how to replace the following code, to get the matrix
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Hi, there! I going to structure the following matrix, which has some regulations.please see the picture below: if n=4;
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/153238/image.png)
if n=5;
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/153239/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/153240/image.png)
My current code is below, which has many 'for'loops.
if true
% code
function P1=transition_probabilityP1(n)
A1={zeros(n^2,n^2)};
for i=1:n-1
for j=i+1
A1{i,j}=diag(0.407*ones(1,n));
end;
end;
for i=2:n
for j=i-1
A1{i,j}=diag(0.298*ones(1,n-1),-1);
end
end
for i=1:n
for j=i
e=diag(0.295*ones(1,n-1),1);
if i==1
f=diag(0.298*ones(1,n));
A1{1,1}=e+f;
A1{1,1}(n,n)=0.593;
elseif i==n
g=diag(0.407*ones(1,n));
A1{n,n}=e+g;
A1{n,n}(1,1)=0.705;
A1{n,n}(n,n)=0.702;
else
A1{i,j}=e;
A1{i,j}(1,1)=0.298;
A1{i,j}(n,n)=0.295;
end;
end
end
ind=cellfun('isempty',A1);
[r,c]=find(ind==1);
for i=1:length(r)
A1{r(i),c(i)}=zeros(n,n);
end
P1=cell2mat(A1);
end
Is there anyone who know how to make the same matrix with less loops, less memory and compute faster. Generally, I want compute the matrix P with dimension larger than 3,200,000 * 3,200,000 ! Thank you!
3 commentaires
Ralf
le 13 Jan 2016
Since the Matrix contains many zeros, consider using the sparse Matrix concepts.
See help sparse.
Réponse acceptée
Brendan Hamm
le 15 Jan 2016
You really just need to make a sparse diagonal matrix. This can very easily be done with spdiags. Use the last syntax: A = spdiags(B,d,m,n), which you would be calling as: A = spdiags(B,d,n^2,n^2);. The issue now is only in setting up B and d which really just requires you to replace a few values in each of the columns of B, but this can be done by finding the appropriate indices to replace (note that not all values in B will be used, so start your counting from the bottom as this function will omit unnecessary values at the top of the matrix B). Put your head to the grindstone and I'm sure you can find the patterns necessary to do this for arbitrary values of n, without loops or cell arrays. I will do it for just for the lowest diagonal:
n = 5; % Change it to 4 and verify the code works in both cases.
B = 0.298*ones(n^2,1); % Use the calculation of wherever 0.298 came from if you wish
d = -(n+1); % The diagonal where the 1st vector of B should appear
B(end-n:-n:1) = 0;
A = spdiags(B,d,n^2,n^2); % This has the lower diagonal
Q = full(A); % For visualizing the correct results.
open Q
Now, just add the columns for the other diagonals to B and add the diagonal identifier to d. Also, never use the line of code:
if true
%code
end
true is always true so the code always runs.
6 commentaires
Stephen23
le 29 Jan 2016
Modifié(e) : Stephen23
le 29 Jan 2016
@Jason: when you click the {} Code button without selecting some text first then it creates this placeholder text:
if true
% code
end
However this is simply placeholder text: you do not need to keep it, it is just supposed to show where the code text should go. It is not markup! The best way to format code on this forum is simply to (in this order):
- select the code text
- click the {} Code button
Alternatively simply put two space characters at the beginning of every code line.
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