Filter löschen
Filter löschen

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

8 Ansichten (letzte 30 Tage)
dila suay
dila suay am 24 Feb. 2022
Kommentiert: Kaitlin Wang am 24 Feb. 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 Kommentare
Walter Roberson
Walter Roberson am 24 Feb. 2022
Your rgbImage is a handle to a deleted image() object -- not an array of data.
dila suay
dila suay am 24 Feb. 2022
Hi Walter, thank you for your quick reply. I uploaded a .fig version. I hope it is useful

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 24 Feb. 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 Kommentare
dila suay
dila suay am 24 Feb. 2022
Thank you so much for your quick replies Walter. It says
Unrecognized function or variable 'c'.
What is c?
Walter Roberson
Walter Roberson am 24 Feb. 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);

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Image Analyst
Image Analyst am 24 Feb. 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 Kommentar
Kaitlin Wang
Kaitlin Wang am 24 Feb. 2022
sir you are my hero. i have been looking at your answers for weeks and learning so much. thank you

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by