Question

第一个矩阵表1包含标准化值。我需要对此矩阵执行一些操作，我必须获得第二个矩阵，如表2所示。

table2的对角元素应该通过给每个值赋予秩（序数值）来获得。这意味着最高元素被赋予第5等级，接下来第4等等第3等等。

通过给从4开始的每个值赋予等级而得到的表的剩余元素，因为只剩下4个元素。因为我们已经给对角元素赋予了等级。零应该保持原样。假设将table1作为矩阵，table2作为B矩阵。

  OPERATION1:For diagonal elements

  B(1,1)=5(first largest element)
  B(2,2)=1(5th largest element)
  B(3,3)=4(4th largest element)
  B(4,4)=2(2nd largest element)
  B(5,5)=3(3rd largest element)

 OPERATION2:For remaining elements
  B(1,2)=3(2nd largest element)
  B(1,3)=4(1st  largest element)
  B(1,4)=1(4th largest element)
  B(1,5)=2(3rd largest element)

等等..

Table1:
  1.0000    0.2727      0.3182     0.0455    0.2727
  0.2727   0.2727         0            0         0
  0.3182     0          0.4545     0.1818        0
   0.0455     0          0.1818     0.2727     0.0909
   0.2727     0              0     0.0909    0.3636

  Source code:
   for i = 1:5
   for j = 1:5
     if i == j
   [~, ii] = sort(diag(y));
   [~, jj] = sort(ii);
   table2 = diag(jj);
    elseif y(i,j)==0
    table2(i,j)=0;
    else
        [~, ij] = sort(y);
        [~, ij] = sort(ij);
         table2 = ij;
           end
       end
   end

I got this output for above code 
    5     0     0     0     0
    0     1     0     0     0
    0     0     4     0     0
    0     0     0     2     0
    0     0     0     0     3

但我需要低于输出。表2：

   5  3  4  1  2
   4  1  0  0  0
   4  0  4  3  0
   2  0  4  2  3
   4  0  0  3  3

Answer 1

这是一个解决方案（没有任何循环）：

E = logical(eye(size(table1))); % create a mask for the two different rules
% rule 1: diagonal elements first
table2 = zeros(size(table1)); % create result matrix
[~,jj] = sort(table1(E));
[~,ii] = sort(jj);
table2(E) = ii; % assign rank of diagonal elements
% rule 2: rest of the matrix
E = ~E;
B = reshape(table1(E),size(table1,1)-1,size(table1,2))'; % B is the matrix of table1 without diagonal elements
[~,jj] = sort(B,2); % sort along column dimension, 
[~,ii] = sort(jj,2);
table2 = table2'; % matlab is column-major, so you have to transpose the dest matrix before putting in the elements
table2(E) = reshape(ii',[],1);
table2 = table2'; % transpose back, done.
% treat zeros apart: 0 has rank 0
table2(table1==0) = 0;

某些元素可能与您的预期不同：它们共享相同的值;所以他们的排名是相当随意选择的。 Matlab具有稳定的排序，因此第一个元素的排名高于后面具有相同值的元素。

在matlab中使用sort命令进行矩阵操作

1 个答案: