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Using output of the fit function in further calculations

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Hannah
Hannah 2023년 11월 27일
댓글: Hannah 2023년 11월 28일
Hi,
I am trying to correct for the change in intensity of an image due to the lighting position. I thought that if I found the polynomial of a fitted surface I could use that to correct the image. I have used the fit function to fit a surface and have managed to generate a polynomial along with the variable values. My problem is that I can't figure out how to calculate val(x,y) for each pixel in the image using the generated polynomial and variables (which I then plan to subtract from my original image). How can I get the formula and variables back into usable code, please?
Many thanks.
Here is the code that I have been using:
backGround = im2double(imread('E3W.JPG'));
subset = backGround(1:50:3000,1:50:4096);
B = reshape(subset,[],1);
si = size(subset);
[x,y] = ndgrid(1:si(2),1:si(1));
x = reshape(x,[],1);
y = reshape(y,[],1);
sf = fit([x, y],B,'poly33');
coefficientNames = coeffnames(sf);
coefficientValues = coeffvalues(sf);
curve = formula(sf);
curve_string = convertCharsToStrings(curve);
curve_symbol = str2sym(curve_string);
curve_usable = str2func(curve_string);

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Steven Lord
Steven Lord 2023년 11월 27일
If all you want to do is evaluate the surface fit there's no need to break it into its components parts and re-assemble it. I'll show an example with a curve fit, but the principle is the same for surfaces.
load census
% Create fit using data from census.mat
F = fit(cdate, pop, 'poly2', 'Normalize', 'on')
F =
Linear model Poly2: F(x) = p1*x^2 + p2*x + p3 where x is normalized by mean 1890 and std 62.05 Coefficients (with 95% confidence bounds): p1 = 25.18 (23.58, 26.79) p2 = 75.43 (74.04, 76.83) p3 = 61.74 (59.7, 63.79)
% Evaluate fit for year 1955
p1955 = F(1955)
p1955 = 168.4030
% Plot original data, fit, and point for year 1955
plot(F);
hold on
plot(cdate, pop, 'ko', 1955, p1955, 'bx')
I'd say that fitted curve fits the data reasonably well and that the evaluated point is on the fitted curve.
  댓글 수: 1
Hannah
Hannah 2023년 11월 28일
Thanks for your help and clear explanation, that is what I needed

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Mathieu NOE
Mathieu NOE 2023년 11월 27일
hello
why not use the smoothed (low pass filtered) image as a new background and you remove that from the original image. In fact this is the same as applying a 2D high pass filter
for the smoothed image I used smoothn available here : smoothn - File Exchange - MATLAB Central (mathworks.com)
you can change the amount of smoothing if you want
backGround = im2double(imread('E3W.JPG'));
backGround2 = smoothn(backGround);
backGround3 = backGround- backGround2;
figure(1),
subplot(1,3,1),imagesc(backGround);
subplot(1,3,2),imagesc(backGround2);
subplot(1,3,3),imagesc(backGround3);

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