MATLAB: Very long execution time for loop “end” command

execution-timefor loopMATLABtime

MATLAB profile viewer showed me that over 20% of my run-time is spend on a loop "end" command. I understand that this line is called millions of times, but there are other lines inside that loop that take less time although they actually do something! Help please.

Best Answer

If NInj is a large number, the lines 70 and 71 are needlessly repetitive and therefore constitute a very inefficient code. If you have room for a temporary additional (NStg-1)-by-NInj matrix (D), the following code should be much faster:

   D = zeros(NStg-1,NInj);
   P = 2*NPrd+(j-1)*NInj;
   k = (1-X)/sqrt(2);
   for i = 1:NInj
     t = 0;
     for n = 2:NStg
       t = t*X+(I(n,i)+I(n-1,i))*k;
       D(n-1,i) = t;
     end
   end
   H(P+(1:NInj),P+(1:NInj)) = D.'*D;

Related Solutions

MATLAB: Avarage running time problem

Here are three ways to use tic/toc to measure the timing of a loop. Each method has pros and cons.

Method 1: Least overhead & resolution

The tic() & toc() functions require processing time, albeit a very small amount. Storing the results of toc() also requires processing time. This method contains the least overhead but you don't get the single-iteration times.

start = tic(); 
for i=1:100
    1+1;
    2+2;
end
k= toc(start)/100;

Method 2: Single iterations times

This method gives you the single-iteration times but the processing time is added by tic/toc and storing the toc output.

t = nan(1, 100); 
for i=1:100
    start = tic(); 
    1+1;
    2+2;
    t(i)=toc(start);
end
k=mean(t)

Method 3: Single iterations times, less overhead

This method removes tic() from the loop which doesn't reduce the overhead by much but it's worth sharing. The single-iteration times are computed outside of the loop.

t = nan(1, 101); 
start = tic(); 
t(1) = toc(start); 
for i=1:100
    1+1;
    2+2;
    t(i+1)=toc(start);
end
k=mean(diff(t))
% Compare that to (t(end)-t(1))/100

Reasons your timing was off

tic() returns an integer (uint64) that has meaning only for the toc function. Example: t = tic(); t = 1943576784277
toc() returns the elapsed time since the previous tic() which is very quick in your loop so you're subracting a tiny amount from a large, meaningless integer.
The 1+1 and 2+2 lines display the result of each iteration in the command window since those lines are not supressed with a semicolon (;). That takes a lot of time.
The t variable was not pre-allocated so the length of t grows with each iteration. That also takes a lot of time.

Since the computations are so simple, you'll find a wide range of mean(t) if you run that script many times. Consider using the timeit function.

MATLAB: What Is the Meaning of This Syntax

W.' is the tranpose of W, so the whole thing, B=W.'*A, means multiply the transpose of W with A.

Best Answer

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