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;

댓글 수: 5

Jan
Jan 2013년 3월 18일
편집: Jan 2013년 3월 18일
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 2013년 3월 19일
편집: sangeeta 2013년 3월 19일
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 2013년 3월 20일
편집: Spandan Tiwari 2013년 3월 20일
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 2013년 3월 21일
ok, Thanks Spandan
Image Analyst
Image Analyst 2013년 3월 21일
편집: Image Analyst 2013년 3월 21일
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().

댓글을 달려면 로그인하십시오.

 채택된 답변

Image Analyst
Image Analyst 2013년 3월 18일

0 개 추천

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;

댓글 수: 10

sangeeta
sangeeta 2013년 3월 19일
its still the same. Kindly find out the loophole.
Jan
Jan 2013년 3월 19일
Pleas explain, what is still the same. I do not get, what you are exactly asking for.
Image Analyst
Image Analyst 2013년 3월 19일
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 2013년 3월 20일
Thanks alot. Its working. Removed comments, placed image in current directory, removed some other irrelevant code...
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 2013년 3월 20일
Yes, I mixed up x and y. Thankyou.
rubina naz
rubina naz 2018년 10월 22일
thanks alot it is working...
RAKESH KUCHANA
RAKESH KUCHANA 2021년 6월 21일
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.
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

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Muhammad Nauman Arshad
Muhammad Nauman Arshad 2020년 2월 19일

0 개 추천

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;

카테고리

도움말 센터File Exchange에서 Particle & Nuclear Physics에 대해 자세히 알아보기

질문:

2013년 3월 18일

댓글:

2021년 6월 22일

Community Treasure Hunt

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

Start Hunting!

Translated by