# How can I color an image?

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MByk 2023 年 3 月 28 日
コメント済み: Image Analyst 2023 年 3 月 29 日
Hello, I'm trying to segment an image using the DBScan algorithm, but the result is black and white. How can I convert the result into a colored image? I think, I need to use the reshape function, but I couldn't figure it out. I have little image processing knowledge. Thanks for the help.
imshow(rgbImage);
rChannel = rgbImage(:, :, 1);
gChannel = rgbImage(:, :, 2);
bChannel = rgbImage(:, :, 3);
Img = double([rChannel(:), gChannel(:), bChannel(:)]);
% m = Idx, n = Indicator for core points -> (X, Epsilon, MinPTS, Distance)
[Idx, n] = dbscan(Img, 0.3, 50,'Distance','chebychev');
imshow(Idx);

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### 採用された回答

Image Analyst 2023 年 3 月 28 日
I don't think kmeans is a good way to segment a color image. I attach a demo as proof. You'd be better off using discriminant analysis. See attached demo "Classify_RGB_Image.m". I'm not sure about dbscan but I think it might even be worse than kmeans.
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MByk 2023 年 3 月 29 日

Thanks again sir. The results of this technique seems better than DA. Do we just intuitively understand that the image is well segmented? For example, the quality of the clustering can be measured with metrics such as Calinski-Harabasz. Is there a metric used for this?
Image Analyst 2023 年 3 月 29 日
I'm not sure any mathematical metric would necessairly tell you one method is better than another. Sure, one method may give a "better" value than another but does that necessarily mean it segmented the colors more like how a real human would do it? I think in the end it comes down to a judgment call.

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### その他の回答 (1 件)

Ergin Sezgin 2023 年 3 月 28 日
Hello,
Reshape function only changes the size of an array but keeps the number of elements same. For a color image in RGB color space, you need to constitute three color channels, same sized matrices with your grayscale image. The problem is that when you convert your color image into a grayscale image, the color information is permanently lost and it is not possible to automatically recolorize. You can manually colorize it with a photo editing software but even in that case it will not be the same.
However, in your case the output doesn't seem to be an image. Instead it's an index array, representing which cluster or segment the pixels in your image belong to.
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MByk 2023 年 3 月 28 日

Cevap için teşekkürler. Bad news. The original picture is a picture of a flower. It is possible to see a black and white silhouette of a flower. However, it will be difficult to define the objects, if there were more. I just found a code for K-means clustering based image segmentation. I will try to modify it for my problem.
[m, n] = kmeans(Img, nClasses);
m = reshape(m, size(rgbImage, 1), size(rgbImage, 2));
n = n/255;
clusteredImage = label2rgb(m, n);
imshow(clusteredImage);
Image Analyst 2023 年 3 月 28 日
Like I said in my answer below, kmeans, and probably dbscan, is not a good color segmentation method. Maybe you should try rgb2ind() if you want something simple.

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