how to separate the channels of an RGB image in emgu?
I need to separate an RGB image into 3 channels. In the other words i need a code to do the following.
Image<Bgr, Byte> imgBgr = CvInvoke.Imread("Im.jpg").ToImage<Bgr, Byte>();
Mat blue = imgBgr.BlueChannel;
Mat green = imgBgr.GreenChannel;
Mat red = imgBgr.RedChannel;
Thanks!
image-processing emgucv
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
add a comment |
I need to separate an RGB image into 3 channels. In the other words i need a code to do the following.
Image<Bgr, Byte> imgBgr = CvInvoke.Imread("Im.jpg").ToImage<Bgr, Byte>();
Mat blue = imgBgr.BlueChannel;
Mat green = imgBgr.GreenChannel;
Mat red = imgBgr.RedChannel;
Thanks!
image-processing emgucv
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
add a comment |
I need to separate an RGB image into 3 channels. In the other words i need a code to do the following.
Image<Bgr, Byte> imgBgr = CvInvoke.Imread("Im.jpg").ToImage<Bgr, Byte>();
Mat blue = imgBgr.BlueChannel;
Mat green = imgBgr.GreenChannel;
Mat red = imgBgr.RedChannel;
Thanks!
image-processing emgucv
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
I need to separate an RGB image into 3 channels. In the other words i need a code to do the following.
Image<Bgr, Byte> imgBgr = CvInvoke.Imread("Im.jpg").ToImage<Bgr, Byte>();
Mat blue = imgBgr.BlueChannel;
Mat green = imgBgr.GreenChannel;
Mat red = imgBgr.RedChannel;
Thanks!
image-processing emgucv
image-processing emgucv
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
asked Nov 14 '18 at 19:09
D.GhiaseddinD.Ghiaseddin
355
355
add a comment |
add a comment |
1 Answer
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There are 2 ways to do that.
Use
imgBgr.Split(). It returns an array of 3 gray images that each image represents a single color channel of the original image.Use
imgBgr.Sub(color). It will subtract the color from the original image. For example, if you want to get red color only, remove green and blue,imgBgr.Sub(new Rgb(0, 255, 255))and so on.
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are 2 ways to do that.
Use
imgBgr.Split(). It returns an array of 3 gray images that each image represents a single color channel of the original image.Use
imgBgr.Sub(color). It will subtract the color from the original image. For example, if you want to get red color only, remove green and blue,imgBgr.Sub(new Rgb(0, 255, 255))and so on.
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
add a comment |
There are 2 ways to do that.
Use
imgBgr.Split(). It returns an array of 3 gray images that each image represents a single color channel of the original image.Use
imgBgr.Sub(color). It will subtract the color from the original image. For example, if you want to get red color only, remove green and blue,imgBgr.Sub(new Rgb(0, 255, 255))and so on.
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
add a comment |
There are 2 ways to do that.
Use
imgBgr.Split(). It returns an array of 3 gray images that each image represents a single color channel of the original image.Use
imgBgr.Sub(color). It will subtract the color from the original image. For example, if you want to get red color only, remove green and blue,imgBgr.Sub(new Rgb(0, 255, 255))and so on.
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
There are 2 ways to do that.
Use
imgBgr.Split(). It returns an array of 3 gray images that each image represents a single color channel of the original image.Use
imgBgr.Sub(color). It will subtract the color from the original image. For example, if you want to get red color only, remove green and blue,imgBgr.Sub(new Rgb(0, 255, 255))and so on.
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
answered Nov 15 '18 at 1:49
Ha BomHa Bom
1,2302518
1,2302518
add a comment |
add a comment |
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