MATLAB: Does the attempt to plot keep resulting in a “Vectors must be the same lengths” error

eigenvalueseigenvectorsmodelingplot

I have tried writing this code a couple of ways and keep producing the same error: "Plot: Vectors must be the same lengths." I am relatively new to Matlab.

    Pop0 = [0.2 0.42 1.0 1.4 5 6 7 8]
 format shortg
 for i = 1:length(Pop0)
    A = [0 0 Pop0(i); .6 0 0; 0 .75 .95]
    [v,lambda] = eig(A)
    lambdavector = max(lambda);
    for b1 = 1:length(lambdavector)
        if max(max(abs(lambda))) == abs(lambdavector(b1))
            loc = b1;
            deig = lambda(b1,b1);
        end
    end
    normv(:,i) = v(:,loc)/sum(v(:,loc));
 end
 a1 = normv(1,:);
 b1 = normv(2,:);
 c1 = normv(3,:);
 plot(Pop0,a1, 'r', Pop0,b1, 'g', Pop0,c1, 'b')
 xlabel('Adult Fecundity')
 ylabel('Equilibrium Structure')

Best Answer

Set a breakpoint at the plot line, and hover your cursor over Pop0, a1, b1, and c1. You'll see that they're not all the same length. If you don't know what I'm talking about, see this excellent link: http://blogs.mathworks.com/videos/2012/07/03/debugging-in-matlab/

That said, I copied and pasted your code and it ran fine - no error message whatsoever !

Related Solutions

MATLAB: Error using plot Vectors must be the same lengths.

Maybe some previous result are interfering with your code, try this

clear
t=[0 .3 .5 .75 .95 .99];
for i = 1:length(t)
  A = [0.0 0.0 .42;.6 0.0 0.0;0.0 0.75 t(i)];
  [v,lambda] = eig(A);
  lambdavector = max(lambda);
  for y = 1:numel(lambdavector)
      if max(abs(lambda(:))) == abs(lambdavector(y))
          loc = y;
          deig = lambda(y,y);
      end
  end
  normv(:,i) = v(:,loc)/sum(v(:,loc));
end
c = normv(1,:);
y = normv(2,:);
a = normv(3,:);
figure
plot(t,abs(c),'b.-',t,abs(y),'g.-',t,abs(a),'r.-')
xlabel('Adult Survival Rate')
ylabel('Equilibrium Population')
title('Adult Survival Changed')
legend('Calves','Yearlings','Adults')

Some of your data are complex numbers, I replaced them by their modulus

You can also replace max(max(A)) by max(A(:))

MATLAB: Vectors must be the same length (leslie matrix)

fprintf('Adult      Dominant       Equilibrium Structure    \n')
fprintf('Fecundity  Eigenvalue  Calves  Yearlings  Adults\n')
fprintf('------------------------------------------------------\n')
 f = [0.2, 0.42, 1 1.4];
% Start the loop 
for i = 1:length(f) % loop thru length of f vector
      % Construct Leslie matrix with appropriate adult fecundity
      A = [   0   0    f(i) ;
            0.6   0    0 ;
              0  0.75  0.95 ];
      % Get eigenvalues and eigenvectors
      [v,lambda] = eig(A);
  % Find dominant eigenvalue in lambda % and make note of position in lambda matrix l
  lambdavector = max(lambda); % array of eigenvalues 
  for j = 1:length(lambdavector) 
      if max(max(abs(lambda))) == abs(lambdavector(j)) 
          loc = j; % position of dom eig in lambda matrix 
          deig = lambda(j,j); % dominant eigenvalue

      end
  end
% Normalize eigenvector associated with dominant eigenvalue 
normv(:,i) = v(:,loc)/sum(v(:,loc)); % This creates a matrix called normv in which the
% iˆth column corresponds to the dominant eigenvector
      % Print these values
      fprintf('   %3.1f   ',f(i)) % adult fecundity value
      fprintf('     %5.3f   ',deig) % dominant eigenvalue
      fprintf('   %5.3f  ',normv(1,i)) % calf proprotion at eq
      fprintf('     %5.3f  ',normv(2,i)) % yearling proprotion at eq
      fprintf('   %5.3f  ',normv(3,i)) % adult proprotion at eq
      fprintf('\n') % new line    
  end
  % Generate graph 
  c = normv(1,:); % vector of calf proportions at eq 
  y = normv(2,:); % vector of yearling proportions at eq 
  a = normv(3,:); % vector of adult proprotions at eq 
  plot(f,c,'r.-',f,y,'g.-',f,a,'b.-')
  legend('calves','yearlings','adults') 
  xlabel('Adult Fecundity')
  ylabel('Equilibrium Structure')

Best Answer

Related Solutions

MATLAB: Error using plot Vectors must be the same lengths.

MATLAB: Vectors must be the same length (leslie matrix)

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