The image calculation found that they are all black or all white.
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bozheng
el 26 de Oct. de 2023
Respondida: Image Analyst
el 28 de Abr. de 2024
The image calculation found that they are all black or all white.
my coding
img1= imread('SCM Data111.jpg');
img = (-0.18 / (-0.28 / (45.39 /img1 - 1))+1) * 5.3;
imwrite(img, 'n_.jpg');
Can you tell me the reason why the subsequent pictures are all black and how to solve it?
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Walter Roberson
el 26 de Oct. de 2023
img = (-0.18 / (-0.28 / (45.39 /double(img1) - 1))+1) * 5.3;
When you do calculations with integer datatypes, the results of the calculations are converted to the integer data type.
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Walter Roberson
el 26 de Oct. de 2023
Your formula divides by the input, but parts of the input can be 0 (especially where there is black.) That leads to large output values in places -- but also leads to small output values for a lot of the image because the division by the large-valued components comes out small.
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/1521731/image.jpeg';
img1 = imread(filename);
min(img1(:)), max(img1(:))
img = (-0.18 / (-0.28 / (45.39 /double(img1) - 1))+1) * 5.3;
min(img(:)), max(img(:))
syms x
Q = @(v) sym(v);
imgs = (-Q(0.18) / (-Q(0.28) / (Q(45.39) /x - 1))+1) * Q(5.3)
fplot(imgs, [0 255])
ir = uint8(rescale(img, 0, 255, 'InputMin', 2.5, 'InputMax', 50));
imshow(ir); colorbar
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Image Analyst
el 28 de Abr. de 2024
img1 is an array so you need to use dot division
img = (-0.18 ./ (-0.28 ./ (45.39 ./ img1 - 1)) + 1) * 5.3;
And double check your parentheses to make sure they're correct. For example is
(45.39 ./ img1 - 1)
supposed to be
((45.39 ./ img1) - 1)
or
(45.39 ./ (img1 - 1))
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