There are 395449 ‘Inf’ values in ‘img’ (probably the result of the subtraction in the denominator of the transformation) so those are going to produce either Inf or NaN values in the mean and standard deviation. One way to deal with that is to assign all the Inf values to be NaN and then use the new fillmissing2 function (R2020a does not have it, and I am not certain what to suggest in its absence other than perhaps fillmissing and then hope for the best, since it seems to produce the same result) on each channel of ‘img’.
Then, do the statistics —
img1 = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1523076/image.jpeg');
img = (-0.18 / (-0.28 / (45.39 / double(img1) - 1)) + 1) * 5.3;
size(img)
nans = nnz(isnan(img)) % Check 'NaN' Values
infs = nnz(isinf(img)) % Check 'Inf' Values
img(isinf(img)) = NaN; % Assign 'Inf' Values To 'NaN'
imgz = zeros(size(img));
for k = 1:size(img,3)
img(:,:,k) = fillmissing2(img(:,:,k), 'nearest');
imgz(:,:,k) = fillmissing(img(:,:,k), 'nearest');
end
std_value = std2(img);
mean_value = mean2(img);
fprintf('standard deviation：%.2f\n', std_value);
fprintf('average value：%.2f\n', mean_value);
std_value = std2(imgz);
mean_value = mean2(imgz);
fprintf('standard deviation：%.2f\n', std_value);
fprintf('average value：%.2f\n', mean_value);
This uses the provided image, that includes the borders and colorbar. (I did not crop it.) You will likely get different results from your actual image (that ideally should have been provided).
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