change parameter values in anonymous function

My problem is one function passing to me a Anonymous function handle, for example,

f_handle = function1(a1,a2,....) 
% Assuming f_handle = @(x,y) [a*(x+y),b*(x-y)], a and be are unknown parameters

As I only get this handle, (and I know there are two unknown parameters of a and b), now I want to change (set) the parameter values of a and b, I don't know how to do it.

b=1; a=1:10;
x = 1; y =2;
for i=1:10
    % how to passing a(i) and b in the anonymouos function
    mm{i} = f_handle(x,y);
end

Should I redefine a anonymous function or by other way?

Answer 1

[madhan ravi]

 b = 1;
 a = 1:10;
 x = 1;
 y = 2;
 mm = cell(1, 10); 
 f_handle = @(x,y,a,b) [a*(x+y),b*(x-y)];
 for ii = 1:10
     mm{ii} = f_handle(x,y,a(ii),b);
 end
 celldisp(mm)

[madhan ravi]

https://www.mathworks.com/help/matlab/math/parameterizing-functions.html

[zongxian]

I'm sorry to bother you. I read the web page, but I still confusing. Your meaning is redefine a anonymous function which contains (x,y,a,b)? Can you give me a sample? Thank you.

[madhan ravi]

 b = 1;
 a = 1:10;
 x = 1;
 y = 2;
 mm = cell(1, 10); 
 f_handle = @(x,y,a,b) [a*(x+y),b*(x-y)];
 for ii = 1:10
     mm{ii} = f_handle(x,y,a(ii),b);
 end
 celldisp(mm)

[zongxian]

Maybe I didn’t make it clear, I only know f_handle (other funciton pass to me, it is not a fixed form, only know that there are two unknown parameters of a and b), now I set values for a and b, then calculation based on this f_handle, but your codes rewrite the f_handle (I think it should be auto redefine).

[madhan ravi]

 doc matlabFunction % I can’t help further  you are making a simple thing really hard to understand

[zongxian]

@madhan ravi thank you.

[madhan ravi]

 n = 1; % predefined
 f = @(x) x^n
 F = func2str(f);
 str = regexprep(F,'.*\)','');
 SYM = str2sym(str);
 "Input in the order of" + char(symvar(SYM))
 F_handle = matlabFunction(SYM)
 F_handle(2,3) % 2 assigned to n

[zongxian]

I also tried to convert it to a string, and then to an anonymous function, but it was not successful. Your way is good, but there is a problem that all variables change to input variables.

for example:

m=1;n=2;
f = @(x) x^n+m    ===>    

f= @(m,n,x)m+x.^n

If I just want to add a parameter? other parameter sill keep its values

m=1;n=2;
f = @(x) x^n+m    ===>    

m=1;
f= @(n,x)m+x.^n

Anyway, I think your method has helped me a lot.

[madhan ravi]

 n = 1;
 m = 2;
 save m
 f = @(x) x^n+m
 F = func2str(f);
 str = regexprep(F,'.*\)','');
 SYM = str2sym(str);
 syms(symvar(SYM))
 clear m
 load m
 subs(subs(SYM, {x,n},{2,3}))

Answer 2

[Steven Lord]

Once you've created an anonymous function handle that "remembers" a particular value, you cannot change that "remembered" value. You would need to recreate the function handle to change it.

n = 2;
f = @(x) x.^n; % n is locked into the value 2 at this point
y1 = f(3) % 9

n = 4;
y2 = f(3) % still 9 not 3^4 = 81

f = @(x) x.^n; % Recreating f "remembers" the new value of n
y3 = f(3) % 81

If you have a value that you want to be able to change, you should probably have the anonymous function accept it. You could, if you need a function with a specific signature, write an adapter. In the example below g accepts both x and n, while f2 and f3 fix specific values for n but make use of g to compute their results.

g = @(x, n) x.^n;
f2 = @(x) g(x, 2);
f3 = @(x) g(x, 3);

[zongxian]

If we have a anonymous function handle like your exmple (n =3; f = @(x) x.^n;), then we can not change the value of n in this handle. We can only redefine anonymous function handle after a predetermined value of n, (i.e. n = 4; f = @(x) x.^n;). Thank you.

[madhan ravi]

Not true. See https://www.mathworks.com/matlabcentral/answers/564617-change-parameter-values-in-anonymous-function#comment_936908

[Steven Lord]

In your comment you defined a new function handle F_handle. You didn't modify the existing function handle f.

[madhan ravi]

Right. I misread OPs comment. What I meant by that was there’s an alternative to overcome that problem.

[Mark Wilson]

I just found this thread while searching for help on some surprising results from odextend(sol,...) and thought I would post my experience here so others can benefit. I was running code similar to this, using ode23 with an anonymous function handle to pass extra parameters, then changing those parameters, and continuing the solution:

aux = [a b];
sol1 = ode23(@(t,y) myfun(t,y,aux),[t0 t1],0);
aux = [c d];
sol2 = odextend(sol1,[],t2);

(Here, [] as the second argument tells odextend to use the same ode function as the original solution.)

However, odextend was internally using aux values of [a b] instead of [c d]. @Steven Lord's answer here taught me that changing the values of the extra parameters in the workspace doesn't affect the function handle stored in sol.extdata.odefun and used by default by odextend(). Once I redefined the anonymous function with the updated aux, it worked.

sol2 = odextend(sol1,@(t,y) myfun(t,y,aux),t2);

This was not clear to me from the odextend documentation.