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

Xen
2015 10 11
1 个回答
回答已采纳
更新时间：2015 10 13
23 次查看（30 天）

0 个投票

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
Image Analyst 2015-10-12
That's a matter of definition. You could also say the perimeter is 2 or 4 or 0. What's the length of that object? 3 or 2? It's a matter of definition.
Xen
Xen 2015-10-12
Well, I see only one definition that gives a perimeter of 8 for those connected 3 pixels, which is not consistent with a perimeter of 2 or 4 or 0 or 58.3! That is, we consider the side of a pixel has a length of 1.
Image Analyst
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
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。

 采纳的回答

David Young
David Young 2015-10-12
编辑：David Young 2015-10-12

0 个投票

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
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。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Region and Image Properties 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by