MATLAB: Syntax of handle: f = @(X)find(X);

function handlehandles

Hi everyone~

I'm learning about use handle. The usual syntax is h = @FuncName

But I also saw an example of f = @(X)find(X)

from here:

http://www.mathworks.com/help/matlab/matlab_prog/calling-a-function-using-its-handle.html

Can anyone please explain what the difference is? Can we use f = @find instead?

Thank you~

Best Answer

Yes, you can use

f = @find

instead in this case. For example:

y = zeros(10,1);
y(3:end) = 3:10;
f(y)

The way the example defines it gives you more flexibility in using find() for example

f = @(x)find(x>5)
f(y)

Related Solutions

MATLAB: Do vector fields mapping R^2 to R^2 have to be coded in separate function files, or can I define anonymous functions for them directly in the script file

Can MATLAB anonymous function return two outputs?

Yes, for example

>> f = @(x) deal(x, x^2);
>> [a, b] = f(2)
a =
     2
b =
     4

But then you will always need to call it with two output arguments. The following will not work

>> a = f(3)
Error using deal (line 37)
The number of outputs should match the number of inputs.

So it is better to write a function in seperate file to handle such cases.

"anonymous function can only have one "executable" statement"

It means that you cannot do something like the following in an anonymous function

f = @(x) y=1, x+y; % not allowed

f = @(x) if x=1, y=2, end; % not allowed

"if the function file name is "nonlinear_equations", then in the script file, I store a variable f = @nonlinear_equations, and then use f instead."

Yes, you can directly use nonlinear_equations instead of f; there will be no difference. However, it is not entirely useless. For example, instead of nonlinear_equations, you need to use some other function. In the current code, you will only need to change the line: f = @new_function; however, in the other case, you will need to change it everywhere.

MATLAB: Predict y values from x values

Hi,

many options to do this - here you have 3 of them. All options start with your known code and want to know the x-value for y = 35.0048:

x=[0 1 2 3 4 5]; 
y=[0 20 60 68 77 110];
coeffs = polyfit(x,y,1);
y_val = 35.0048;

#1 - Using the elementary Matlab function roots

coeffs_new = coeffs;
coeffs_new(2) = coeffs_new(2) - y_val;
result1 = roots(coeffs_new);

leads to:

result1 =
      1.5000

#2 - If you have Symbolic Math Toolbox you can use the finverse function:

syms f(x) f(y)
f(x) = coeffs(1) * x + coeffs(2)
f(y) = finverse(f)

which gives:

f(x) =
(729*x)/35 + 79/21
f(y) =
(35*x)/729 - 395/2187

To calculate values with this you will need a function handle:

f_y = matlabFunction(f(y))

which is:

f_y =
    function_handle with value:
      @(x)x.*(3.5e1./7.29e2)-1.80612711476909e-1

Then you can calculate the x-value belonging to 35.0048:

>> result2 = f_y(y_val)
result2 =
      1.5000

*#3 - Use fsolve* (from Optimization Toolbox) to solve the problem:

result3 = fsolve(@(x)coeffs(1)*x+coeffs(2)-y_val,0)

which results in:

Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
result3 =
      1.5000

Best regards

Stephan

Best Answer

Related Solutions

MATLAB: Do vector fields mapping R^2 to R^2 have to be coded in separate function files, or can I define anonymous functions for them directly in the script file

MATLAB: Predict y values from x values

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