I want to calculate the mean of the neighbors of the pixels in an image. I want to do it for all pixels not only the internal ones. That means for example that I want to calculate the mean of neighbors of the pixels (1,1) or (1,size(image,2)) or (size(image,1),size(image,2)) which means I cannot use a matrix divided by 8 as a kernel for all these pixels because for example neighbors of (1,1) are only 3 not 8. Does anyone have an idea how to do it without using 8 ifs?

 Akzeptierte Antwort

Johannes Korsawe
Johannes Korsawe am 14 Okt. 2016

3 Stimmen

Let A be your matrix.
% use the help of a bigger matrix
B=nan(size(A)+2);
B(2:end-1,2:end-1)=A;
% pre-define memory for result
result = 0*A;
% calculate!
for i=2:size(A,1)+1,
for j=2:size(A,2)+1,
tmp=B(i-1:i+1,j-1:j+1);
tmp(2,2)=nan;
result(i-1,j-1)=mean(tmp(~isnan(tmp)));
end
end

Weitere Antworten (3)

Guillaume
Guillaume am 14 Okt. 2016

1 Stimme

Well, you can certainly use a convolution for the central part. I would just use smaller convolution kernels for the edges so:
img = reshape(1:200, 10, 20); %demo image
meanimg = [mean2(img(1:2, 1:2)), conv2(img(1:2, :), ones(2,3)/6, 'valid'), mean2(img(1:2, end-1:end)); ...
conv2(img(:, 1:2), ones(3,2)/6, 'valid'), conv2(img, ones(3)/9, 'valid'), conv2(img(:, end-1:end), ones(3,2)/6, 'valid'); ...
mean2(img(end-1:end, 1:2)), conv2(img(end-1:end, :), ones(2,3)/6, 'valid'), mean2(img(end-1:end, end-1:end))]
That is one convolution for the central part, 1 convolution for each edge and just mean2 for each corner.
Image Analyst
Image Analyst am 14 Okt. 2016

1 Stimme

I'd do it a different way. I'd do a full convolution so that I can get the sums and pixel counts at each window location. Then I'd crop off the outer layer (to give an output of the same size as the original) and finally divide them. Here's my demo, with extensive comments.
% Read in sample image.
grayImage = imread('cameraman.tif');
% Make an image of 1's so we can count how many
% neighbors there are at each pixel location.
binaryImage = ones(size(grayImage));
% Define a kernel to do the summing of the images at each location.
kernel = ones(3);
% Get sum of gray levels at each window location.
% Use 'full' option so we can let the window slide out and count neighbors of edge pixels.
sumImage = conv2(double(grayImage), kernel, 'full');
% Count the pixels at each window location.
countImage = conv2(double(binaryImage), kernel, 'full');
% Get the mean by dividing the sum by the pixel count.
% but ignore the outer 1-pixel-wide layer.
meanImage = sumImage(2:end-1, 2:end-1) ./ countImage(2:end-1, 2:end-1);
Don't be afraid - the actual code is only 5 lines long.
Rose Mahmudi
Rose Mahmudi am 15 Apr. 2019

0 Stimmen

hello guys
I need help with the same question but a little diffrent.
I want to obtain all 8 neighborhood connectivity for each pixles in an image.
so after i read the image and convert it to gray level image what can I do for obtaining 8neighbor-c???
and I have another problem ... I want to use first row as neighbor for the last row and vice versa. also I want to do same for columns.
could you help me figure out pleaseeeee.
thank you very much

8 Kommentare

Image Analyst
Image Analyst am 15 Apr. 2019
It doesn't make any sense. Say what you want as an output. Is it another image? Is it an N-by-16 matrix/list of (x,y) or (row, column) coordinates for every 8 neighbors for every pixel in the image (though that seems useless)? What do you really want to do?
Rose Mahmudi
Rose Mahmudi am 15 Apr. 2019
I want a matrix 3 by 3 of each pixels in output. infact the value of graylevel 8-neighborhood for each pixle with the center pixle in a matrix 3by3.
I want to use this output for an input of fuzzy system.
why it doesn't make sens???
Image Analyst
Image Analyst am 15 Apr. 2019
It was confusing for the reasons I gave - multiple ways to interpret that. "8 neighborhood connectivity" is a concept for how you process pixels -- it's not any specific kind of set of numbers, like their gray levels, their coordinates, their histogram, their mean, or whatever.
See the attached demo. It gives you an N-by-10 matrix: row, column, then the 8 neighbor pixel gray levels. Adapt as needed. If it doesn't work, attach it with your image to a new question.
Rose Mahmudi
Rose Mahmudi am 26 Apr. 2019
thank you it works for me .
now I have a nother problem.
you know I wanted to use this output in a fuzzy system (a .fis) .the problem is when I run the code in a .m seprated file it works fast. but I need to do this code as a seprate function and then use the uotput of fuzzy system in my first code but this time it tooks about 400sec I don't know why ???
how could I fix this?is the problem with fuzzy system??? or what else?
appreciate your help :)
Image Analyst
Image Analyst am 27 Apr. 2019
I have no idea. My code should run in pretty close to the same amount of time regardless if it's a script or a function. Are you sure it's MY code you're talking about and not your fuzzy code?
Rose Mahmudi
Rose Mahmudi am 27 Apr. 2019
no your code is great . I know it's my evalfis that took so much time but I don't know why it run in a sec when it's not a function but when it's a function it takes 400 sec.
Image Analyst
Image Analyst am 27 Apr. 2019
I don't know. I don't have the Fuxxy toolbox and never run evalfis(). Good luck though.
Rose Mahmudi
Rose Mahmudi am 28 Apr. 2019
thank you very much for your code and your help. I'll try to figure it out some how.
:) best regards

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