Is there a way to find the neighbourhood pixels, given a cordinate point of the pixel?
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I need to find out the neighbourhood pixel values, given a particular coordinate point in an image. the pixel point is user defined.
6 件のコメント
Image Analyst
2016 年 4 月 22 日
Sorry, I haven't worked at that task before, and I don't have any demos. I'd have to delve into it and learn it just as you would.
shikha mangal
2016 年 4 月 22 日
Ohh,well ok..i am still thankful to you for your help.If anyone find something related to the above problem please let me know it will be much helpful to me thanks.
採用された回答
Image Analyst
2014 年 9 月 5 日
neighbors(1) = img(r-1,c-1); % Upper left. r = row, c = column.
neighbors(2) = img(r-1,c); % Upper middle. r = row, c = column.
neighbors(3) = img(r-1,c+1); % Upper right. r = row, c = column.
neighbors(4) = img(r,c-1); % left. r = row, c = column.
neighbors(5) = img(r,c+1); % right. r = row, c = column.
neighbors(6) = img(r+1,c+1); % Lowerleft. r = row, c = column.
neighbors(7) = img(r+1,c); % lower middle. r = row, c = column.
neighbors(8) = img(r+1,c-1); % Lower left. r = row, c = column.
Order them in whatever order is convenient for you.
7 件のコメント
Image Analyst
2022 年 2 月 5 日
The only other way I know is
neighbors = img(r-1:r+1, c-1:c+1); % Get 3x3 array.
neighbors(5) = []; % Delete the center pixel itself.
Of course if you just want to process the array somehow, unless you're doing something really unusual, you can just use imfilter(), nlfilter(), or blockproc(). Demoa attached. Using those you don't need to manually scan the image and get the neighbors yourself since it's handled by the function.
DGM
2022 年 2 月 5 日
You should be able to do it any number of ways. This extracts the pixel and its neighbors:
inpict = randi([0 1],10);
pt = [5 5]; % row, col
imshow(inpict); hold on
plot(pt(2),pt(1),'yx','linewidth',2)
% find the sub-array
s = size(inpict);
idy = max(pt(1)-1,1):min(pt(1)+1,s(1));
idx = max(pt(2)-1,1):min(pt(2)+1,s(2));
nhood = inpict(idy,idx)
Note that the output is truncated if the selected point is on the edge of the image.
It should be asked why this is being done. If this is part of a filter process, then edge handling wouldn't be done inside the loop. If the edge handling is done elsewhere and the neighborhood is guaranteed to be 3x3, then this can be simplified:
% another way
s = size(inpict);
idx = pt.' + [-1 0 1];
nhood = inpict(idx(1,:),idx(2,:))
% or if only the neighbors are needed
nhood = nhood([1:4 6:9])
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