anonymous function with a variable number of input variables

9 Ansichten (letzte 30 Tage)
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!

Akzeptierte Antwort

Ameer Hamza
Ameer Hamza am 10 Mai 2020
Bearbeitet: Ameer Hamza am 11 Mai 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 Kommentare
dario
dario am 10 Mai 2020
Thanks Ameer. Very Helpful.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

per isakson
per isakson am 10 Mai 2020
Bearbeitet: per isakson am 10 Mai 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 Kommentare
per isakson
per isakson am 10 Mai 2020
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?
dario
dario am 10 Mai 2020
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

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by