0
我来分析一组图像，而这些都是我需要执行如下操作：图片操纵 - 强度和TIF图像
- 总和另一组图像（称为开放光束中的代码） ，计算中位数并旋转90度;
- 加载一组图像，列在文件“list.txt”中;
- 图像已被收集在一组3组。对于每个组，我想要生成的图像的强度值是图像中位数高于某个阈值的3倍，否则等于强度值的总和;
- 对于每组三个图像，减去从所述合成图像（在3计算）
考虑TIFS之一的开放光束中位数（在1计算）使用上述方法生产的，我最大值是65211，这不是3 *相应组的三个图像的中位数（我考虑像素位置检查）。你有什么建议，为什么会发生这种情况，以及我如何解决它？
该代码报告如下。谢谢！
%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