remove boundary contours from the image

11 vues (au cours des 30 derniers jours)
Mammadbaghir Baghirzade
Mammadbaghir Baghirzade le 19 Jan 2020
Hi, I have an attached image, and I want to delete everything outside the middle circle.
From its binary data, I tried some implementations by replacing some 255 values with 0.
And I want to know if there is a specific way of removing specified contours.
Thank you for your time in advance.
Best regards
MB

Réponses (1)

Image Analyst
Image Analyst le 19 Jan 2020
Do you know the diameter of the inner circle? If so, just use the FAQ to create a circle mask, and mask it away
mask(circleMask) = false;
Can I see the original image that you made this edge image from? Because it might be possible to get a mask using thresholding instead of edge detection. Edge detection is usually NOT the first thing you want to do to an image. For most images thresholding is the way to do, with phase contrast and DIC microscopy images being one exception where you might want to do edge detection.
  35 commentaires
Image Analyst
Image Analyst le 24 Jan 2020
Yes. The default spacing, if nothing is specified otherwise, is 1 pixel. It interpolates it because chances are that the locations will not fall exactly on pixel centers (unless the line is perfectly along a row or column).
Mammadbaghir Baghirzade
Mammadbaghir Baghirzade le 26 Jan 2020
I understand thank you very much.
To make sure that I count the pixels along the horizontal line, I run the following script.
[x,y] = ginput(2);
xline = [x(1), x(2)];
yline = [y(1), y(1)];
w=improfile(I, xline, yline);
hold on
imshow(I)
hold on
%show a line
line ([xline(1), xline(2)],[yline(1), yline(2)]);
figure(1);line([0 1],[0 1]);
Because of curiosity I run both w=improfile(I, xline, yline) and w=improfile(I, x, y).
And it gave me the same number of pixels for xline and x from w=improfile(I, xline, yline) and w=improfile(I, x, y).
If going with w=improfile(I, xline, yline) is not correct, do you have any suggestion that wil count the pixels only along horizontal line.

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