Can matlab not perform 115880 iterations fast?its taking hours to save the result of all iterations..how to iterate over all combinations and get results in minimum time

Question

summyia qamar 2017-1-12

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https://ww2.mathworks.cn/matlabcentral/answers/320156-can-matlab-not-perform-115880-iterations-fast-its-taking-hours-to-save-the-result-of-all-iterations

回答： Steven Lord 2017-1-13

I have an array all_comb_ofRoutes=<1x64 cell> and onother array allCell_array=<1x1806 cell> I am doing multiplication of each elements using this code

    [r2,c2]=size(all_comb_ofRoutes)
    count=1;
    for m=1:c2
        for n=1:numel(allCell_array)
            movement=(all_comb_ofRoutes{m})*allCell_array{n};
            movement=movement>0;
            total_movement=sum((sum(movement,2))-1);
            count=count+1;
            total(m:n)=total_movement
         end
    end
    minValue=min(total)
    [all_comb_ofRoutes(m) combination_array(n) allCell_array(n) minValue]

I want to store values of all iterations to get the minimum out of them but its taking very long time. how can I speed up this code? anyone please suggest me improvements

the complete code is given below

    tic;
      no_of_machines=7;
      no_of_cells=3;
      No_of_Parts=6;
      P1=[1  0  0  1  0  1  1];
      P2=[0  1  1  1  0  0  1];
      P3=[1  0  0  1  1  0  0];
      P4=[1  0  0  0  1  0  1];
      P5=[1  1  0  0  0  1  0;1  1  0  0  1  0  1];
      P6=[0  1  0  0  0  1  1;1  1  0  1  0  1  0];
      P=vertcat(P1,P2,P3,P4,P5,P6);
      n= 2^2-1;
      all_comb_ofRoutes = zeros(No_of_Parts,size(P1,2),n+1);
      for i=0:n
          all_comb_ofRoutes(:,:,i+1) = P((1:1:6)+(dec2bin(i,No_of_Parts)-'0'),:);
      end
      all_comb_ofRoutes = cell(1,n+1);
      for j=0:n
          all_comb_ofRoutes{j+1} = P((1:1:6)+(dec2bin(j,No_of_Parts)-'0'),:);
      end
      C = zeros(no_of_cells^no_of_machines,no_of_machines);
      t = 0;
      for k = 0:(no_of_cells^no_of_machines)-1
          s = dec2base(k,no_of_cells,no_of_machines);
          if length(unique(s))==no_of_cells
              t = t+1;
              C(t,:) = s-'0'+1;
          end
      end
      C = C(1:t,:);
      combination_array=num2cell(C,2);% all possible combinations
      %generating machine/cell matrix for above combinations
      [r1,c1]=size(combination_array)
      for l=1:r1
          R=1:numel(combination_array{l});
          Z = zeros(R(end),max(combination_array{l}));
          Z(sub2ind(size(Z),R,combination_array{l})) = 1;
          allCell_array{l}=Z; % machine cell matrix array of whole population
      end   
      %calculating total movements
      [r2,c2]=size(all_comb_ofRoutes)
      count=1;
      for m=1:c2
          for n=1:numel(allCell_array)
              movement=(all_comb_ofRoutes{m})*allCell_array{n};
              movement=movement>0;
              total_movement=sum((sum(movement,2))-1);
              count=count+1;
              total(m:n)=total_movement
           end
      end
      minValue=min(total)
      [all_comb_ofRoutes(m) combination_array(n) allCell_array(n) minValue]
      toc;

I need help in changing code for last potion % calculating total movements

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Answer 1

Jan 2017-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/320156-can-matlab-not-perform-115880-iterations-fast-its-taking-hours-to-save-the-result-of-all-iterations#answer_250462

在 MATLAB Online 中打开

Omit the useless "count=count+1;"
Do you really mean total(m:n)=total_movement, or total(m, n) = ...?
Is total pre-allocated? If not, search in this forum for "pre-allocation".

total = zeros(c2, numel(allCell_array));
for m = 1:c2
  for n = 1:numel(allCell_array)
    movement    = all_comb_ofRoutes{m} * allCell_array{n};
    total(m, n) = sum(sum(movement > 0, 2)) - size(movement, 1);
  end
end

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summyia qamar 2017-1-13

count I used to check number of iterations I want to combine the results of all iterations in one. but sir this is still taking long time I have to stop that by Ctrl+C

summyia qamar 2017-1-13

编辑：summyia qamar 2017-1-13

sir this SUM((SUM(MOVEMENTS>0,2)-1) is my objective function which means 1 minus all the movements why you replaced it with size(movement,1)? have to run this code in your software?

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Answer 2

Steven Lord 2017-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/320156-can-matlab-not-perform-115880-iterations-fast-its-taking-hours-to-save-the-result-of-all-iterations#answer_250481

Run your code in the MATLAB Profiler using a subset of the data from your larger cell arrays -- for instance, run your code using the first 10 or 20 cells in each cell array. Use the Profiler output (and any suggestions provided by Code Analyzer in the Editor window) to identify the bottlenecks in your code and potentially ways to improve those bottlenecks.