在matlab

Question

我有一个用于计算sigma的等式，其中i是1 to N的索引，*表示卷积运算，Omega是图像域。

我想用matlab代码实现它。目前，我有三个选项来实现上述等式。你能看看我的方程并对我说哪一个是正确的？我花了很多时间来看看方法之间有什么不同但我找不到。提前致谢

方法1和方法2之间的差异是方法1计算循环后的西格玛，但方法2在循环中计算它。

sigma(1:row,1:col,1:dim) = nu/d; 

是否会给出相同的结果？

=========== Matlab代码==============

方法1

     nu = 0;
     d = 0;
     I2 = I.^2;
     [row,col] = size(I);
      for i = 1:N
          KuI2 = conv2(u(:,:,i).*I2,k,'same');
          bc = b.*(c(:,:,i));
          bcKuI = -2*bc.*conv2(u(:,:,i).*I,k,'same');
          bc2Ku = bc.^2.*conv2(u(:,:,i),k,'same');

          nu = nu + sum(sum(KuI2+bcKuI+bc2Ku));
          ku = conv2(u(:,:,i),k,'same');
          d = d + sum(sum(ku));
      end
      d  = d + (d==0)*eps; 
      sigma(1:row,1:col,1:dim) = nu/d;

方法2：

  I2 = I.^2;
  [row,col] = size(I);
  for i = 1:dim
      KuI2 = conv2(u(:,:,i).*I2,k,'same');
      bc = b.*(c(:,:,i));
      bcKuI = -2*bc.*conv2(u(:,:,i).*I,k,'same');
      bc2Ku = bc.^2.*conv2(u(:,:,i),k,'same');
      nu = sum(sum(KuI2+bcKuI+bc2Ku));
      ku = conv2(u(:,:,i),k,'same');
      d =  sum(sum(ku));
      d  = d + (d==0)*eps;
      sigma(1:row,1:col,i) = nu/d;
  end

方法3：

  I2 = I.^2;
  [row,col] = size(I);
  for i = 1:dim
      KuI2 = conv2(u(:,:,i).*I2,k,'same');
      bc = b.*(c(:,:,i));
      bcKuI = -2*bc.*conv2(u(:,:,i).*I,k,'same');
      bc2Ku = bc.^2.*conv2(u(:,:,i),k,'same');

      ku = conv2(u(:,:,i),k,'same');
      d  = ku + (ku==0)*eps;
      sigma(:,:,i) = (KuI2+bcKuI+bc2Ku)./d;
  end
  sigma = sigma + (sigma==0).*eps;

Answer 1

我认为方法1假设sigma1=sigma2=...sigman因为你是在循环函数之外计算的

sigma(1:row,1:col,1:dim) = nu/d;

其中nu和d是每次迭代的累积和。

虽然，方法2显示sigma1 !=sigma 2 !=..sigman，因为每个西格玛都是在循环函数中计算的

希望有所帮助