Question

我必须分析一组图像，这些是我需要执行的操作：

求和另一组图像（在代码中称为开放光束），计算中位数并将其旋转90度;
加载一组图片，列在文件“list.txt”;
图像以3组为单位收集。对于每组，我想生成一个图像，其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;
对于每组三幅图像，从组合图像中减去开放光束中值（以1计算）（以3计算）。

考虑到使用上述过程产生的一个tif，我得到的最大值是65211，这不是相应组的三个图像的中位数3 *（我检查了像素位置）。你有什么建议，为什么会发生这种情况，以及我如何解决它？

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

Answer 1

uint16数据类型可以表示的最大整数是

>> a=100000; uint16(a)

ans =

  65535

要绕过此限制，您需要将数据重新调整为double类型，并在保存为{{uint16之前调整范围（图像对比度）以符合uint16数据类型施加的限制1}}。

使用Matlab进行图像处理 - 强度和tif图像

1 个答案: