How can I apply a logical mask to an image variable?

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dila suay
dila suay el 24 de Feb. de 2022
Comentada: Kaitlin Wang el 24 de Feb. de 2022
Hello,
I am trying to apply a logical mask to an image variable, however I couldnt manage to do it so far.
I have tried
maskedRgbImage = bsxfun(@times, rgbImage, cast(mask, 'like', rgbImage));
maskedRgbImage = rgbImage.*mask
Also, I've tried to apply mask to the rgbImage.CData directly. All of them are giving me errors. What else I can try?
Thank you so much
  2 comentarios
Walter Roberson
Walter Roberson el 24 de Feb. de 2022
Your rgbImage is a handle to a deleted image() object -- not an array of data.
dila suay
dila suay el 24 de Feb. de 2022
Hi Walter, thank you for your quick reply. I uploaded a .fig version. I hope it is useful

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Respuesta aceptada

Walter Roberson
Walter Roberson el 24 de Feb. de 2022
rgb = rgbImage.CData;
maskedImage = rgb .* cast(repmat(c, [1 1 size(rgb,3)]),'like',rgb);
This cannot be an image() object unless you want to do something like
maskedRgbImage = copyobj(rgbImage, gca);
rgb = rgbImage.CData;
maskedImage.CData = rgb .* cast(repmat(c, [1 1 size(rgb,3)]),'like',rgb);
... but that would be directly on top of the old image.
  2 comentarios
dila suay
dila suay el 24 de Feb. de 2022
Thank you so much for your quick replies Walter. It says
Unrecognized function or variable 'c'.
What is c?
Walter Roberson
Walter Roberson el 24 de Feb. de 2022
rgb = rgbImage.CData;
maskedImage = rgb .* cast(repmat(mask, [1 1 size(rgb,3)]),'like',rgb);
or
maskedRgbImage = copyobj(rgbImage, gca);
rgb = rgbImage.CData;
maskedImage.CData = rgb .* cast(repmat(mask, [1 1 size(rgb,3)]),'like',rgb);

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Más respuestas (1)

Image Analyst
Image Analyst el 24 de Feb. de 2022
You don't need to do both of these:
maskedRgbImage = bsxfun(@times, rgbImage, cast(mask, 'like', rgbImage));
maskedRgbImage = rgbImage.*mask
All you need is the first one. Don't do the second one. It's not right and would need "fixing".
  1 comentario
Kaitlin Wang
Kaitlin Wang el 24 de Feb. de 2022
sir you are my hero. i have been looking at your answers for weeks and learning so much. thank you

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