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

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John Stahura
John Stahura on 20 Sep 2019
Edited: madhan ravi on 20 Sep 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?

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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.

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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

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