1 个投票

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 个评论
显示 3更早的评论 隐藏 3更早的评论

Jan
Jan 2013-3-18
编辑：Jan 2013-3-18
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 2013-3-19
编辑：sangeeta 2013-3-19
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 2013-3-20
编辑：Spandan Tiwari 2013-3-20
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 2013-3-21
ok, Thanks Spandan
Image Analyst
Image Analyst 2013-3-21
编辑：Image Analyst 2013-3-21
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().

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Image Analyst
Image Analyst 2013-3-18

0 个投票

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 个评论
显示 8更早的评论 隐藏 8更早的评论

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

请先登录，再进行评论。

更多回答（1 个）

0 个投票

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;

类别

在 帮助中心 和 File Exchange 中查找有关 Particle & Nuclear Physics 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