simplyfing computations in arrays

2 vues (au cours des 30 derniers jours)
Sara
Sara le 4 Juil 2012
int = zeros(1,length(x));
for n = 3:length(x)
y1 = x(2:n) ;
t1 = n*dt - (2*dt:dt:n*dt) ;
y2 = x(1:n-1) ;
t2 = n*dt - (dt:dt:(n-1)*dt) ;
int(n) = sum (t1.^(k-1)/factorial(k-1).*y1 ...
+ t2.^(k-1)/factorial(k-1).*y2)*dt/2 ;
end
I want to write t1= ((n-2)*dt:dt:0dt) is it okay? and also t2=((n-1)dt:dt:dt).
Is it possible? I sew also some errors...

Réponse acceptée

Walter Roberson
Walter Roberson le 4 Juil 2012
Modifié(e) : Walter Roberson le 4 Juil 2012
Is t1= ((n-2)*dt:dt:0dt) intended to go to 0 ? If so then because n-2 is positive so you are going backwards from positive towards 0, you will need
(n-2)*dt: -dt : 0
  2 commentaires
Sara
Sara le 4 Juil 2012
Walter, Do you have any suggestion, how can I replace 'for' loop in this code to another data type e.g. vector? (New to MATLAB) Thanks in Advance
Walter Roberson
Walter Roberson le 5 Juil 2012
You cannot replace the "for" loop with a different data structure. Your loop index, "n", is used as the endpoint for a colon index, so you are using expressions with different numbers of terms for different values of "n", and expressions with differing numbers of terms depending on the loop variable, cannot be vectorized.
To have any hope of vectorization, you will need to figure out the symbolic difference between the results for adjacent values of "n", and hope that that difference consists only of constants and individual array elements.

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Plus de réponses (1)

F.
F. le 4 Juil 2012
Hello,
t1 = ( (n-2):-1:0 ) *dt ;
t2 = t1 + dt ;
no ???
  1 commentaire
Sara
Sara le 4 Juil 2012
Modifié(e) : Sara le 5 Juil 2012
for both lines , your replacements for t1 and t2 are fine

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