I called Matlab's genetic algorithm solver, ga, in the following way:
[theta,fval,exitflag] = ga(smmobj,26,[],[],[],[],LB,UB,[],[]);
where theta is a 26-by-1 column vector that needs to be optimized.
So in the main function, it goes like this:
clc
clear
global var1 var2...
load ('abcd.mat')
theta0=[1 2 3....];
LB=[26-by-1 row vector];
UB=[26-by-1 row vector];
[theta,fval,exitflag] = ga(smmobj,26,[],[],[],[],LB,UB,[],[]);
The fitness function, smmobj, is defined as:
function [obj]=smmobj(theta)
global var1 var2...
But when I run it, it always says:
Error using smmobj (line 4)
Not enough input arguments.
Error in SMMga (line 32)
[theta,fval,exitflag] = ga(smmobj,26,[],[],[],[],LB,UB,[],[]);
But I run the fitness function by itself, it works.
[theta,fval,exitflag] = ga(smmobj,26,[],[],[],[],LB,UB,[],[]);
will throw that error because you are essentially calling the function smmobj (with no parameters) instead of passing your ga function a function handle to your objective function. You do this using the # symbol so
[theta,fval,exitflag] = ga(#smmobj,26,[],[],[],[],LB,UB,[],[]);
Incidentally, I would recommend that you do not use global in order to pass the extra parameters of var1 and var2 to your objective function but rather use anonymous functions:
[theta,fval,exitflag] = ga(#(x)(smmobj(x,var1,var2)),26,[],[],[],[],LB,UB,[],[]);
and then change smmobj's definition to
function [obj]=smmobj(theta,var1,var2)
Related
It was suggested in this comment that there is a difference between how Matlab and Python pass around functions. From what I can tell by looking and using the two, there is no difference between the two, but maybe I'm missing something?
In Matlab, you would create a quick function handle like this:
fun = #(x) x.^2 + 1;
In Python, using a lambda function, you could create a similar function like this:
def fun(x):
return x^2
In both languages, it's possible to send the term 'fun' to another function as an argument - but the commenter I linked to insinuated that they are not the same and/or need to be used differently.
What am I missing?
The first comment seems to simply reiterate the idea that you can pass a MATLAB function handle as an argument (although the answer didn't state anything that would make me think otherwise). The second comment seemed to interpret this to mean that the first commenter thought that you couldn't do this in Python and responded to state that you can use either a lambda or pass the function directly.
Regardless, assuming that you use them correctly, a function handle in MATLAB is functionally equivalent to using either a lambda or function object as an input argument in Python.
In python, if you don't append the () to the end of the function, it doesn't execute the function and instead yields the function object which can then be passed to another function.
# Function which accepts a function as an input
def evalute(func, val)
# Execute the function that's passed in
return func(val)
# Standard function definition
def square_and_add(x):
return x**2 + 1
# Create a lambda function which does the same thing.
lambda_square_and_add = lambda x: x**2 + 1
# Now pass the function to another function directly
evaluate(square_and_add, 2)
# Or pass a lambda function to the other function
evaluate(lambda_square_and_add, 2)
In MATLAB, you have to use a function handle because MATLAB attempts to execute a function even if you omit the ().
function res = evaluate(func, val)
res = func(val)
end
function y = square_and_add(x)
y = x^2 + 1;
end
%// Will try to execute square_and_add with no inputs resulting in an error
evaluate(square_and_add)
%// Must use a function handle
evaluate(#square_and_add, 2)
Could you please help me with the following issue: I have the following function handle:
r1 = #(lambda) b + lambda*(r - b); % r and b are vectors of return data
I want to find the optimal lambdas that set me a mean function to zero, for a given set of powers within that function. What I tried to do and didn't work, as it returns me an error for undefined operators for input arguments of type 'function_handle' is:
lambda0 = 0.3;
for a = 2:10 %power coefficient
S1(a) = fzero(mean((r - b)*r1.^(1/(a - 1))),lambda0);
end
Any suggestion as to how to go about this problem is highly appreciated! Thank you in advance.
fzero accepts a function handle as the first input. As you currently have it, you're trying to pass a statement as the first input. This statement can't even be properly evaluated because you are trying to perform numerical operations on a function handle (more on this in a bit).
You need to instead do something like this where we create a new function handle that evaluates the original function handle and performs the other operations you need.
S1(a) = fzero(#(lambda)mean((r - b)*r1(lambda).^(1/(a - 1))),lambda0);
Further Explanation
Performing operations on a function handle is not the same as performing them on the result.
So for example, if we had a function handle:
func = #(x)2*x;
If we evaluation this, by calling it with an input value for x
func(2)
4
This works as we would expect. If now we really want the value (2*x)^2, we could try to write it the way that you wrote your statement in your question
func2 = func^2;
We will get an error!
Undefined operator '^' for input arguments of type 'function_handle'.
This does not work because MATLAB attempts to apply the ^ operation to the function handle itself and not the value of the evaluated function handle.
Instead, we would need to create a new function handle that essentially wraps the other one and performs any additional options:
func2 = #(x)func(x)^2;
func2(2)
16
Bringing it Full-Circle
So if we go back to your question, you defined your anonymous function r1 like this.
r1 = #(lambda) b + lambda*(r - b); % r and b are vectors of return data
This all looks great. You have one input argument and you reference r and b from the parent workspace.
Now when you call fzero you try to perform operations on this function handle in hopes of creating a new function handle.
mean((r - b)*r1.^(1/(a - 1)))
Like we just showed this will result in a very similar error
Undefined operator .^ for input arguments of type 'function_handle'
So we need to wrap this into a new function.
newfunc = #(lambda)mean((r - b)*r1(lambda).^(1 / (a - 1)));
Now we can safely pass this to fzero.
result = fzero(newfunc, lambda0);
I'd like to call a certain function from within an anonymous function, as in
#(){fooBar(baz)}
Trouble is, fooBar has no outputs, which makes the anonymous function complain. Is there a way around this besides making the fooBar function return a dummy output?
The problem is in your anonymous function definition. By enclosing your function foobar(baz) between the characters {...}, you are programming a function which has to :
evaluate the expression foobar(baz)
Place the result of this expression into a cell
return the cell
Obviously in step (2) Matlab cannot place the result of the expression (1) in a cell because there is no output from (1).
So simply define your function without the curly braces:
myFunction = #() fooBar(baz)
and everything should work ok.
To demonstrate with an example, let's define the function fooBar by doing something which does not produce an output (change an axe limits for example):
fooBar = #(axlim) set(gca,'XLim',axlim)
I can now call fooBar([0 20]) and the current axes will directly have its axes limits set to [0 20]
If there is an axis span which I use often ([-5 5] for example), I could be tempted to define a new function which will always call fooBar with the same (often used) parameters:
fooBarPrefered = #() fooBar([-5 5])
Now every time I call fooBarPrefered(), my axes X limits are directly set to [-5 5].
To further prove the point, since calling fooBar([-5 5]) does not produce an output, Matlab will indeed complain if I define my function with curly braces:
fooBarPrefered = #() {fooBar([-5 5])} ;
>> fooBarPrefered()
One or more output arguments not assigned during call to "set".
Error in #(axlim)set(gca,'XLim',axlim)
Error in #(){fooBar([-5,5])}
But note that this is the same error than if you were trying to assign the output of fooBar to a variable directly in the workspace:
a = fooBar([0 20])
One or more output arguments not assigned during call to "set".
Error in #(axlim)set(gca,'XLim',axlim)
Bottom line: If a function does not have an output, do not try to redirect this output to a variable or an expression.
I want to call a function in Matlab using another one, which has the same number of inputs and outputs. In fact, those inputs and outputs have the same name.
Example:
function [a,b] = gettwo(matrix,string,varargin)
[a,b] = getone(matrix,string,varargin{:});
end
This code produces the following error:
Error in getone(line 3)
aux = 'matrix(varargin{:})';
Output argument "b" (and maybe others) not assigned during
call to "C:\Users\baister\Documents\MATLAB\soft\getone.m>getone".
Error in results (line 4)
[a,b] = getone(matrix,string,varargin{:});
How should I wrap getone?
(The definitive function will have more lines than those shown in this post.)
Thanks.
The general wrapping for variable number of outputs should work like this:
function [varargout] = gettwo(matrix,string,varargin)
[varargout{1:nargout}] = getone(matrix,string,varargin{:});
end
You'll get the same error as above though, in case you do
[a,b] = gettwo(...);
and getone returns only 1 argument.
I created an anonymous function inside a script and I can't get MATLAB to run the fminsearch? Here's an what I have so far:
V=x(1);
f=x(2);
q=#(x) (pi.*D.*L)./(1000.*V.*f);
fminsearch(#q,x);
The variables D and L are defined, but MATLAB gives me the following error:
Error: File: Testing.m Line: 51 Column: 17
"q" was previously used as a variable, conflicting with its use here as the name of a function or command.
See "How MATLAB Recognizes Command Syntax" in the MATLAB documentation for details.
q is not mentioned before this command. What am I doing wrong?
Another thing that could solve my problem is to get my script to write a function file, but how to do that?
Remove the second #:
V=x(1);
f=x(2);
q=#(x) (pi.*D.*L)./(1000.*V.*f);
fminsearch(q,x);
q is a function handle. fminsearch expects a function handle. You can create a function handle out of a function using an # (e.g. #min), but you don't need to do that here.
You can also write the anonymous function inline with the search command:
V=x(1);
f=x(2);
fminsearch(#(x) (pi.*D.*L)./(1000.*V.*f),x);
UPDATE (credits to #wakjah)
For your code to do anything sensible, you should use the argument x of the anonymous function:
x0 = [initialV, initialF];
fminsearch(#(x) (pi.*D.*L)./(1000.*x(1).*x(2)), x0);
#function creates a function handle for an existing function.
q = #(x) whatever... creates a function handle called q.
But, you can't create a function handle for a function handle, only for a function.
See this:
>> fones = #ones
fones =
#ones
>> ffones = #fones
Error: "fones" was previously used as a
variable,
conflicting with its use here as the name
of a function or command.
See MATLAB Programming, "How MATLAB
Recognizes Function Calls That Use
Command Syntax" for details.
In Matlab, a function handle is a kind of a pointer to a function and is distinct from a function (unlike in some other languages where the a function identifier can be passed and stored as any other variable).
It's important to note that calling a function and a function handle results in the same behaviour. Except for the case where the identifier is used without any parentheses following it:
>> ones
ans =
1
>> fones
fones =
#ones
>> ones(2)
ans =
1 1
1 1
>> fones(2)
ans =
1 1
1 1