simplyfing computations in arrays

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...

 采纳的回答

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 个评论

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
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.

请先登录,再进行评论。

更多回答(1 个)

Hello,
t1 = ( (n-2):-1:0 ) *dt ;
t2 = t1 + dt ;
no ???

1 个评论

Sara
Sara 2012-7-4
编辑:Sara 2012-7-5
for both lines , your replacements for t1 and t2 are fine

请先登录,再进行评论。

类别

帮助中心File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by