Close Contours of Image Segments

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Dimitris M
Dimitris M on 28 May 2012
Hello
Working on a segmentation project a have fell upon a serious problem. Sometimes when my segmentation algorithm runs do not produces close contour boundaries, as it considers that the edge of the image is already a boundary but does not creates a boundary there. Is there a way to introduce this information to the segmented image ?
My problem is more complicated as I need to process one by one all the image segments so I only need the part of the boundary that creates a close contour with the segment each time.
Every boundary is exported using bwboundaries() and saved in a binary image (here occurs the problem)
Please if you have any idea how to overcome this problem let me know.
Thank you

Answers (1)

Image Analyst
Image Analyst on 28 May 2012
I've never heard of that, but you can try padarray() so that your object edges that are on the edges of the image are no longer on the edges of the image.
  3 Comments
Image Analyst
Image Analyst on 28 May 2012
Let's back up. What do you really want to do? Can you post your grayscale image, and post your binary image? Then tell me what you want to find. Because I don't see any reason at all to find boundaries of things that are already 1 pixel wide lines.
Dimitris M
Dimitris M on 28 May 2012
1)So this is my Image:
http://imm.io/qM7F
2) This is the segmentation result (on the RGB)
http://imm.io/qM89
3) This is the superimposed
http://imm.io/qM8v
As you can see the boundaries are not closed (creating polygons) as there are some gaps in the boundaries. I am know trying to export each boundary individually in an image using bwboundaries() and plot the locations on a black image. In this way I want to create polygons and calculate parameters for them.
The problem is that if I do not have closed polygons I can not calculate parameters for them. I think by superimposing the original image borders I can work this out as then I will have polygons in every case.
Am I clear or should I provide you more information ???
Thank you again for your time and patience !

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