Determining symbolic function values without using matlabFunction

Hello all, I am trying to evaluate a symbolic matrix at a certain point. The symbolic functions are as follows:

clc, clear all
x=sym("x",[1 3]);
f(1,1)=x(1)^2 + 2*x(2) +x(3);
f(2,1)=2*x(1) + x(2)^3 + x(2)^2;
f(3,1)=x(1) + x(2) + 2*x(3)^2;

% Finding the jacobian
for i=1:length(f)
    for j=1:length(f)
        jac(i,j)=diff(f(i), x(j)); 
    end
end

Now, I would like to evaluate the jac at x=[1 2 3].

x=[1 2 3]; % point where jac will be evaluated

It can be done by several ways.

First method:

fv1=tic;
jac_n=matlabFunction(jac);
x_n=num2cell(x);
J1=jac_n(x_n{:})
t_fv1=toc(fv1)

time taken = 0.2255 s.

Second method:

fv2=tic;
J2=[2*x(1) 2 1; 2 3*x(2)^2 2*x(3); 1 1 2] % jac matrix is written manually
t_fv2=toc(fv2

time taken = 4.2230e-04 s

First method takes way longer time than the second method. Also, the first method may not work if all x variables are not present in jac. Now, I want to use the second method as it takes very less time. But I have to write the expressions of jac manually. Is there any way that I can write the expressions of jac automatically? Also, it would be great if anybody can suggest any method faster than those two mentioned. Thank you.

Answer 1

[Star Strider]

You are creating ‘jac_n’ inside the timing block in the code you posted, and creating a cell array from ‘x’.

If you create the function first, and define the ‘x’ values as a vector (they appear in order as ‘in1’ in ‘jac_n’ in my code here), the code is much faster:

jac_n = matlabFunction(jac, 'Vars',{[x]})
x=[1 2 3];
fv1 = tic
J1 = jac_n(x)
fv2 = toc(fv1)

with:

fv2 =
     4.436000000000000e-04

and:

in1 = x;
fv3 = tic
out = reshape([in1(:,1).*2.0,2.0,1.0,2.0,in1(:,2).*2.0+in1(:,2).^2.*3.0,1.0,1.0,0.0,in1(:,3).*4.0],[3,3]);
fv4 = toc(fv3)

with:

fv2 =
     1.782000000000000e-04

Function calls are always going to be a bit slower than running the code ourside the function. The times I get are nowhere near as disparate as yours.

Running this:

jac_n = matlabFunction(jac, 'Vars',{[x]})
x=[1 2 3];
for k = 1:100
    fv1 = tic
    J1 = jac_n(x)
    fv2 = toc(fv1)
    
    in1 = x;
    fv3 = tic
    out = reshape([in1(:,1).*2.0,2.0,1.0,2.0,in1(:,2).*2.0+in1(:,2).^2.*3.0,1.0,1.0,0.0,in1(:,3).*4.0],[3,3]);
    fv4 = toc(fv3)
    ratio(k) = fv2 / fv4;
end
ratio_stats = [mean(ratio) std(ratio)]

produces:

ratio_stats =
     3.146795126810789e+00     8.189169834987492e+00

so the direct code is about three times faster than the function, however there is wide variation.

.

[M Al Mamun]

Thank you very much for your so quick answer. Actually, creating the matlabFunction in the very first line of your code takes loger time of 0.225. How can I get rid of matlabFunction and solve the problem?

[M Al Mamun]

@Star Strider, anyway, your code helped me a lot. Now I have solved the problem in my original code. Thanks a lot.

[Star Strider]

As always, my pleasure!

Creating the function using matlabFunction does take time, however it only needs to be done once.

Answer 2

[Walter Roberson]

without having done the x=[1 2 3];

xv = [1 2 3];
subs(jac, x, xv)

time about 0.0157 on my system.