tangent tangent correlation calculation in matrix form MATLAB - matlab

Question

Consider the following calculation of the tangent tangent correlation which is performed in a for loop
v1=rand(25,1);
v2=rand(25,1);
n=25;
nSteps=10;
mean_theta = zeros(nSteps,1);
for j=1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j)=mean(theta);
end
plot(mean_theta)
How can matlab matrix calculations be utilized to make this performance better?

Answer 1

There are several things you can do to speed up your code. First, always preallocate. This converts:
theta = [];
for i = 1:(n-j)
%...
theta = [theta acosd(d/n1/n2)];
end
into:
theta = zeros(1,n-j);
for i = 1:(n-j)
%...
theta(i) = acosd(d/n1/n2);
end
Next, move the normalization out of the loops. There is no need to normalize over and over again, just normalize the input:
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
%...
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
This does change the output very slightly, within numerical precision, because the different order of operations leads to different floating-point rounding error.
Finally, you can remove the inner loop by vectorizing that computation:
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
Timings (n=25):
Original: 0.0019 s
Preallocate: 0.0013 s
Normalize once: 0.0011 s
Vectorize: 1.4176e-04 s
Timings (n=250):
Original: 0.0185 s
Preallocate: 0.0146 s
Normalize once: 0.0118 s
Vectorize: 2.5694e-04 s
Note how the vectorized code is the only one whose timing doesn't grow linearly with n.
Timing code:
function so
n = 25;
v1 = rand(n,1);
v2 = rand(n,1);
nSteps = 10;
mean_theta1 = method1(v1,v2,nSteps);
mean_theta2 = method2(v1,v2,nSteps);
fprintf('diff method1 vs method2: %g\n',max(abs(mean_theta1(:)-mean_theta2(:))));
mean_theta3 = method3(v1,v2,nSteps);
fprintf('diff method1 vs method3: %g\n',max(abs(mean_theta1(:)-mean_theta3(:))));
mean_theta4 = method4(v1,v2,nSteps);
fprintf('diff method1 vs method4: %g\n',max(abs(mean_theta1(:)-mean_theta4(:))));
timeit(#()method1(v1,v2,nSteps))
timeit(#()method2(v1,v2,nSteps))
timeit(#()method3(v1,v2,nSteps))
timeit(#()method4(v1,v2,nSteps))
function mean_theta = method1(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j) = mean(theta);
end
function mean_theta = method2(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta(i) = acosd(d/n1/n2);
end
mean_theta(j) = mean(theta);
end
function mean_theta = method3(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
end
mean_theta(j) = mean(theta);
end
function mean_theta = method4(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
mean_theta(j) = mean(theta);
end

Answer 2

Here is a full vectorized solution:
i = 1:n-1;
j = (1:nSteps).';
ij= min(i+j,n);
a = cat(3, v1(i).', v2(i).');
b = cat(3, v1(ij), v2(ij));
d = sum(a .* b, 3);
n1 = sum(a .^ 2, 3);
n2 = sum(b .^ 2, 3);
theta = acosd(d./sqrt(n1.*n2));
idx = (1:nSteps).' <= (n-1:-1:1);
mean_theta = sum(theta .* idx ,2) ./ sum(idx,2);
Result of Octave timings for my method,method4 from the answer provided by #CrisLuengo and the original method (n=250):
Full vectorized : 0.000864983 seconds
Method4(Vectorize) : 0.002774 seconds
Original(loop) : 0.340693 seconds