I want to draw a circle with centre at centre of given image. Image get displayed but circle is missing. Without using hold on/off, a separate figure window shows circle, but i want the circle on the image. Plz correct the following code.
imshow(PICpng);
centx = x / 2;
centy = y / 2;
r = 10;
hold on;
theta = 0 : (2 * pi / 10000) : (2 * pi);
pline_x = r * cos(theta) + centx;
pline_y = r * sin(theta) + centy;
k = ishold;
plot(pline_x, pline_y, '*');
hold off;

5 commentaires

Jan
Jan le 18 Mar 2013
Modifié(e) : Jan le 18 Mar 2013
I have formatted youz code. Please read the instructions and use the "{} Code" button, Thanks.
What are x and y? How large is your image?
sangeeta
sangeeta le 19 Mar 2013
Modifié(e) : sangeeta le 19 Mar 2013
Thankyou Jan for the correction.
x and y are length and width of image (500, 400) Image can be of varying sizes, the one I m using is of 5ookb.
Spandan Tiwari
Spandan Tiwari le 20 Mar 2013
Modifié(e) : Spandan Tiwari le 20 Mar 2013
FYI, there's a function in the Image Processing Toolbox called VISCIRCLES which can be used for drawing circles. The basic syntax takes the centers and radii of the circles and draws them.
sangeeta
sangeeta le 21 Mar 2013
ok, Thanks Spandan
Image Analyst
Image Analyst le 21 Mar 2013
Modifié(e) : Image Analyst le 21 Mar 2013
Spandan, does the help's "See also" connect to other toolboxes? If so, then the help for rectangle(), which people usually use to draw circles, should mention viscircle().

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 Réponse acceptée

Image Analyst
Image Analyst le 18 Mar 2013

0 votes

It was working. Your x and y were probably messed up. Try this:
PICpng = imread('peppers.png');
[rows columns numberOfColorChannels] = size(PICpng)
x = columns/2
y = rows/2
imshow(PICpng);
centx = x / 2;
centy = y / 2;
r = 60;
hold on;
theta = 0 : (2 * pi / 10000) : (2 * pi);
pline_x = r * cos(theta) + centx;
pline_y = r * sin(theta) + centy;
k = ishold;
plot(pline_x, pline_y, 'r-', 'LineWidth', 3);
hold off;

10 commentaires

sangeeta
sangeeta le 19 Mar 2013
its still the same. Kindly find out the loophole.
Jan
Jan le 19 Mar 2013
Pleas explain, what is still the same. I do not get, what you are exactly asking for.
Image Analyst
Image Analyst le 19 Mar 2013
It's DEFINITELY not the same. My code works. Your code doesn't. Did you copy and paste my code? Evidently not. So let's see your new code. Did you actually make any changes to it? Like specifying values for x and y?
sangeeta
sangeeta le 20 Mar 2013
Thanks alot. Its working. Removed comments, placed image in current directory, removed some other irrelevant code...
Image Analyst
Image Analyst le 20 Mar 2013
So you did change/break my code. You mixed up x and y with rows and columns. x=columns, and y = rows. You have it the opposite way. Try
[y, x]=size(pngPIC);
sangeeta
sangeeta le 20 Mar 2013
Yes, I mixed up x and y. Thankyou.
rubina naz
rubina naz le 22 Oct 2018
thanks alot it is working...
RAKESH KUCHANA
RAKESH KUCHANA le 21 Juin 2021
Hello Image Analyst, I saw the code of drawing circle on image and it worked out. Can you explain how to identify whether the red circled area contains all white pixel area or not? Please provide code for it.
The sample images are provided for your reference.
Image Analyst
Image Analyst le 22 Juin 2021
Once you have a mask for the inside the red circle (call poly2mask() if you need to), you can do
pixelsInside = binaryImage(circleMask);
if all(pixelsInside)
% All values in mask are true/white/1
else
% At least one pixel is false/black/0.
end

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Plus de réponses (1)

Muhammad Nauman Arshad
Muhammad Nauman Arshad le 19 Fév 2020

0 votes

ICpng = imread('peppers.png');
[rows columns numberOfColorChannels] = size(PICpng)
x = columns/2
y = rows/2
imshow(PICpng);
centx = x / 2;
centy = y / 2;
r = 60;
hold on;
theta = 0 : (2 * pi / 10000) : (2 * pi);
pline_x = r * cos(theta) + centx;
pline_y = r * sin(theta) + centy;
k = ishold;
plot(pline_x, pline_y, 'r-', 'LineWidth', 3);
hold off;

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