Simple Piece of Code for improvement
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Hi. It is known that MATLAB works slow with for loop. I have tried to vectorize the following code without success. Perhaps I am wrong with the implementation.
for I = NS2:-1:1
A = 0;
for J=1:8
A = A + KS2(J,I)*FA(J);
end
S2 = S2 + ( SS2(1,I)*sin(A) + SS2(2,I)*cos(A) );
end
where:
NS2 = 25
FA is a matrix 1x8
KS2 is a matrix 8x25
SS2 is a matrix 2x25
A is a scalar
S2 is a scalar
I try to improve it in this way:
A = 0;
J = 1:8;
for I = NS2:-1:1
A = FA(1,J)*KS2(J,I);
S2 = S2 + ( SS2(1,I)*sin(A) + SS2(2,I)*cos(A) );
end
However, the runtime for this improvement is similar to the original code.
1 Comment
Fangjun Jiang
on 23 Aug 2011
Please see my comparison post. There is a solution without any for-loop and it is much faster.
Accepted Answer
Fangjun Jiang
on 23 Aug 2011
I am amazed by Jan's insight. He seems to know without tic/toc.
t1=0;t2=0;t3=0;
for k=1:1000 % repeated run
%common data
FA=rand(1,8);KS2=rand(8,25);SS2=rand(2,25);
NS2=25;S2=0;S4=0;
%Original
tic;
for I = NS2:-1:1
A = 0;
for J=1:8
A = A + KS2(J,I)*FA(J);
end
S2 = S2 + ( SS2(1,I)*sin(A) + SS2(2,I)*cos(A) );
end
t1=t1+toc;
%Fangjun's method
tic;
A=FA*KS2;
S3=sum(sum(SS2.*[sin(A);cos(A)]));
t2=t2+toc;
if abs(S3-S2)>1e-10,error('incorrect'),end
%Jan's method
tic;
FAJ = FA(1, 1:8);
KS2J = KS2(1:8, :);
SS21 = SS2(1, :);
SS22 = SS2(2, :);
for I = NS2:-1:1
A = FAJ * KS2J(:, I);
S4 = S4 + (SS21(I) * sin(A) + SS22(I) * cos(A));
end
t3=t3+toc;
if abs(S4-S2)>1e-10,error('incorrect'),end
end
fprintf('Original time: %f\n',t1);
fprintf('Fangjun''s time: %f\n',t2);
fprintf('Jan''s time: %f\n',t3);
Original time: 0.409646
Fangjun's time: 0.022221
Jan's time: 0.136472
2 Comments
Jan
on 23 Aug 2011
Thanks, Fangjun. A nice counter-example for the "vectorization is not better than FOR loops" theory. +1
More Answers (3)
Jan
on 23 Aug 2011
Move the indexing out of the loop:
A = 0;
FAJ = FA(1, 1:8);
KS2J = KS2(1:8, :);
SS21 = SS2(1, :);
SS22 = SS2(2, :);
for I = NS2:-1:1
A = FAJ * KS2J(:, I);
S2 = S2 + (SS21(I) * sin(A) + SS22(I) * cos(A));
end
How much does this help?
In general "X(1:8)" is faster than "J=1:8; X(J)", because for the first case the boundaries are checked for the first and last element only.
3 Comments
Jan
on 23 Aug 2011
@Julian: You are right, for this case KS2J is not very helpful at first sight. I've shown this for educational reasons only. The actual acceleration is coming from using "KS2J(:, I)" instead of "KS2(J, I)" and I've limited the 1st dimension of KS2J to allow to use the colon even if the original KS2 is larger.
Fangjun Jiang
on 23 Aug 2011
See if the following gives correct result and fast speed.
A=FA*KS2;
S2=sum(sum(SS2.*[sin(A);cos(A)]));
William
on 23 Aug 2011
To speed things up make an array of zeros
zeros( however big you need it)
then perform operation on each member of the array. So basically you want to allocate space for the loop prior to doing any operations.
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