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How to prevent overwriting in loop

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lily
lily am 9 Apr. 2015
Bearbeitet: Stephen23 am 11 Apr. 2015
Hi, I have the following loop. Does anyone know how to avoid the output to be overwritten in each iteration, and instead get x1, x2, ..., x5 as output?
N=5
T=5
for n=1:N
x=NaN*ones(T,1)
c=0.3
sigma=1
rho=0.98
x(1)=c/(1-rho)
for t=1:T
x(t+1)=c+rho*x(t)+sigma*randn(1)
end
end

Akzeptierte Antwort

Adam
Adam am 9 Apr. 2015
Bearbeitet: Adam am 9 Apr. 2015
You can move a lot of that outside of your loop as a first step as follows ( never define constants in a loop ):
N = 5;
T = 5;
c = 0.3;
sigma = 1;
rho = 0.98;
x = NaN(T,1);
x(1) = c / ( 1 - rho );
for t = 1:T
x(t+1) = c+rho*x(t)+sigma*randn(1);
end
That is not the equivalent of what you did though because it isn't obvious what you are trying to do in the outer loop once all the constants get moved outside the loop.
If you want a 2-d array then you can, as a second step, create something like:
x = NaN(T,N);
randvals = randn(T-1,N);
x(1,:) = c / ( 1 - rho );
x(2:T,:) = c + rho * x(1,1) + sigma * randvals;
instead using a vectorised approach instead of a loop.
  6 Kommentare
Adam
Adam am 10 Apr. 2015
That looks sufficiently similar to your original that you should still be able to apply the vectorisation I gave above, with minor modifications possibly.
lily
lily am 10 Apr. 2015
OK I understand now. As you probably guessed, I am quite new to MATLAB and I'm struggling a fair deal. Anyway, it is working, thank you very much for your help.

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Weitere Antworten (1)

Stephen23
Stephen23 am 10 Apr. 2015
Bearbeitet: Stephen23 am 11 Apr. 2015
Here is a fully vectorized version. First define the parameters:
N = 100;
T = 10;
sigma = 1;
Generate all random steps in an array S, then force the first row of S to be all zero, and finally use cumsum to get the "paths" of each column of S:
S = sigma*randn(T,N);
S(1,:) = 0;
x = cumsum(S,1);
Then plot it just like before:
plot(x)
to produce this figure:
This produces, accepting that the "random" steps are of course different, the same thing as what the loop does:
  1 Kommentar
lily
lily am 10 Apr. 2015
Thank you very much Stephen, this really helps me understand what is going on.

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