intlut() alternative for LUT consisting of doubles

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Stefan
Stefan il 15 Ago 2012
Is there a fast alterative to intlut() accepting an array of doubles as LUT?
Problem is, I want to convert grayscale values (uint8) of an image using a LUT consisting of 256 doubles.
Example:
lut = (1:256)+0.1; % upshift by 0.1
A = uint8(randi(256,50));
% current approach
lutMat = repmat(reshape(lut, 1, 1, numel(lut)), ...
size(A));
idx = sub2ind( ...
size(lutMat), ...
repmat((1:size(A, 1))', 1, size(A, 2)), ...
repmat((1:size(A, 2)) , size(A, 1), 1), ...
double(A));
B = lutMat(idx);
The above code works fine, but takes a lot of time and needs too much memory
  3 Commenti
Stefan
Stefan il 20 Ago 2012
I forgot to add the line
lut = repmat(reshape(lut, 1, 1, numel(lut)), ...
size(A));
just after
% current approach
I changed the code in the original question, accordingly.
Image Analyst
Image Analyst il 20 Ago 2012
You're still having the problem? Didn't my code do what you want?

Accedi per commentare.

Risposte (2)

Jan
Jan il 18 Ago 2012
I'm not sure if I understand the question. Does this help:
B = lut(A);

Image Analyst
Image Analyst il 18 Ago 2012
Modificato: Image Analyst il 18 Ago 2012
If you're just adding a constant value to the lut-transformed values, like 0.1, you can just do this:
lut = uint8(1:256); % upshift by 0.1
A = uint8(randi(256,50));
B = double(intlut(A, lut)) + 0.1;
If it's not some constant added value, then you can do it like this:
lut = 1000 * rand(256, 1); % A Random reassignment between 0 and 1000.
A = uint8(randi(256,50));
B2 = zeros(size(A), class(lut));
for inputValue = uint8(0) : uint8(255)
indexes = (A == inputValue);
B2(indexes) = lut(inputValue + 1);
end

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