Performance – What is the fastest way to apply functions on each line of matrix?

So far I have searched, and I know there are several ways (1, 2, 3, 4 ) So far I have used the following code:

Fv_calc(:,2) = arrayfun(@(n) MaxPositiveRoot2DegreePolynomial(QuadraticCoefficients(n,:)), 1:size(QuadraticCoefficients,1));

Among them MaxPositiveRoot2DegreePolynomial is the following function:

function root = MaxPositiveRoot2DegreePolynomial(c)
 d = c./c(1);
 root = eig([0 -d(3);1 -d(2)]);
 condition = root == abs(root);
 root = root(condition);
 if isempty(root)
 root = 0;
 end
 root = max(root);
end

Among them, QuadraticCoefficients is a 62271×3 matrix, each row contains the coefficients a, b, c of the general quadratic equation. ax ^ 2 bx c
About the problem I am solving, all the roots are real, Therefore I use a modified function to find the root so as not to waste time checking whether the root is true.
By analyzing the code, I found that the function MaxPositiveRoot2DegreePolynomial takes about 0.000047 seconds for each run, and about 2.914925371 seconds for the 62271 run. But the code< /p>

Fv_calc(:,2) = arrayfun(@(n) MaxPositiveRoot2DegreePolynomial(QuadraticCoefficients(n,:)), 1:size(QuadraticCoefficients,1));

It takes 3.198597803 to run. So about 0.283672431 is created by a This is done by rrayfun. I want to see if there is a way to reduce this time?

This is a perfect example, it shows where you should continue to use loops. The vectorized solution I want is ultimately better than your loop method Slow.

But it may depend on your Matlab version, because I have used Matlab R2015b with optimzed JIT-Compiler, in the old version, vectorized The method may be faster.

[n,~] = size(C);
%// division
d = bsxfun(@rdivide,C ,C(:,1));
%// create blocks for eigen matrix block procession
X = ([zeros(n,1) ones(n,1) -d(:,3) -d(:,2)] ).'; %'
Y = reshape(X,2,[]);
%// use block processing to get eigenvalues ​​of blocks
rt = blockproc(Y,[2 2],@(x) eig(x.data) ).'; %'
%// check if eigen values ​​are real positives
condition = rt == abs (rt);
%// output filter
out = rt(condition);

benchmark

function [t] = bench()
 %// load your data
 DATA = load('matlab');
 C = DATA.QuadraticCoefficients;

% functions to compare< br /> fcns = {
 @() thewaywewalk(C);
 @() sepideh(C);
 };

% timeit
 t = zeros(2,1);
 for ii = 1:10;
 t = t + cellfun(@timeit, fcns);
 end
end

function out = thewaywewalk(C) %thewaywewalk
 [n,~] = size(C);
 d = bsxfun(@rdivide,C,C(:,1));
 X = ([zeros(n,1) ones(n ,1) -d(:,3) -d(:,2)] ).'; %'
 Y = reshape(X,2,[]);
 rt = blockproc(Y, [2 2],@(x) eig(x.data) ).'; %'
 condition = rt == abs(rt);
 out = rt(condition);
 end
function out = sepideh(C) %sepideh
 [n,~] = size(C);
 out = zeros(n,1);
 for ii = 1 :n
 c = C(ii,:);
 d = c./c(1);
 rt = eig([0 -d(3);1 -d(2 )]);
 condition = rt == abs(rt);
 rt = rt(condition);
 if isempty(rt)
 rt = 0;
 end
 rt = max(rt);
 out(ii) = rt;
 end
end

ans =

 12.2086 %// thewaywewalk
 9.2176 %// sepideh

I still don’t see the point of the if condition.