Creating a vector from nested for loops

Hi

So I basically want to create a column vector from two nested for loops

Initially, i planned my code to be like

for i =1:n
    for j=1:m
        P=(j-1)*n+i;
        b(P,1)=T((i-1)*dx,(j-1)*dy); %%T is an anonymous function T=@(x,y) 20*cos(x)*sin(y);
    end
end

But when I run that I get a singular row vector (rank 1) and I can't use that for later on when I need to divide it by a matrix A.

What am I doing wrong over here?

Many thanks

Answer 1

[Cris LaPierre]

Two ideas come to mind.

for i =1:n
    for j=1:m
        b(i,j)=...
    end
end
b=b(:);

Another way might be to add the new result to the end of b

b=[];
for i =1:n
    for j=1:m
        b=[b; T((i-1)*dx,(j-1)*dy)]; 
    end
end

[Maaz Madha]

I ran these two codes but they aren't working as I need mine to be 20301*1 and these aren't. Also, I need to implement the P function as that is needed.

[Cris LaPierre]

I can't debug without more info from you. What is n? m? Does nxm=20301?

You can still create P. You just don't need it for creating your column vector b.

[Maaz Madha]

%%Initialisation 1 
alpha=0.1;
beta=0.1;
dx=(2*pi)/100;
dy=dx;
dt=0.01; %%timeskip
nu=0.1;
H=2*pi;
L=4*pi;
t_values=[0,0.1,0.5,1,2,5];
T_end=5;


%Mesh 
n=round((L/dx)+1);%%lecture notes 21/1/2020. Converts rectangle to a mesh
m=round(((H/dy)+1));
t=round((T_end/dt)+1);
x=[0:dx:L]';
y=[0:dy:H]';
v=@(x,t) 5*(1-(x-2*pi)^2/(4*pi^2)).*cos(pi*t)*sin(x);  
T=@(x,y) 20*cos(x)*sin(y);
b=zeros(n*m,1);
A=zeros(n*m);

n is 201 and m is 101 so if you multiply them you get 20301. The P is a pointer function which helps map 2D matrix into a 1D one

[Cris LaPierre]

Try this. You calculation of P was incorrect for this purpose

for i =1:n
    for j=1:m
        P=(i-1)*m + j;
        b(P,1)=T((i-1)*dx,(j-1)*dy);
    end
end

[Maaz Madha]

I tried your method of P and it now shows Warning: Matrix is singular to working precision. I would also like to ask how you got that value of P as I would like to know your method of reasoning

[Cris LaPierre]

Let's walk through the code.

i=1

j=1

store the current value of T in b(1,1)

i=1

j=2

store the current value of T in b(2,1)

...

i=1

j=m

store the current value of T in b(m,1)

i=2

j=1

What row of b should the value of T be stored in? It should be m+1, right?

Plug the current values of i and j into my equation for P.

P=(2-1)*m + 1 which equals m+1

Test this for the first case (i=1,j=1).

P=(1-1)*m + 1 or 1.

The outer loop increments one, then the inner loop increments though all it's loops. You wrote P for the opposite case - the outer loop running through all it's values and then the inner loop increments one.

[Cris LaPierre]

As for P, you're doing something somewhere else that is not shown that is causing your error. You can read more about that here, but that's a separate issue that is not related to the question asked in this post.