How to store the results from a for loop into a matrix?

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John Stahura
John Stahura el 20 de Sept. de 2019
Editada: madhan ravi el 20 de Sept. de 2019
clear all
clc
F = 2;
x = [.1:.05:1.9];
for i = x;
theta1 = atan((2-i)/1);
theta2 = atan(1/(2-i));
theta3 = atan(1/i);
A = [-1 -cos(theta2) 0 0 0 0 0 0 0 0; %Fab
0 sin(theta2) 0 0 0 0 0 0 0 0; %Fac
1 0 -1 0 0 0 0 0 0 0; %Fbd
0 0 0 1 0 0 0 0 0 0; %Fbc
0 sin(theta1) 0 0 -1 -cos(theta3) 0 0 0 0; %Fce
0 -cos(theta1) 0 -1 0 -sin(theta3) 0 0 0 0; %Fcd
0 0 1 0 0 cos(theta3) -1 0 0 0; %Dx
0 0 0 0 0 sin(theta3) 0 1 0 0; %Fde
0 0 0 0 1 0 0 0 -1 0; %Ex
0 0 0 0 0 0 0 -1 0 1]; %Ey
b = [0 F 0 F/2 0 0 0 0 0 0]';
Reactions = A\b
end
How do i store the "Reactions" results from each iteration of the loop into one matrix?

Respuesta aceptada

thoughtGarden
thoughtGarden el 20 de Sept. de 2019
Editada: thoughtGarden el 20 de Sept. de 2019
Note that there area some poor practices in place here, but with the most minimal change to what you already have is to create a blank matrix Reactions and then add each column to that matrix as they are developed.
clear all
clc
F = 2;
x = [.1:.05:1.9];
Reactions = [];
for i = x;
theta1 = atan((2-i)/1);
theta2 = atan(1/(2-i));
theta3 = atan(1/i);
A = [-1 -cos(theta2) 0 0 0 0 0 0 0 0; %Fab
0 sin(theta2) 0 0 0 0 0 0 0 0; %Fac
1 0 -1 0 0 0 0 0 0 0; %Fbd
0 0 0 1 0 0 0 0 0 0; %Fbc
0 sin(theta1) 0 0 -1 -cos(theta3) 0 0 0 0; %Fce
0 -cos(theta1) 0 -1 0 -sin(theta3) 0 0 0 0; %Fcd
0 0 1 0 0 cos(theta3) -1 0 0 0; %Dx
0 0 0 0 0 sin(theta3) 0 1 0 0; %Fde
0 0 0 0 1 0 0 0 -1 0; %Ex
0 0 0 0 0 0 0 -1 0 1]; %Ey
b = [0 F 0 F/2 0 0 0 0 0 0]';
Reactions = [Reactions, A\b];
end
a better solution would be to loop over index value (instead of the values of x) and preallocate the matrix.
clear all
clc
F = 2;
x = [.1:.05:1.9];
Reactions = zeros(10,size(x,1));
for ii = 1:length(x)
theta1 = atan((2-x(ii))/1);
theta2 = atan(1/(2-x(ii)));
theta3 = atan(1/x(ii));
A = [-1 -cos(theta2) 0 0 0 0 0 0 0 0; %Fab
0 sin(theta2) 0 0 0 0 0 0 0 0; %Fac
1 0 -1 0 0 0 0 0 0 0; %Fbd
0 0 0 1 0 0 0 0 0 0; %Fbc
0 sin(theta1) 0 0 -1 -cos(theta3) 0 0 0 0; %Fce
0 -cos(theta1) 0 -1 0 -sin(theta3) 0 0 0 0; %Fcd
0 0 1 0 0 cos(theta3) -1 0 0 0; %Dx
0 0 0 0 0 sin(theta3) 0 1 0 0; %Fde
0 0 0 0 1 0 0 0 -1 0; %Ex
0 0 0 0 0 0 0 -1 0 1]; %Ey
b = [0 F 0 F/2 0 0 0 0 0 0]';
Reactions(:,ii) = A\b;
end
disp(Reactions)
Both are valid, but the later is more efficient and clear to future readers.
  1 comentario
madhan ravi
madhan ravi el 20 de Sept. de 2019
Editada: madhan ravi el 20 de Sept. de 2019
Preallocation is preferred over appending datas.

Iniciar sesión para comentar.

Más respuestas (1)

madhan ravi
madhan ravi el 20 de Sept. de 2019
Editada: madhan ravi el 20 de Sept. de 2019
F = 2;
x = .1:.05:1.9;
Reactions = cell(numel(x),1);
b = [0 F 0 F/2 0 0 0 0 0 0]'; % there's no point in defining it over and over again inside the loop.
for ii = 1:numel(x);
theta1 = atan((2-x(ii))/1);
theta2 = atan(1/(2-x(ii)));
theta3 = atan(1/x(ii));
A = [-1 -cos(theta2) 0 0 0 0 0 0 0 0; %Fab
0 sin(theta2) 0 0 0 0 0 0 0 0; %Fac
1 0 -1 0 0 0 0 0 0 0; %Fbd
0 0 0 1 0 0 0 0 0 0; %Fbc
0 sin(theta1) 0 0 -1 -cos(theta3) 0 0 0 0; %Fce
0 -cos(theta1) 0 -1 0 -sin(theta3) 0 0 0 0; %Fcd
0 0 1 0 0 cos(theta3) -1 0 0 0; %Dx
0 0 0 0 0 sin(theta3) 0 1 0 0; %Fde
0 0 0 0 1 0 0 0 -1 0; %Ex
0 0 0 0 0 0 0 -1 0 1]; %Ey
Reactions{ii} = A\b;
end
celldisp(Reactions)
cat(2,Reactions{:}) % if you want a matrix

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