TIF image being displayed incorrectly
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I am trying to do a very simply task; reading R, G, B, bands from a file (in TIF), and then combining them to produce a colored TIF image. When I read the bands, they appear to be of type uint16. Therefore, I convert them to uint8 before combining them (otherwise the image appears black). But after that, the combined image appears white. Why does this happen? This is my code
[red, rmap] =imread('red.tif'); %uint16
[green, gmap] =imread('green.tif'); %uint16
[blue, bmap] =imread('blue.tif'); %uint16
subplot(2,2,1); imshow(red, rmap)
subplot(2,2,2); imshow(green,gmap)
subplot(2,2,3); imshow(blue,bmap)
rgb = cat(3, uint8(red), uint8(green), uint8(blue));
subplot(2,2,4); imshow(rgb); title('Combined')
This is a sample output:
![sample output.png](https://www.mathworks.com/matlabcentral/answers/uploaded_files/206127/sample%20output.png)
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Respuestas (2)
Jan
el 27 de Feb. de 2019
uint8 converts all values above 255 to 255. So use:
rgb = im2uint8(cat(3, red, green, blue));
which scales the data accordingly.
4 comentarios
Jan
el 28 de Feb. de 2019
You can prefer double or uint8 as you like. Using doubles will allow to keep the information of uint16, while this is lost with uint8. The main idea is to avoid troubles with different types and/or index or RGB images. If you mix the types, problems like saturated images (all white or all black) can be expected. So decide for a standard format and convert all input to it:
function Img = myIMRead(File, varargin)
[Img, Map] = imread(File, varargin{:});
if ~isempty(Map) % If it is an indexed image, convert it to 3D RGB array
Img = ind2rgb(Img, Map);
end
Img = im2double(Img); % Convert it to double (or im2uint8, if you like)
end
Image Analyst
el 28 de Feb. de 2019
Did you try it without a map but using [], like
imshow(red, []); % Use bracket bracket to scale actual data to display's range without altering the data variable.
or do you require the stored pseudocolor color map? If that works, then no need to use ind2rgb() or im2uint() or im2double() or anything else.
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