Question

请考虑以下在for循环中执行的切线正切相关性的计算

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)

如何利用matlab矩阵计算来提高性能？

Answer 1

您可以采取几种措施来加快代码速度。首先，始终preallocate。这将转换：

theta = [];
for i = 1:(n-j)
    %...
    theta = [theta acosd(d/n1/n2)];
end

进入：

theta = zeros(1,n-j);
for i = 1:(n-j)
    %...
    theta(i) = acosd(d/n1/n2);
end

接下来，将规范化移出循环。无需一遍又一遍地标准化，只需将输入标准化：

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,:)));

这确实会在数值精度内稍微改变输出，因为不同的运算顺序会导致不同的浮点舍入误差。

最后，您可以通过对计算进行矢量化来删除内部循环：

i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));

时间（n=25）：

原始：0.0019 s
预分配：0.0013秒
标准化一次：0.0011 s
矢量化：1.4176e-04 s

时间（n=250）：

原始时间：0.0185 s
预分配：0.0146 s
标准化一次：0.0118 s
矢量化：2.5694e-04 s

请注意，矢量化代码是唯一一个其时序与n不线性增长的代码。

计时代码：

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

这是一个完整的矢量化解决方案：

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);

我的方法的八度计时的结果，方法4来自@CrisLuengo提供的答案和原始方法（n=250）：

Full vectorized    : 0.000864983 seconds
Method4(Vectorize) : 0.002774 seconds
Original(loop)     : 0.340693 seconds

矩阵形式的正切正切相关性计算MATLAB

2 个答案: