Saving all output of for loop

for i=1:1500
   X1=J(i)-sqrt((D(i)^2)/((sqrt(A(i))+1)))
   Y1=I(i)-A(i)*(J(i)-X1)
   X2=J(i)+sqrt((D(i)^2)/((sqrt(A(i))+1)))
   Y2=I(i)+A(i)*(X2-J(i))
end

coord=[X1(:),Y1(:),X2(:),Y2(:)]

%coord
writematrix(coord,'Coord_15.csv')

I am trying to save the coordinates (X1, X2, Y1,Y2) from a for loop as shown in the code above. All inputs (I, J, A, D) are columns vectors , but i only the last iteration is saved. I will appreciate any help.

Answer 1

[Sulaymon Eshkabilov]

Hi,

Simply add (i) after your output variables, viz. X1, Y1, X2, Y2:

for i=1:1500
   X1(i)=J(i)-sqrt((D(i)^2)/((sqrt(A(i))+1)))
   Y1(i)=I(i)-A(i)*(J(i)-X1)
   X2(i)=J(i)+sqrt((D(i)^2)/((sqrt(A(i))+1)))
   Y2(i)=I(i)+A(i)*(X2-J(i))
end
coord=[X1(:),Y1(:),X2(:),Y2(:)]
%coord
writematrix(coord,'Coord_15.csv')

Good luck

[madhan ravi]

What is preallocation and why is it important for beginners to learn that?

[Tchilabalo]

Thanks for your answer. I have tried your suggestion but i am getting this error message: "Unable to perform assignment because the left and right sides have a different number of elements".

[madhan ravi]

Then why did you accept the answer if it doesn’t solve the problem??

[Sulaymon Eshkabilov]

Before commenting any point verify what you have stated. That is the correct answer. Vectorization is another solution.

Answer 2

[madhan ravi]

You don't need a loop , this is straight forward just vectorize your code:

Note: The other answer doesn‘t show the importance of preallocation.

X1=J-sqrt((D.^2)./((sqrt(A)+1)));
Y1=I-A.*(J-X1);
X2=J+sqrt((D.^2)./((sqrt(A)+1)));
Y2=I+A.*(X2-J);
coord=[X1(:),Y1(:),X2(:),Y2(:)];
writematrix(coord,'Coord_15.csv') 

[Stephen23]

+1 simple and efficient

[madhan ravi]

Thank you Stephen.