what does mean: "@(x) ....."?

Hi everbody,

i want to understand some code, which I need for my own project.

This ist an function handle. But i can understand this:

@(x) xValue - polyval([x(1:end-1),0],yValue,[],[0,x(end)]);

What does mean x(end) or x(1:end-1) ?? How can I realize this codeline in function?

Thanks for the comming answer.

Answer 1

[Stephan]

execute this code:

x = 1:10        % create a vector with elements from 1...10
a = x(end)      % access last element of vector and write to a
b = x(end-1)    % access 2.last element of vector and write to b
c = x(1:5)      % access elements 1...5
d = x(1:end)    % access all elements
e = x(1:end-1)  % access all elements but not the last one
f = x(1:end-3)  % access all elements but not the last three
g = x(1:2:end)  % access every second element

[Furkan Kocaman]

But this is a example with an Array in Matlab.. The question is, how can I do this steps by anaounmus function? The x in

x(1:end-1)

is from the function.. Where are the x-Values???

[Stephan]

It are the x-values that are given to this function handle:

fun = @(x) 3.*x
x = 1:10
a = fun(x)
b = fun(x(1:3))
c = fun(x(end))

once you understood it for an array, you also know what it means in a function handle

[Furkan Kocaman]

Ok, the hole line in Matlab are:

p1 = lsqnonlin(@(x) xValue-polyval([x(1:end-1), 0], yValue, [], [0, x(end)]), [p(1:end-1), mu(2)] , [], [], opts);

And now, how can I understand this??

[p(1:end-1), mu(2)]  // x-values that are given to this function handle???

My goal is it to write this codeline in Python..

[Walter Roberson]

Those variables would have already been assigned to before that line. Initial guess about the solution is being constructed as all except the last p vector, followed by the second mu entry. You could also phrase this like.

Temp = p ;
Temp(end) = mu(2);

and then pass Temp in place of the construct.

[Furkan Kocaman]

Thanks.

What do Matlab / lsqnonlin with the x0 Parameter?? Matlab Help say, that is the Initial Point..

Ok, I would like to programming this in Python and use Eigen. I can pass only x- and y-Values to the LM-Solver in Eigen.. How can I create the x- and y-Values??

[Walter Roberson]

Eigen for Python appears to be bindings for a C++ implementation.

https://stackoverflow.com/questions/18509228/how-to-use-the-eigen-unsupported-levenberg-marquardt-implementation

The values such as zInit in the examples there correspond to x0.