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?

Solution 1

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".

Solution 2

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).

Solution 3

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()

Solution 4

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((b[:len(b)//2],b[len(b)//2+1:] ))
    return n

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

Solution 5

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)

View more solutions

Comments

[Joe Kington]

Just for whatever it's worth, for regularly-gridded data, a quadtree is not ideal. Indexing the grid that the data is on gives you a much faster lookup for neighbors.

[James]

What does b refer to here?

[Tommaso Guerrini]

b is not defined, as James said

[Cris Luengo]

Please edit your answer so it explains how it answers OP’s question. Code dumps are not considered useful answers on Stack Overflow.

[Michael Baudin]

Please add a comment before your code in order to clarify how this solves the problem, separate the code from its output and add a comment after the result to analyse what is specifically useful in the result.

[Omphemetse]

I tried to explain my thought process if there is any issue or confusion I'd be happy to address.

[Angel]

This method is slower than my custom own method using for loops