1

from skimage.util import view_as_windows
window_shape = (3, 3, 3)
neighborhoods = view_as_windows(img[200:220], window_shape) 

Here is one D if you need:

>>> A = np.arange(10)
>>> A
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> window_shape = (3,)
>>> B = view_as_windows(A, window_shape)
>>> B.shape
(8, 3)
>>> B
array([[0, 1, 2],
       [1, 2, 3],
       [2, 3, 4],
       [3, 4, 5],
       [4, 5, 6],
       [5, 6, 7],
       [6, 7, 8],
       [7, 8, 9]]

reference :https://scikit-image.org/docs/stable/api/skimage.util.html#skimage.util.view_as_windows

Answer 2

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.

Answer 3

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).

Answer 4

We first init our matrix of interest using numpy

import numpy as np

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

print(x)

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

Our neighbors is a function of distance for instance we might be interested in neighbors of distance 2 this tells us how should we pad our matrix x. We choose to pad with zeros but you can pad with whatever you like might be mean,mode,median of a row/column

d = 2

x_padded = np.pad(x,d,mode='constant')

print(x_padded)

[[0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0]
 [0 0 1 2 3 0 0]
 [0 0 4 5 6 0 0]
 [0 0 7 8 9 0 0]
 [0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0]]

We use x_padded matrix to get neighbors of any value in matrix x. Let (i,j) and (s,t) be indexes of x and x_padded respectively. Now we need to translate (i,j) to (s,t) to get neighbors of (i,j)

i,j = 2,1
s,t = 2*d+i+1, 2*d+j+1

window = x_padded[i:s, j:t]

print(window)

[[0 1 2 3 0]
 [0 4 5 6 0]
 [0 7 8 9 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]

Please Note!!! the indexes (i,j) point to any value you wish to get its neighbors in matrix x

One might wish to iterate over each point in matrix x, get its neighbors and do computation using the neighbors for instance in Image Processing, the convolution with a kernel. One might do the following to get neighbors of each pixel in an image x

for i in range(x.shape[0]):
    for j in range(x.shape[1]):
        i,j = 2,1
        s,t = 2*d+i+1, 2*d+j+1
        window = x_padded[i:s, j:t]

Answer 5

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()

Answer 6

I agree with Joe Kingtons response, just an add to the footprints

import numpy as np
from scipy.ndimage import generate_binary_structure
from scipy.ndimage import iterate_structure
foot = np.array(generate_binary_structure(2, 1),dtype=int)

or for bigger/different footprints for ex.

np.array(iterate_structure(foot , 2),dtype=int)

Answer 7

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 :)

---

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".

Answer 8

Possibly use a KDTree in SciPy ?