anonymous function with a variable number of input variables

Hi!
I have an anonymous function:
ht = @(A1,A2,A3,....) ...
meaning that I don't know in advance how many variables Ai are included in the definition of ht. Then I need to minimise ht in respect of the variables Ai. For example, for 2 indipendent variables I get a result with:
ht2 = @(A)ht(A(1),A(2));
A0 = [1,1];
[res,fval] = fminunc(ht2,A0)
How can I define ht2 dinamically with a variable number of unknown Ai?
Thanks!

 Respuesta aceptada

Ameer Hamza
Ameer Hamza el 10 de Mayo de 2020
Editada: Ameer Hamza el 11 de Mayo de 2020
See nargin()
f = @(x,y,z) x+y+z;
n = nargin(f)
Output
n =
3
Edit 2:
To avoid helper functions, you can use the following code.
Create the ht2 with all variables.
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
ht2 = @(x) ht(struct('x', num2cell(x)).x);
x_sol = fmincon(ht2, rand(1,3))
Result:
x_sol =
1.0e-08 *
-0.7647 -0.6963 -0.3199
Create ht2 for single variable
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
i = 2;
xs = [1 2];
ht2 = @(x) ht(struct('x', num2cell(xs(1:i-1))).x, x, struct('x', num2cell(xs(i:end))).x);
x_sol = fmincon(ht2, rand(1))
Result:
>> x_sol
x_sol =
-1.1660e-09
Edit 1:
To create ht2 on runtime, you will need to define a helper function like this
function y = helperFun(x, fun_handel)
x = num2cell(x);
y = fun_handel(x{:});
end
Then you can use it to create ht2 which is compatible for optimization functions, e.g., fmincon
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
ht2 = @(x) helperFun(x, ht);
x_sol = fmincon(ht2, rand(1,3))
Result
x_sol =
1.0e-08 *
-0.4592 -0.8045 0.0830
If you want to optimize a particular variable (say x2), then you can modify the helper function like this. You will need to specify the value of variables you don't want to optimize
function y = helperFun(x, fun_handel, i, val)
% x: optimization variable
% fun_handel: function to optimize
% i: variable number to optimize, e.g., i=2 if x2 is optimization
% variable
% val: values of variables other then i-th variable
val = num2cell(val);
y = fun_handel(val{1:i-1}, x, val{i:end});
end
Then run it like this
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
ht2 = @(x) helperFun(x, ht, 3, [1 2]);
x_sol = fmincon(ht2, rand(1))
Result
x_sol =
-5.3326e-09

7 comentarios

dario
dario el 10 de Mayo de 2020
Editada: dario el 10 de Mayo de 2020
Thanks Ameer.
Referring to my example, if nargin(ht) returns 4, how can I define ht2 runtime?
To create ht2 on runtime, you will need to define a helper function like this
function y = helperFun(x, fun_handel)
x = num2cell(x);
y = fun_handel(x{:});
end
Then you can use it to create ht2 which is compatible for optimization functions, e.g., fmincon
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
ht2 = @(x) helperFun(x, ht);
x_sol = fmincon(ht2, rand(1,3))
Result
x_sol =
1.0e-08 *
-0.4592 -0.8045 0.0830
If you want to optimize a particular variable (say x2), then you can modify the helper function like this. You will need to specify the value of variables you don't want to optimize
function y = helperFun(x, fun_handel, i, val)
% x: optimization variable
% fun_handel: function to optimize
% i: variable number to optimize, e.g., i=2 if x2 is optimization
% variable
% val: values of variables other then i-th variable
val = num2cell(val);
y = fun_handel(val{1:i-1}, x, val{i:end});
end
Then run it like this
ht = @(x1,x2,x3) x1.^2+x2.^2+x3.^2;
ht2 = @(x) helperFun(x, ht, 3, [1 2]);
x_sol = fmincon(ht2, rand(1))
Result
x_sol =
-5.3326e-09
Thanks, it's what I was looking for. Is there any way round to avoid the helper function?
It is possible, but it can get a bit messy. Check the code in my updated answer.
Thanks Ameer. Very Helpful.
I am glad to be of help.

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Más respuestas (1)

per isakson
per isakson el 10 de Mayo de 2020
Editada: per isakson el 10 de Mayo de 2020
"anonymous function with a variable number of input variables"
Something like this?
>> ht(1,2,3)
ans =
6
>> ht(1,2,3,4)
ans =
10
>> ht(1,2,3,4,5)
ans =
15
>>
where
ht = @(varargin) dario( varargin{:} );
and
function out = dario( varargin )
out = sum([ varargin{:} ]);
end

3 comentarios

Thaks.
What if:
ht(A1, A2)
or
ht(A1, A2, A3)
and I need to find the values of Ai that minimise ht?
I assume that ht shall returns a scalar numerical output. That output could readily be calculated in dario, since all inputs are available. The variables, Ai, what type and size are they?
Ai are scalar and yes, ht returns a scalar. I'm calculating Ai as
ht2 = @(A)ht(A(1),A(2));
A0 = [1,1];
[res,fval] = fminunc(ht2,A0)
but I need to know in advance how many Ai are included in the definition of ht2

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Preguntada:

el 10 de Mayo de 2020

Editada:

el 11 de Mayo de 2020

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