accelerating code in matlab - for loop

Question

Sara 2012-7-5

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Hi Guys,

    int = zeros(1,length(x));
    if k <= 0
        int = x ;
        return
    end
    for n = 3:length(x)
          y1 = x(2:n) ;
          t1 = ((n-2):-1:0)*dt ;
         y2 = x(1:n-1) ;
          t2 = t1 +dt; 
          int(n) = sum (t1.^(k-1)/factorial(k-1).*y1 + t2.^(k-1)/factorial(k-1).*y2)*dt/2 ;
    end

is there any suggestion to write this part of code more optimal? It is used for computing Integration. The most time consuming line is int(n)= ... Whether I can change and remove the for loop here, and implement the for functionality in another way, e.g. vectors? Any suggestions would be appreciated ...

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Ryan 2012-7-5

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for n = 3:length(x ) 
    y1 = x(2:n) ; 
    t1 = ((n-2):-1:0)*dt ; 
    y2 = x(1:n-1) ; 
    t2 = t1 +dt; 
    int(n) = sum (t1.^(k-1)/factorial(k-1).*y1 + t2.^(k-1)/factorial(k-...
                 1).*y2)*dt/2 ;
end

Jan 2012-7-11

@Sara: You can delete your question using the "Delete" button left beside the question. If such a button does not appear although you are logged in, please contact an editor or files@mathworks.com and ask for deleteing the thread. Duplicate posts waste the time of the ones, who want to help you.

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Answer 1

Jan 2012-7-11

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At first I've clean up your code a bit:

if k <= 0
   int = x ;
   return
end
int = zeros(1, length(x));
fk1 = factorial(k - 1);
dt2 = dt / 2;
for n = 3:length(x)
  y1 = x(2:n);
  t1 = ((n-2):-1:0) * dt;
  y2 = x(1:n-1);
  t2 = t1 + dt; 
  int(n) = sum(t1 .^ (k-1) / fk1 .* y1 + t2 .^ (k-1) / fk1 .* y2) * dt2;
end

Now some improvements:

lenx = length(x);
fk1 = factorial(k - 1);
t1Vec = (((lenx - 2):-1:0) * dt) .^ (k-1) * dt / fk1;
int = zeros(1, lenx);
dt2 = dt / 2;
for n = 3:lenx
  y1 = x(2:n);
  t1 = t1vec(lenx-a-n:lenx-b);  % **see comment**
  y2 = x(1:n-1);
  t2 = t1 + dt; 
  int(n) = sum(t1 .* y1 + t2 .^ (k-1) / fk1 .* y2) * dt2;
end

I do not have the time to find the right constantd for a and b. Without access to Matlab I cannot simply try it, but you can. The idea is to avoid the repeated expensive calculation of t1 .^(k-1), when all elements except for one have been processed already. The same works for t2. I leave it up to you to elaborate this.

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Sara 2012-7-11

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t1Vec = (((lenx - 2):-1:0) * dt) .^ (k-1) * dt / fk1;

the second dt is required where .. (k-1)times dt...

Jan 2012-7-11

I assume this is correct, Sara. "a" and "b" are used only to replicate the limits of t1 = ((n-2):-1:0) * dt.

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Answer 2

Matt Kindig 2012-7-5

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Could you please format your code properly? It is difficult to tell which lines are commented, etc. Please edit your question to include proper line breaks, and then apply the 'code' tag (see the {} Code icon in the editor window).

Also, as a first observation, you should pre-allocate your variable 'int'--that is the reason that it is taking so long. Prior to the loop, add this line:

int = NaN(length(x), 1);

I think that this change alone will substantially improve your performance.