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

7 views (last 30 days)
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?

Accepted Answer

thoughtGarden
thoughtGarden on 20 Sep 2019
Edited: thoughtGarden on 20 Sep 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.

More Answers (1)

madhan ravi
madhan ravi on 20 Sep 2019
Edited: madhan ravi on 20 Sep 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

Categories

Find more on Numerical Integration and Differential Equations in Help Center and File Exchange

Products


Release

R2016a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!