Inconveniences working with matlabFunction

2 views (last 30 days)
Hello,
I think that I am repeating the same sort of question I bothered you yesterday but I did not expect to run into peoblem again. Consider the following:
>> syms x y
>> g=[x*y y^2;x^3 1-x^2];
>> G=matlabFunction(g)
G =
function_handle with value:
@(x,y)reshape([x.*y,x.^3,y.^2,-x.^2+1.0],[2,2])
I am not happy with this. I would prefer the 'reshape' to be removed. Why? because this is never convenient in letting me to to do vector operations. For instance, if I use the command G([1 1],[1 1]) then matlab throws an erros message. How can I tell matlab PLEASE make the following matlab function for me:
G=(x,y)[x.*y,x.^3,y.^2,-x.^2+1.0];
In my codes G (different G, of course) is the Hessian matrix and I need to use it for my optimization problem in order to get faster and more accurate solutions.
Thanks for your kind help, in advance!
Babak

Accepted Answer

Star Strider
Star Strider on 8 Sep 2022
The reshape argument simply takes the vector of argumentsin the square brackets [] and creates a matrix from them, as required. One option is to use arrayfun. However, it wil then interpret its inputs as —
GL([x(1) x(2)],[y(1) y(2)])
So changing the arguments so you and the function agree on how they should be claculated will be necessary, regardless of the funciton version you end up using —
G = @(x,y)reshape([x.*y,x.^3,y.^2,-x.^2+1.0],[2,2]); % 'matlabFunction' Result
GL = @(xv,yv) arrayfun(G, xv,yv, 'Unif',0);
V = GL([1 2],[3 4])
V = 1×2 cell array
{2×2 double} {2×2 double}
V{1}
ans = 2×2
3 9 1 0
V{2}
ans = 2×2
8 16 8 -3
G2 = @(x,y)[x.*y y.^2; x.^3 1-x.^2]; % Your Original Function (Vectorised)
V = G2([1 2],[3 4])
V = 2×4
3 8 9 16 1 8 0 -3
G2L = @(xv,yv) arrayfun(G2, xv,yv, 'Unif',0);
V2 = G2L([1 2],[3 4])
V2 = 1×2 cell array
{2×2 double} {2×2 double}
V2{1}
ans = 2×2
3 9 1 0
V2{2}
ans = 2×2
8 16 8 -3
Both end up producing the same result.
.
  6 Comments
Mohammad Shojaei Arani
Mohammad Shojaei Arani on 9 Sep 2022
Thanks Star,
Thanks for your kind help!
Babak

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!