How to find the "true" perimeter of objects in a binary image?

Question

Xen 2015-10-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248035-how-to-find-the-true-perimeter-of-objects-in-a-binary-image

评论： Xen 2015-10-13

采纳的回答： David Young

在 MATLAB Online 中打开

Say, there is an object consisting of 3 pixels next to each other (■■■). If I calculate the perimeter using

sum(sum(bwperim(image)))

I get 3. But the actual perimeter (considering all sides of the object) is 8. How do I find this? Thanks.

4 个评论
显示 2更早的评论隐藏 2更早的评论

Image Analyst 2015-10-12

That's too bad. Maybe some day you'll realize and understand the other ways of considering the pixels.

Optically speaking, the pixel center-to-pixel center definition is probably the most accurate, so an image of [0,1,1,1,0] would have a length of 2 rather than 3.

Xen 2015-10-13

Thanks for the info (I guess?). Though I was expecting an explanation as to how the length of an object (which could have an abstract shape) relates to the perimeter of it, which is what I was asking in the first place, and maybe how that could give me the number 8 for that particular example.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

David Young 2015-10-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248035-how-to-find-the-true-perimeter-of-objects-in-a-binary-image#answer_195657

编辑：David Young 2015-10-12

在 MATLAB Online 中打开

One possible way is like this:

% Example binary image with 3 non-zero pixels
img = false(5);  % Use zeros(5) if you prefer
img(3, 2:4) = true; % use 1 instead of true if you prefer
% Get complement of image, with a border round it in case the
% blob is at the boundary
notimg = true(size(img)+2);
notimg(2:end-1, 2:end-1) = ~img;
% Find locations where a non-zero pixel is adjacent to a zero pixel,
% for each cardinal direction in turn
topedges =    img & notimg(1:end-2, 2:end-1);
leftedges =   img & notimg(2:end-1, 1:end-2);
bottomedges = img & notimg(3:end,   2:end-1);
rightedges =  img & notimg(2:end-1, 3:end);
% Sum each set of locations separately, then add to get total perimeter
perim = sum(topedges(:)) + sum(leftedges(:)) + ...
    + sum(bottomedges(:)) + sum(rightedges(:));
% print result
fprintf('Perimeter is %d\n', perim);

This is correct for your example. Before using it you should check it by testing on a wide range of examples. If you find a case for which it is incorrect, please describe it.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Xen 2015-10-12

编辑：Xen 2015-10-12

Many thanks David. This works perfectly for a number of examples I've tried. Of course, if the object is unfilled (with a hole) it includes the "internal perimeter", but that's an easy fix. Thanks again.

请先登录，再进行评论。