Improve the code's time of execution. I want to avoid the for loops and use element by element calculations. Do you guys have something in mind?? Thank you in advance.

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Dennis_Pana 2016-5-7

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评论： dpb 2016-5-8

for i=1:v
    for j=1:v
        for k=1:m
            if dm==k && j<=n && A~=j
                ddii=((design(i,k)-design(j,k))^2);
                summ=summ+ddii; 
            elseif dm~=k && j<=n && A~=j 
                ddii=((design(A,k)-design(j,k))^2);
                summ=summ+ddii;                                  
            end
            k=k+1;j=j;i=i;
        end
        ddcc(j,i,:)=[j i summ];
        j=j+1; i=i; k=1;summ=0;
    end
    cc(i,:)=min(nonzeros(ddcc(:,i,3))); 
    i=i+1;j=1;k=1;
end

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dpb 2016-5-8

What're dm, n and A?

Dennis_Pana 2016-5-8

编辑：Dennis_Pana 2016-5-8

All this variables are integers, for example dm stands for dimensions, n number of rows and A is an integer too. All the integers are greater or equal to 1.

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Answer 1

dpb 2016-5-8

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在 MATLAB Online 中打开

See if the following produces correct output, first; should reduce the time slightly by eliminating redundant tests and shortening the loop on j. Whether it's significant will depend how many cases get missed so is data-dependent.

for i=1:v
  for j=1:n
    if j==A, continue, end
    for k=1:n
      if k==dm 
        ddii=((design(i,k)-design(j,k))^2);
        summ=summ+ddii; 
      else
        ddii=((design(A,k)-design(j,k))^2);
        summ=summ+ddii;                                  
      end
    end
    ddcc(j,i,:)=[j i summ];
    summ=0;
  end
  cc(i,:)=min(nonzeros(ddcc(:,i,3))); 
end

After that, you're walking thru the arrays in the opposite storage as to sequential addressing; whether this is a major issue has to do with stride and cache size, etc., but might try using

design=design.';

before beginning the loop and see if it makes any difference in speed when the first index gets incremented sequentially in the innermost loop.

After that, I'd have to think more than I wish to expend on a Sunday afternoon to see if could do better on the actual computation altho it would seem with some consideration a logical addressing expression and pdist|pdist2 should be able to help.

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Dennis_Pana 2016-5-8

I checked the code you sent me and it gives the right results. The execution time is the same or almost the same with what i have had before. Thank you for your effort.

I do not understand what you mean by using design=design.'; I tried to use it before the loops but there is a size problem during the calculations.

dpb 2016-5-8

在 MATLAB Online 中打开

...
for k=1:n
  if k==dm 
    ddii=((design(i,k)-design(j,k))^2);
    ...

Note that k in both references in the array is the column index and is the innermost loop. This means the stride between every access is size(design,1), not sequential if were design(k,i) instead. I had a brain cramp on the transpose operator by itself, though, you would need to reference the indices in the opposite order as well, and if the array isn't square as I figured was, then you have to rearrange the loop limits as well. But, the big idea is that should iterate over the leftmost array index first to process the array in sequential order.

Still think you should be able to do something with the pdistX functions and logical addressing, though...

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Improve the code's time of execution. I want to avoid the for loops and use element by element calculations. Do you guys have something in mind?? Thank you in advance.

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Improve the code's time of execution. I want to avoid the for loops and use element by element calculations. Do you guys have something in mind?? Thank you in advance.

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