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);
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)
I have been using MATLAB fminunc function to solve my optimization problem. I want to try the minFunc package :
http://www.di.ens.fr/~mschmidt/Software/minFunc.html
When using fminunc, I defined a function funObj.m which gives me the objective value and the gradient at any point 'x'. It also takes in several external inputs say, {a,b,c} which are matrices. So the function prototype looks like :
function [objVal,G] = funObj(x,a,b,c)
I want to use the same setup in the minFunc package. From the examples, I figured this should work :
options.Method='lbfgs';
f = #(x)funObj(x,a,b,c);
x = minFunc(f,x_init,options);
But when I call this way, I get an error as:
Error using funObj
Too many output arguments.
What is the correct way to call minFunc for my case?
**EDIT : Alright, here is a sample function that I want to use with minFunc. Lets say I want to find the minimum of a*(b-x)^2, where a,b are scalar parameters and x being a scalar too. The MATLAB objective function will then look like :
function obj = testFunc(x,a,b)
obj = a*(b-x)^2;
The function call to minimize this using fminunc (in MATLAB ) is simply:
f = #(x)testFunc(x,a,b);
x = fminunc(f,x_init);
This gives me the minimum of x = 10. Now, How do I do the same using minFunc ?
"Note that by default minFunc assumes that the gradient is supplied, unless the 'numDiff' option is set to 1 (for forward-differencing) or 2 (for central-differencing)."
The error is because only one argument is returned by the function. You can either return the gradient as a second argument or turn on numerical differencing.
Agree with Mark. I think the simplest way to solve it is
minFunc(#testFunc, x_init, a, b, c)
In MATLAB temporary function can only have one return value. So f = #(x)testFunc(x,a,b); let your method drop gradient part every time. Because minFunc can accept extra paramters, you can pass a, b and c after x_init. I think this would work.
I have a function for cached evaluation. As one of the arguments, it takes a function handle. Under some circumstances, the function handle is unaccessible, and I don't quite understand why. The example below shows what got me stumped:
>> A.a = #plus; feval(#A.a, 1, 1)
ans =
2
>> clear A
>> A.a.a = #plus; feval(#A.a.a, 1, 1)
Error using feval
Undefined function 'A.a.a' for input arguments of type 'double'.
So, if I have a function handle stored as a structure member, I can pass it along fine if it's one level deep, but not if it's two levels deep. In my real use case, I have a structure D that holds many (117) instances of various classes, so I actually have stct.obj.meth, where stct is a structure, obj is a class instance/object, and meth is a method. Passing #stct.obj.meth fails, but if I assign A = stct.obj, then passing #A.meth succeeds.
Under what conditions can I pass a function handle as an argument, so that it's still accessible down the stack?
Edit: Although in the use case above, I could simply remove the # because #plus is already a function handle. However, consider the situation here:
>> type cltest.m
classdef cltest < handle
methods
function C = mymeth(self, a, b)
C = a + b;
end
end
end
>> A.a = cltest();
>> feval(#A.a.mymeth, 1, 1)
Error using feval
Undefined function 'A.a.mymeth' for input arguments of type 'double'.
>> b = A.a;
>> feval(#b.mymeth, 1, 1)
ans =
2
In this case, I need the # before A.a.mymeth...
Introducing classes was a big deal for MATLAB. So big, in fact, that they still do not work properly today. Your example shows that structure access and class method access conflict, because they had to overload the the meaning of dot '.' and didn't get it to work seamlessly. It all more or less works fine when you are calling class methods explicitly by their name on the MATLAB console, e.g. in your example >> A.a.mymeth(1,1). But when you have any type of indirection, it soon breaks.
You tried getting the function handle by >> #A.a.mymeth, which MATLAB cannot make sense of, probably because it gets confused by the mixed structure/class thing. Trying to work around using str2func doesn't work either. It works, again, only for explicit name access, as shown here. It breaks for your example, e.g. >> str2func('b.mymeth'). It does not even work inside the class. Try other indirections and watch them fail.
Additionally, MATLAB does not like giving you a class method's handles. There's no function for it. There's no way to get all function handles in one go, or even dynamically by a name string.
I see three options here. First, try changing your program, if possible. Do these functions need to sit in a classdef?
Second, follow your or nispio's workaround. They both create a temporary variable to hold a reference to the class instance in order to create a non-mixed access to its member methods. The problem is, they both require explicitly naming the function. You have to explicitly put this code for every function involved. No way to abstract that out.
Third, cheat by giving out your class' method handles from the inside. You can give them out in a structure.
classdef cltest < handle
methods
function C = mymeth(self, a, b)
C = a + b;
end
function hs = funhandles(self)
hs = struct('mymeth', #self.mymeth, ...
'mymeth2', #self.mymeth2);
end
end
end
You can then access the handles by name, even dynamically.
>> A.a = cltest;
>> feval(A.a.funhandles.mymeth, 1, 1);
>> feval(A.a.funhandles.('mymeth'), 1, 1)
ans =
2
But be careful, by using this you can access Access=private methods from outside.
Try this:
feval(#(varargin)A.a.mymeth(varargin{:}),1,1);
It is a little kludgy, but it should work.
EDIT:
The way it works is by creating an Anonymous Function that takes a variable number of arguments, and dumps those arguments into the method A.a.mymeth(). So you are not actually passing a pointer to the function A.a.mymeth, you are passing a pointer to a function that calls A.a.mymeth.
An alternative way of achieving the same thing without using varargin would be:
feval(#(x,y)A.a.mymeth(x,y),1,1);
This creates an anonymous function that accepts two arguments, and passes them along to A.a.mymeth.
<speculation> I think that it must be inherent in the way that the unary function handle operator # works. The Matlab parser probably looks at #token and decides whether token is a valid function. In the case of a.mymeth it is smart enough to decide that mymeth is a member of a, and then return the appropriate handle. However, when it sees A.a.mymeth it may discover that A is not a class, nor does A have a member named a.mymeth and therefore no valid function is found. This seems to be supported by the fact that this works:
A.a.a = #plus; feval(A.a.a,1,1)
and this doesn't:
A.a.a = #plus; feval(#A.a.a,1,1)
</speculation>
You can get around it by introducing a separate function that corrects what # operator is not doing:
function h=g(f)
x = functions(f);
if ~strcmp(x.type, 'anonymous')
h = evalin('caller', ['#(varargin)' x.function '(varargin{:})']);
else
h = f;
end
end
Now for your example:
>> feval(g(#A.a.mymeth), 1, 1)
ans =
2
>> feval(g(#b.mymeth), 1, 1)
ans =
2
I think this will have the smallest impact on your code. You can make it a bit more elegant but less robust and/or readable. The uplus method is not defined for function_handle class so you can create uplus.m in folder #function_handle somewhere in your path with this content:
function h=uplus(f)
x = functions(f);
if ~strcmp(x.type, 'anonymous')
h = evalin('caller', ['#(varargin)' x.function '(varargin{:})']);
else
h = f;
end
end
Now you just need to use +# instead of #. For your examples:
>> feval(+#A.a.mymeth, 1, 1)
ans =
2
>> feval(+#b.mymeth, 1, 1)
ans =
2
I am trying to write a minimal function that can be called with a variable number of arguments but that will not throw a wrong number of arguments error if miscalled.
Here is where I start from :
function varargout=fname(varargin)
% FNAME
% Usage: output=fname(input)
% Arguments check
if(nargin~=1 || nargout~=1)
disp('Function fname requires one input argument');
disp('and one output argument');
disp('Try `help fname`');
varargout(1:nargout)={0};
return;
end
input=varargin{1};
output=input;
varargout(1)={output};
end
However this does not work as I would like it to. Is there a way to write a function that :
never throw a "wrong number of arguments" error (so that the rest of the execution can continue)
accepts variable number of input and output arguments and checks them inside the function
(maybe more tricky) if the number of input / output arguments is not correct, does not replace the value of the provided output arguments (so that any misplaced call does not erase the previous value of the output argument)
I am open to any suggestions / other methods.
Thank you for your help.
UPDATE: thanks to #Amro for his answer, I guess what I miss here is either a call by address of reference for Matlab functions or a way to interrupt a function without returning anything and without stopping the rest of the execution.
Here is one way to implement your function:
function varargout = fname(input,varargin)
%# FNAME
%# Usage: output=fname(input)
%%# INPUT
if nargin<1
varargout(1:nargout) = {[]};
warning('Not enough input arguments.'), return
end
if ~isempty(varargin)
warning('Too many input arguments.')
end
%%# YOUR CODE: manipulate input, and compute output
output = input;
%%# OUTPUT
varargout{1} = output;
if nargout>1
warning('Too many output arguments.')
varargout(2:nargout) = {[]};
end
end
Obviously you can customize the warning messages to your liking...
Also, if you want your function to simply print the message instead of issuing warnings, replace all WARNING calls with simple DISP function calls.
Examples of function call:
fname()
fname(1)
fname(1,2)
x = fname()
x = fname(1)
x = fname(1,2)
[x,y] = fname()
[x,y] = fname(1)
[x,y] = fname(1,2)
The above calls execute as expected (showing warning messages when applicable). One caveat though, in the last three calls, if the variable y already existed in the workspace prior to the calls, it would be overwritten by the empty value y=[] in each...
If I understand your question correctly, then the answer is no. If a caller calls a function like this:
[a, b, c] = fname('foo');
then fname is required to return (at least) three outputs. There's no way to tell MATLAB that it should leave b and c alone if fname only returns one output.
Working on an assignment involving Genetic Algorithms (loads of headaches, loads of fun). I need to be able to test differing crossover methods and differing mutation methods, to compare their results (part of the paper I have to write for the course). As such, I want to just pass the function names into the Repopulate method, as function handles.
function newpop = Repopulate(population, crossOverMethod, mutationMethod)
...
child = crossOverMethod(parent1, parent2, #mutationMethod);
...
function child = crossOverMethod(parent1, parent2, mutationMethod)
...
if (mutateThisChild == true)
child = mutationMethod(child);
end
...
The key point here is like 3, parameter 3: how do I pass mutationMethod down another level? If I use the # symbol, I get told:
"mutationMethod" was previously used as a variable,
conflicting with its use here as the name of a function or command.
If I don't use the # symbol, then mutationMethod gets called, with no parameters, and is quite unhappy.
While I am aware that yes, I could just rewrite my code to make it work differently, I'm now curious as to how to make it actually work.
Any help is greatly appreciated.
Actually just dont use the # symbol, use it when you call the Repopulate function instead.
Example:
function x = fun1(a,m)
x = fun2(a,m);
end
function y = fun2(b,n)
y = n(b);
end
which we call as:
> fun1([1 2 3], #sum)
6
Refer to the documentation for Passing Function Handle Arguments
Note you can check if the argument is a function handle by: isa(m,'function_handle'). Therefore you can make your function Repopulate more flexible by accepting both a function handle and a function name as a string:
function x = fun(a,m)
if ischar(m)
f = str2func(m);
elseif isa(m,'function_handle')
f = m;
else
error('expecting a function')
end
x = fun2(a,f);
end
which now can be called both ways:
fun1([1 2 3], #sum)
fun1([1 2 3], 'sum')