Info
Questa domanda è chiusa. Riaprila per modificarla o per rispondere.
While loop question in row addition sums. Want row to be less then 2.
    4 visualizzazioni (ultimi 30 giorni)
  
       Mostra commenti meno recenti
    
I am generating a matrix such as this,
      1.0739    0.9517    0.1078    0.4077    0.7066    0.7983
      0.6003    1.1885    0.1747    0.3946    0.3541    0.0705
      0.7753    0.0929    0.0649    0.0188    0.0298    0.0954
      0.4378    0.1140    2.8074    0.5060    0.7248    0.2897
      0.1666    0.4086    0.3344    0.0417    0.5798    2.7909
      0.0513    1.0805    0.4291    1.3061    0.1078    0.6336
      0.5828    0.4903    2.0129    0.5368    0.2691    0.6476
      0.1921    0.0626    0.1121    0.1333    0.0121    0.0682
      0.2640    0.1722    0.3182    0.2676    1.0384    0.5087
      0.5586    0.1019    0.1335    0.4800    0.5046    1.1357
      0.2461    0.0407    0.2132    0.7918    0.1956    1.2577
      1.4788    0.0586    0.3212    1.1473    0.3263    0.2079
      0.2761    2.1830    0.6020    0.5124    0.0489    0.0253
      0.0572    0.1084    0.3595    0.1271    0.6155    0.0778
      0.2312    0.1831    0.0423    0.1079    0.6713    0.1707
      0.2744    0.6776    0.3923    1.0708    0.1322    0.1165
I also have a matrix the same size full of zeros. I want to loop trough every row and sum every value until the next value yields a value above 2.
So a row such as [1.0739 0.9517 0.1078 0.4077 0.7066 0.7983 ...] would loop trough every value and if the sum is not 2 yet, it would rewrite it in the other matrix that is now zeros.
initiate_matrix_z = zeros(16,6);
initiate_matrix_p = %the numbered matrix
for r=1:16 %rows
      while (sum <= 2) %note that all sums must be below 2
          for c=1:6    %for all values in row
              sum = initiate_matrix_p(r,c) + sum %take the value and add it to previous sum
          end
      end
At the end of this I want the following:
1.0739    0         0         0         0         0       %second value makes sum of row exceed 2
0.6003    1.1885    0.1747    0         0         0       %fourth value makes sum of row exceed 2
0.7753    0.0929    0.0649    0.0188    0.0298    0.0954  %no value makes sum(row)>2
0.4378    0.1140    0         0         0         0       %third value makes sum of row exceed 2
2 Commenti
  Image Analyst
      
      
 il 9 Apr 2017
				DON'T USE sum AS THE NAME OF YOUR VARIABLE BECAUSE IT'S THE NAME OF A VERY IMPORTANT BUILT IN VARIABLE. Now, do you NEED to use a loop, or can you use cumsum() like the vast majority of good MATLAB programmers would use?
Risposte (2)
  Image Analyst
      
      
 il 9 Apr 2017
        Try this:
m = [...
  1.0739    0.9517    0.1078    0.4077    0.7066    0.7983
  0.6003    1.1885    0.1747    0.3946    0.3541    0.0705
  0.7753    0.0929    0.0649    0.0188    0.0298    0.0954
  0.4378    0.1140    2.8074    0.5060    0.7248    0.2897
  0.1666    0.4086    0.3344    0.0417    0.5798    2.7909
  0.0513    1.0805    0.4291    1.3061    0.1078    0.6336
  0.5828    0.4903    2.0129    0.5368    0.2691    0.6476
  0.1921    0.0626    0.1121    0.1333    0.0121    0.0682
  0.2640    0.1722    0.3182    0.2676    1.0384    0.5087
  0.5586    0.1019    0.1335    0.4800    0.5046    1.1357
  0.2461    0.0407    0.2132    0.7918    0.1956    1.2577
  1.4788    0.0586    0.3212    1.1473    0.3263    0.2079
  0.2761    2.1830    0.6020    0.5124    0.0489    0.0253
  0.0572    0.1084    0.3595    0.1271    0.6155    0.0778
  0.2312    0.1831    0.0423    0.1079    0.6713    0.1707
  0.2744    0.6776    0.3923    1.0708    0.1322    0.1165]
[rows, columns] = size(m);
output = zeros(size(m));
for row = 1 : rows
  cdf = cumsum(m(row, :))
  index = find(cdf < 2, 1, 'last');
  cdf(index+1:end) = 0;
  output(row, :) = cdf;
end
output % Print to command window
0 Commenti
  Walter Roberson
      
      
 il 9 Apr 2017
        temp = cumsum( initiate_matrix_p, 2);
mask = temp < 2;
initiate_matrix_z(mask) = temp(mask);
No loop.
0 Commenti
Questa domanda è chiusa.
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


