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 Commento
Fangjun Jiang
Fangjun Jiang il 23 Ago 2011
Please see my comparison post. There is a solution without any for-loop and it is much faster.

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Risposta accettata

Fangjun Jiang
Fangjun Jiang il 23 Ago 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 Commenti
Jan
Jan il 23 Ago 2011
Thanks, Fangjun. A nice counter-example for the "vectorization is not better than FOR loops" theory. +1
Julián Francisco
Julián Francisco il 2 Nov 2011
@Jan Simon: After seeing the Fangjun's results, it does appears that the vectorization is better than FOR loops theory (almost one order of magnitude).

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Più risposte (3)

Jan
Jan il 23 Ago 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 Commenti
Julián Francisco
Julián Francisco il 23 Ago 2011
@Jan Simon: What is the difference between writing KS2J = KS2(1:8,:) or KS2? I think you are using all elements of KS2 in both cases. My opinion is the same for FA.
Jan
Jan il 23 Ago 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.

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Fangjun Jiang
Fangjun Jiang il 23 Ago 2011
See if the following gives correct result and fast speed.
A=FA*KS2;
S2=sum(sum(SS2.*[sin(A);cos(A)]));

William
William il 23 Ago 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.
  1 Commento
Julián Francisco
Julián Francisco il 23 Ago 2011
@William: Thank you for your answer. I have forgot to mention the matrices and vectors were already allocated prior to for loops.

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