change parameter values in anonymous function

Question

zongxian 2020-7-14

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https://ww2.mathworks.cn/matlabcentral/answers/564617-change-parameter-values-in-anonymous-function

编辑： Mark Wilson 2021-6-21

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?

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Answer 1

madhan ravi 2020-7-14

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https://ww2.mathworks.cn/matlabcentral/answers/564617-change-parameter-values-in-anonymous-function#answer_465401

编辑：madhan ravi 2020-7-14

在 MATLAB Online 中打开

 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)

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zongxian 2020-7-15

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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 2020-7-15

编辑：madhan ravi 2020-7-15

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 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}))

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Answer 2

Steven Lord 2020-7-14

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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);

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madhan ravi 2020-7-15

编辑：madhan ravi 2020-7-15

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

Mark Wilson 2021-6-21

编辑：Mark Wilson 2021-6-21

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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.

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