Rotate Image segment without background
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I want to rotate image like this
I have rotated Image but I have got image like this in matlab
How to fix this problem in matlab Actually I want only rotate leaf part only not whole image. Is it posibble
4 个评论
Gopichandh Danala
2016-9-16
编辑：Gopichandh Danala
2016-9-16
A simple trick is to change all the pixels other than the segment to a white background in your case.
I cropped your main image as u didn't post original image
and sample code is here:
leaf = imread('leaf.jpg');
figure, imshow(leaf, []);
rotateImg = imrotate(leaf,45); % rotate by 45 or 315 degrees
figure,
subplot(1,2,1)
imshow(rotateImg, [min(min(rotateImg(:))) max(max(rotateImg(:)))]);
newrotateImg = rotateImg;
newrotateImg(newrotateImg == 0) = 255; % make all 0 pixels to 255..
subplot(1,2,2)
imshow(newrotateImg, [min(min(newrotateImg(:))) max(max(newrotateImg(:)))]);
Hope it helps,
Image Analyst
2016-9-16
编辑：Image Analyst
2016-9-16
Gopichandh, please post it as an answer below, not a comment, so you might get credit for it. And read this.
And use the 'bbox', 'crop' option of imrotate() to keep the final image the same size.
That method is okay if the background is perfectly uniform gray level and there are no pixels of that gray level inside the leaf. Otherwise it's not 100% robust, since it rotates the background as well as the leaf, and sets black pixels that may be inside the leaf to white.
And see my attached demo if you want to find the farthest points so you know the angle you need to rotate by.
Gopichandh Danala
2016-9-16
Thanks image analyst i will do as u suggested from now..
Malan Jayanka
2016-9-18
采纳的回答
Gopichandh Danala
2016-9-16
A simple trick is to change all the pixels other than the segment to a white background in your case.
I cropped your main image as u didn't post original image
and sample code is here:
leaf = imread('leaf.jpg');
figure, imshow(leaf, []);
rotateImg = imrotate(leaf,45,'crop'); % rotate by 45 or 315 degrees
figure, %imshow(rotateImg);
subplot(1,2,1)
imshow(rotateImg, [min(min(rotateImg(:))) max(max(rotateImg(:)))]);
newrotateImg = rotateImg;
newrotateImg(newrotateImg == 0) = 255; % make all 0 pixels to 255..
subplot(1,2,2)
imshow(newrotateImg, [min(min(newrotateImg(:))) max(max(newrotateImg(:)))]);
Hope it helps.
12 个评论
Image Analyst
2016-9-17
编辑：Image Analyst
2016-9-17
imshow(rotateImg, [min(min(rotateImg(:))) max(max(rotateImg(:)))]);
is more simply done as
imshow(rotateImg, []);
And like I said in my first comment, you can see some artifacts with this approach due to the line
newrotateImg(newrotateImg == 0) = 255;
which gives rise to funny colored spots in the output image. Let me know if you want to avoid that.
Malan Jayanka
2016-9-18
Image Analyst
2016-9-18
I don't think it was correct. Did you see the colored artifacts I mentioned? Anyway, the leaf will be the same size. There is no magnification or demagnification when doing a simple rotation on an image with square pixels.
Malan Jayanka
2016-9-18
编辑：Image Analyst
2016-9-18
Gopichandh Danala,
Thank you for your answer. Code was worked perfectly on rotation which I want to do.
But there is a small issue. When all the black pixels turn into white using this code rotateImg(rotateImg == 0) = 255;* then some leaf parts were also changed the colour.
So is there way to convert only background black pixels only?
The before image
The after image:
This is the original image:
Image Analyst
2016-9-18
编辑：Image Analyst
2016-9-18
Setting black to white won't be totally robust either. What if the stem has black in it? What if the leaf has black in it? What if the stem touches the border, so that setting only black touching the border to white is changed? That would also set the stem to white. You need to define parameters. For example, it appears that the background can now be blue instead of white, and it can have a gradient in color across the image. That will have to be taken into account. You could just find the mean color of the background and set the corner triangles to that color, but you'll have noticeable triangles in the corner because the color won't match the edges everywhere. But it's a good start.
You hopefully looked at my other leaf code in other posts. I'd probably convert to HSV color space and segment out the green by thresholding the saturation, and possibly the hue channel. Then you might need to combine it with a threshold on the V channel to get the dark brown stem. Then combine the two binary images to get a mask of the leaf and stem together. You might have to call imfill() to get rid of any internal holes in the binary image. Then get the mean RGB color of the non-leaf mask so you can fill in the triangles.
So there's a bit more to it if you want to be robust for all kinds of leaves with different colors. Ideally you'd have control over the lighting and snap a background picture. Even better, have an x-rite Color Checker chart in each photo to do color correction on it. You'll need to call functions like imfill(), imclearborder(), bwareafilt(), mean(), etc. A robust program is not going to be some simple 20 line program.
One question, it looks like you're removing the background. Do you want to keep the blue/white shaded background, or just mask it out and make it all white?
Perhaps Gopichandh Danala will do all that for you, and I don't want to jump in if he's going to do it since you asked him specifically. If not, you could open up your question to others.
Malan Jayanka
2016-9-18
编辑：Malan Jayanka
2016-9-18
Gopichandh Danala
2016-9-19
Hi Malan,
I am also a student and i would love to try what Image Analyst suggested above and will post answer when i figure it out
Gopichandh Danala
2016-9-19
编辑：Gopichandh Danala
2016-9-19
Here is the code i wrote for this It works for your requirement without changing the inside pixels
cd 'D:\MATLAB ANSWERS';
leaf = imread('leaf.jpg');
rotateImg = imrotate(leaf,45,'crop'); % rotate by 45 or 315 degrees
figure,imshow(rotateImg);
r = rotateImg(:,:,1);
g = rotateImg(:,:,2);
b = rotateImg(:,:,3);
newrotateImg = rotateImg;
[rows cols] = size(leaf(:,:,1));
for i = 1:rows
for j = 1:cols
if( r(i,j) == 0 && g(i,j) == 0 && b(i,j) == 0)
newrotateImg(i,j,1) = 255;
newrotateImg(i,j,2) = 255;
newrotateImg(i,j,3) = 255;
end
end
end
figure, imshow(newrotateImg,[]);
The only suggestion i give to you is the quality and size of the images if they are all not constant the imrotate function angle cannot be fixed..
Let me know if this is ok for your requirement
Malan Jayanka
2016-9-21
Gopichandh Danala
2016-9-21
Happy to hear that it works for your requirement..
Thanks and you too
Busy Bee
2018-2-10
what if the image segment that I want to rotate is itself made of black pixels?
Image Analyst
2018-2-10
What if you showed it to us?
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