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How to compare the the values of a 3-d dimensional array

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Pallov Anand
Pallov Anand le 20 Jan 2023
Commenté : Pallov Anand le 20 Jan 2023
Suppose I have a 3-D array, lets say, p(1,n,k) where n varies from 0 to t_span and t_span = 0:dt:80 (dt = 0.01) and k varies from 1 to 8.
Lets say I have an energy level denoted by E and decreasing at a constant rate in the following way:
E(1) = 6000;
E_lower = 4500;
E(n+1) = E(n) + dt * (-100);
What I want is, at the instant when E < E_lower, I want to compare the values between p(1,n,1) to p(1,n,8) and I want to know the index 'k' where the lowest value of p(1,n,k) is present at that particular instant "n".
Can anyone help me in this, how to proceed.
Thanks in advance.
  2 commentaires
Dyuman Joshi
Dyuman Joshi le 20 Jan 2023
As an index n must be a natural number, which is not the case here.
The smallest value of n or the value of n at the instant E < E_lower is, n = 1502, from the relationship you wrote.
Now we have to find the minimum of p(1,n,1:8). You can use the min for that, and obtain both the minimum value and the corresponding index.
Pallov Anand
Pallov Anand le 20 Jan 2023
Thanks for your reply.
In this case, it might be 1502. But I want to write a code in such a way that, as soon as E < E_lower, then I need to compare the values between p(1,n,1:8) and get the minimum value as well as the index of the corresponding minimum value.

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Réponse acceptée

Matt J
Matt J le 20 Jan 2023
Modifié(e) : Matt J le 20 Jan 2023
n=find( 6000-100*t_span < E_lower,1);
[~,kmin]=min(p(:,n,:),[],3);
  3 commentaires
Matt J
Matt J le 20 Jan 2023
Modifié(e) : Matt J le 20 Jan 2023
Yes, it will. Did you check it?
Pallov Anand
Pallov Anand le 20 Jan 2023
It worked.....thanks a lot again.

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