Pad a matrix with additional rows and concatenate another column

Question

Bill Symolon 2018-5-3

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/398757-pad-a-matrix-with-additional-rows-and-concatenate-another-column

评论： Walter Roberson 2018-5-3

Hello, all.

I asked this question a couple of days ago and you provided some excellent help. I've taken your code and scaled it up to my full data set, but it's still not working quite right. The goal is to end up with a matrix that is 7776x7.

The code runs through without errors, but the resulting matrix is 7776x6. The 7th column (time) isn't being concatenated properly. I hoping you can tell me what I'm missing. Here is my full code.

if true
% Define initial conditions
mu = 398600.4415;
Alt = 400:400:1200;
Ree = 6378;
a = Alt + Ree;
e = 1.7E-16;
i = 20:20:60;
RAAN = 0:60:300;
omega = 0:60:180;
nu = 0:60:300;
% Calculate the orbital period in minutes
C = unique(a);
for z = 1:length(C)
  if C(z) == 6778.14
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  elseif C(z) == 7178.14
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  else
      T(z) = (sqrt((4*pi.^2)/(mu) * C(z).^3))/60;     % Period
  end
end
% Divide the orbital period into 6 equal time steps
T_step = zeros(length(T), 6);
T_step(1,:) = [0:T(1)/5:T(1)];
T_step(2,:) = [0:T(2)/5:T(2)];
T_step(3,:) = [0:T(3)/5:T(3)];
T_step = T_step';
% Nested for-loops
index = 1;
for j = 1:length(a)
  for k = 1:length(i)
      for l = 1:length(RAAN)
          for o = 1:length(omega)
              for p = 1:length(nu)
                  parameters(index,:) = [a(j) e i(k) RAAN(l) omega(o) nu(p)];
                  index = index + 1;
              end
          end
      end
  end
end
% Add the time steps to the parameters matrix
R=length(parameters);
for index=1:R
  temp = padarray(parameters(index,:), [m-1 0], 'replicate','pre');
    if parameters(index,1) == 6778.14
        temp = [temp, T_step(:,1)];
    elseif parameters(index,1) == 7178.14
        temp = [temp, T_step(:,2)];
    elseif parameters(index,1) == 7578.17
        temp = [temp, T_step(:,3)];
    end
    if(index == 1)
        M = temp;
    else
        M = [M; temp];
    end
end
parameters = M;
  end

Thanks again for any help you can provide!

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Answer 1

Walter Roberson 2018-5-3

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/398757-pad-a-matrix-with-additional-rows-and-concatenate-another-column#answer_318391

Those comparisons on the parameter values will probably never be true.

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

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Bill Symolon 2018-5-3

在 MATLAB Online 中打开

Thanks, Walter. I modified the code based on the FAQ and I'm still getting the same result. I'm sure it's operator error.

if true
% Add the time steps to the parameters matrix
[m n] = size(T_step);
R=length(parameters);
tol1 = eps(6778.14);
tol2 = eps(7178.14);
tol3 = eps(7578.14);
for index=1:R
  temp = padarray(parameters(index,:), [m-1 0], 'replicate','pre');
    if ismembertol(parameters(index,1), 6778.14, tol1)
        temp = [temp, T_step(:,1)];
    elseif ismembertol(parameters(index,1), 7178.14, tol2)
        temp = [temp, T_step(:,2)];
    elseif ismembertol(parameters(index,1), 7575.14, tol3)
        temp = [temp, T_step(:,3)];
    end
    if(index == 1)
        M = temp;
    else
        M = [M; temp];
    end
end
parameters = M;
  end

I tried two versions. "if ismembertol(parameters(index,1), 6778.14, tol1)" and "if ismembertol(parameters(index,1), tol1)"

They both ran without errors, but I'm still getting a 7776x6 matrix. I'd appreciate any other thoughts you might have.

Walter Roberson 2018-5-3

在 MATLAB Online 中打开

I suggest you debug with

vals = [6778.14, 7178.14, 7575.14];
for K = 1 : length(vals)
  [~, idx] = min( abs(parameters(:,1)-vals(K)) );
  fprintf('%closest to %g is at index %d and is %g which is a difference of %g\n', vals(K), idx, parameters(idx,1), parameters(idx,1)-vals(K));
end

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