Array indices must be positive integers or logical values
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I have this MatLab code where I want to run a metropolis algorithm.
In my s rho is a matrix 12x12 and L is a matrix 12x12. m0 is a vector 1x12.
When I run my code I got the error 'Array indices must be positive integers or logical values' in hfuncval(1) where I want to compute s in the starting point.
How can come throught that?
Thank
s = -0.5.*(rho + L);
% (4) Metropolis algorithm
K = 10000; %Number of samples
pts = zeros(length(m0),K); %Array with sample output points
hfuncval = zeros(K); %Array with function value outputs
pts(:,1) = m0; %Set starting points
hfuncval(1) = s(pts(:,1)); %Compute function in starting point
step = 0.5; %Set step length
%Start sampling
for k = 1:K
[ptpert] = pts(:,k-1) + (2*rand(2,1)).*step; % propose perturbed point
hfuncpert = hfunc(ptpert); % Compute function in perturbed point
u = rand; % Generate random number in [0,1]
if u < hfuncpert/hfuncval(k-1) % accepting new points
% if u < exp(log(hfuncpert)-log(hfuncval(k-1)))
pts(:,k) = ptpert;
hfuncval(k) = hfuncpert;
else % Rejecting new points
pts(:,k) = pts(:,k-1);
hfuncval(k) = hfuncval(k-1);
end
end
The theory of the algorithm is:

6 件のコメント
Rik
2020 年 1 月 8 日
What is the value of m0? That call should also return a vector, which will not fit in your scalar index. And do you intend your output matrix to be K by K? That seems very large, especially for code that you are currently writing.
Jonas Damsbo
2020 年 1 月 8 日
Jonas Damsbo
2020 年 1 月 8 日
Rik
2020 年 1 月 9 日
You're using the values in the first column of pts as indices into s. So m0 contains one or more values you can't use as indices. Once you fix that you will notice this line is trying to store 12 values to a single position. Note that s is not a function, it is an array.
Jonas Damsbo
2020 年 1 月 9 日
Rik
2020 年 1 月 9 日
Look at the values of ptpert at that point. Use the debugging tools to step through your code line by line.
回答 (1 件)
the cyclist
2020 年 1 月 8 日
In this line:
hfuncval(k) = hfuncval(k-1);
in the first iteration of the for loop, k == 1, so you are attempting to access the "zeroth" element of the array hfuncval, which does not exist.
9 件のコメント
Jonas Damsbo
2020 年 1 月 8 日
the cyclist
2020 年 1 月 8 日
This is not valid MATLAB syntax:
(2*rand(2,1)1)
I'm not sure what you intended there, so I can't offer a solution.
Jonas Damsbo
2020 年 1 月 8 日
the cyclist
2020 年 1 月 8 日
What did you expect from the line of code I pasted, specifically?
Maybe you intended
(2*rand(2,1)-1)
and left out the minus sign?
Jonas Damsbo
2020 年 1 月 8 日
Jonas Damsbo
2020 年 1 月 8 日
Jonas Damsbo
2020 年 1 月 8 日
Jonas Damsbo
2020 年 1 月 9 日
the cyclist
2020 年 1 月 9 日
I have to admit I have not reviewed your code closely.
But the error makes it pretty clear, right? ptpert is not a positive integer, so it cannot be used as an index into a vector. For example, if ptpert is 1.5, what does sigma(1.5) represent?
It looks like the line
ptpert = pts(:,k-1) + (2*rand(12,1)-1)*step;
is where you go wrong. That doesn't look like it is going to be an integer.
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