0
我來分析一組圖像，而這些都是我需要執行如下操作：圖片操縱 - 強度和TIF圖像
- 總和另一組圖像（稱爲開放光束中的代碼） ，計算中位數並旋轉90度;
- 加載一組圖像，列在文件「list.txt」中;
- 圖像已被收集在一組3組。對於每個組，我想要生成的圖像的強度值是圖像中位數高於某個閾值的3倍，否則等於強度值的總和;
- 對於每組三個圖像，減去從所述合成圖像（在3計算）
考慮TIFS之一的開放光束中位數（在1計算）使用上述方法生產的，我最大值是65211，這不是3 *相應組的三個圖像的中位數（我考慮像素位置檢查）。你有什麼建議，爲什麼會發生這種情況，以及我如何解決它？
該代碼報告如下。謝謝！
%Here we calculate the average for the open beam
clear;
j = 0;
for i=1:5
s = sprintf('/Users/Alberto/Desktop/Midi/17_OB_2.75/midi_%04i.fits',i);
j = j+1;
A(j,:,:) = uint16(fitsread(s));
end
OB_median = median(A,1);
OB_median = squeeze(OB_median);
OB_median_rot=rot90(OB_median);
%Here we calculate, for each projection, the average value from the three datasets
%Read list of images from text file
fid = fopen('/Users/Alberto/Desktop/Midi/list.txt', 'r');
a = textscan(fid, '%s');
fclose(fid);
%load images
j = 0;
for i = 1:1:42 %556 entries; 543 valid values
s = sprintf('/Users/Alberto/Desktop/Midi/%s',a{1,1}{i,1});
j = j+1;
A(j,:,:) = uint16(fitsread(s));
end
threshold = 80 %This is a discretional number. I put it after noticing
%that we get the same number of pixels with a value >100 if we use 80 or 50.
k = 0;
for ii = 1:3:42
N(1,:,:) = A(ii,:,:);
N(2,:,:) = A(ii+1,:,:);
N(3,:,:) = A(ii+2,:,:);
median_N = median(N,1);
median_N = squeeze(median_N);
B(:,:) = zeros(2160,2592);
for i = 1:1:2160
for j = 1:1:2592
RMS(i,j) = sqrt((double(N(1,i,j).^2) + double(N(2,i,j).^2) + double(N(3,i,j).^2))/3);
if RMS(i,j) > threshold
%B(i,j) = 30;
B(i,j) = 3*median_N(i,j);
else
B(i,j) = A(ii,i,j) + A(ii+1,i,j) + A(ii+2,i,j);
%B(i,j) = A(ii,i,j);
end
end
end
k = k+1;
filename = sprintf('/Users/Alberto/Desktop/Midi/Edited_images/Despeckled_images/despeckled_image_%03i.tif',k);
%Now we rotate the matrix
B_rot=rot90(B);
imwrite(B_rot, filename);
%imwrite(uint16(B_rot), filename);
%Now we subtract the OB median
B_final_rot = double(B_rot) - 3*double(OB_median_rot);
filename = sprintf('/Users/Alberto/Desktop/Midi/Edited_images/Final_image/final_image_%03i.tif',k);
imwrite(uint16(B_final_rot), filename);
end