Azzera filtri
Azzera filtri

simplyfing computations in arrays

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Sara
Sara il 4 Lug 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...

Risposta accettata

Walter Roberson
Walter Roberson il 4 Lug 2012
Modificato: Walter Roberson il 4 Lug 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 Commenti
Sara
Sara il 4 Lug 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 il 5 Lug 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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Più risposte (1)

F.
F. il 4 Lug 2012
Hello,
t1 = ( (n-2):-1:0 ) *dt ;
t2 = t1 + dt ;
no ???
  1 Commento
Sara
Sara il 4 Lug 2012
Modificato: Sara il 5 Lug 2012
for both lines , your replacements for t1 and t2 are fine

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