Matlab - 没有得到预期的结果 [英] Matlab - Not getting the expected result

问题描述

• 输入两个集合


• 使用 pdist2 取两个集合之间的距离，其中代码显示为这里。 在开始的时候会在两组之间取距离。然后，对于 second 集合，在每次迭代时，它将（i，j）位置设置为 0 。当进入下一次迭代时，它应该将下一个位置值更改为0，同时将之前设置为0的值返回到其原始值。


• 请注意， pdist2 的结果最初会以矩阵的形式返回，但是为了比较，我总结了矩阵的值比较。
  
  
  

  基于此，我写了下面的函数（注意可以使用 pdist2。 m 功能来自

   函数m = pixel_minimize_distance（x = svn81& r = 81rel =nofollow> here ） ，y）
   sum1 = 0; 
   sum2 = 0; 
   [r c] = size（y）; 
   d1 = pdist2（x，y）; 
   [r1 c1] = size（d1）; 
   for i = 1：r1 
   for j = 1：c1 
   sum1 = sum1 + d1（i，j）; 
   end 
   end 
   maximum = sum1; 
   
   for i = 1：r 
   for j = 1：c 
   o = y（i，j）
   y（i，j）= 0; 
   d2 = pdist2（x，y）; 
   [r2 c2] = size（d2）; 
   for i = 1：r2 
   for j = 1：c2 
   sum2 = sum2 + d2（i，j）; 
   end 
   end 
   $ b如果sum2> =最大值
  如果o〜= 0 
   maximum = sum2; 
   m = o; 
   end 
   
   end 
   
  如果sum2 <=最大值
   maximum = maximum; 
   end 
   y（i，j）= o; 
   end 
   end 
   end 
    

  现在，作为测试运行：
  
  

   >> A = [1 2 3; 6 5 4]; 
  >> B = [4 5 3; 7 8 1]; 
  >> pixel_minimize_distance（A，B）
   
  o = 
   
   4 
   
   
  o = 
   
   4 
   
   
  o = 
   
   1 
   
   
  o = 
   
   7 
   
   
  o = 
   
   7 
   
   
  o = 
   
   0 
   
   
   ans = 
   
   7 
    

  请看这里的答案是 7 （向下滚动，如果你看不见），而我手动计算这个期望值应该是 3 ，因为当我们将它设置为 0 距离的总和将为 142 。

  
  
  

  任何想法在代码中可能是错的？我想应该是在设置 o = y（i，j）的代码中的位置，其中 o 表示原始值 ，但真的无法解决这个问题。

  
  
  

  谢谢。
  

  解决方案

  我认为你的代码中有很多冗余的命令。我只是删除他们，没有别的。我将 m 作为 3 的值。我使用MATLAB的 pdist2 函数和欧氏距离的平方（因为这是您提供的函数的默认值）。我没有得到142的距离。

  
  
  

  以下是代码：
  
  

    function m = pixel_minimize_distance（x，y）
   [rc] = size（y）; 
   maximum =（sum（sum（pdist2（x，y））））。^ 2; ％下面解释
   
  为i = 1：r 
  为j = 1：c 
   o = y（i，j）; 
   y（i，j）= 0 
   sum2 =（sum（sum（pdist2（x，y））））^ 2; 
   
   
  如果sum2> =最大
  如果o〜= 0 
   maximum = sum2; 
   m = o; 
   end 
   end 
   
   y（i，j）= o; 
   end 
   end 
   end 
    

  输出结果为： p>
  
  

  y =
  
  

    0 5 3 
   7 8 1 
    

  
  
  
  
  

    4 0 3 
   7 8 1 
    

  y =

    4 5 0 
   7 8 1 
    

  
  
  $ =
  
  

    4 5 3 
   0 8 1 
    code> 

  y =

    4 5 3 
   7 0 1 
    

  y =



    4 5 3 
   7 8 0 
    

  m =



    3 
    

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%
说明：
您已经写下了以下代码片段：



    d2 = pdist2（x，y）; 
   [r2 c2] = size（d2）; 
   for i = 1：r2 
   for j = 1：c2 
   sum2 = sum2 + d2（i，j）; 
   end 
   end 
    

  这只是计算两组之间的距离使用 pdist2 并总结整个距离矩阵，得到一个存储在 sum2 中的值。
让我们来看看我的代码：

$ $ $ $ $ $ $ $ $ $ $ $ $ $ sum $（sum（sum（pdist2（x，y））） 。）^ 2;


  pdist2 会给出距离。第一个 sum 命令将沿着行进行求和，然后第二个将沿着列进行求和以给出矩阵中所有值的总和（这是您对两个用于循环）。现在，。^ 2 背后的原因是：
  在你的链接中原来的 pdist2 已经提供，你可以从下面的代码片段中看到：
  
  

    if（nargin< 3 || isempty（metric））;度量= 0;结束; 
   
   switch metric 
   case {0，'sqeuclidean'} 
    

  平方欧几里得是默认距离，而在MATLAB中，欧几里得距离是默认的。所以，我已经把这个词汇平分了。
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%

  
  

  I have written a function that is supposed to do the following:

  • Take as an input two sets
  • Take the distance between the two sets using pdist2 which code is shown here.
  • It will take the distance between the two sets at the beginning. Then, for the second set, at each iteration, it will set the (i,j) location to 0 and calculate the distance with this change. An, when it goes to the next iteration, it should change the next location value to '0' while at the same time return the preceding value which was set to '0' to its original value.
  • Note that the result from pdist2 originally return as a matrix, but for the comparison, I sum up the matrix values to use them for comparison.

  Based on that, I have written the following function (note that you can use the pdist2.m function from the link here):

     function m = pixel_minimize_distance(x,y)
      sum1=0;
      sum2=0;
      [r c] = size(y);
      d1 = pdist2(x,y);
      [r1 c1] = size(d1);
      for i=1:r1
      for j=1:c1
          sum1=sum1+d1(i,j);
      end
  end
  maximum = sum1;
  
  for i=1:r
  for j=1:c
  o = y(i,j)
  y(i,j) = 0;
  d2 = pdist2(x,y);
  [r2 c2] = size(d2);
  for i=1:r2
      for j=1:c2
          sum2=sum2+d2(i,j);
      end
  end
  
  if sum2 >= maximum
      if o ~= 0
      maximum = sum2;
      m = o;
      end
  
  end
  
  if sum2 <= maximum
  maximum = maximum;
  end
  y(i,j)=o;
  end
  end
  end
  

  Now, this is what I have run as a test:

  >> A=[1 2 3; 6 5 4];
  >> B=[4 5 3; 7 8 1];
  >> pixel_minimize_distance(A,B)
  
  o =
  
       4
  
  
  o =
  
       4
  
  
  o =
  
       1
  
  
  o =
  
       7
  
  
  o =
  
       7
  
  
  o =
  
       0
  
  
  ans =
  
       7
  

  See the the answer here is 7 (scroll down if you cannot see it), while the expected value when I calculate this manually should be 3, as since when we set it to 0 the sum of the distance will be 142.

  Any idea what could be wrong in the code? I think it would be in the location in the code of setting o = y(i,j) where o denotes original value, but really couldn't figure a way of solving that.

  Thanks.

  解决方案

  I think you have many redundant commands in your code. I just removed them, nothing else. I am getting value of m as 3. I used MATLAB's pdist2 function with squared euclidean distance (since that is the default in the function you provided). I did not get 142 as the distance.

  Here is the code:

  function m = pixel_minimize_distance(x,y)
  [r c] = size(y);
  maximum = (sum(sum(pdist2(x,y)))).^2; %explained below
  
  for i=1:r
      for j=1:c
          o = y(i,j);
          y(i,j) = 0
          sum2 = (sum(sum(pdist2(x,y)))).^2;
  
  
          if sum2 >= maximum
              if o ~= 0
                  maximum = sum2;
                  m = o;
              end
          end
  
          y(i,j)=o;
      end
  end
  end
  

  and output is:

  y =

   0     5     3
   7     8     1
  

  y =

   4     0     3
   7     8     1
  

  y =

   4     5     0
   7     8     1
  

  y =

   4     5     3
   0     8     1
  

  y =

   4     5     3
   7     0     1
  

  y =

   4     5     3
   7     8     0
  

  m =

   3
  

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Explanation: You have written the following snippet of code:

  d2 = pdist2(x,y);
  [r2 c2] = size(d2);
  for i=1:r2
      for j=1:c2
          sum2=sum2+d2(i,j);
      end
  end
  

  what this simply does is calculates the distance between two sets using pdist2 and sums up the entire distance matrix to come up with one value stored in sum2 in your case. Lets look at the my code:

  sum2 = (sum(sum(pdist2(x,y)))).^2;
  

  pdist2 will give the distance. First sum command will sum along the rows and then the second one will sum along the columns to give you a total of all values in the matrix (This is what you did with two for loops). Now, the reason behind .^2 is: In the original pdist2 function in the link which you have provided, you can see from the following snippet of code:

  if( nargin<3 || isempty(metric) ); metric=0; end;
  
  switch metric
    case {0,'sqeuclidean'}
  

  that squared Euclidean is the default distance, whereas in MATLAB, Euclidean distance is default. Therefore, I have squared the term. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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