How to call a function with vector input in the fit type function?

Question

Shaily_T 2022-5-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1730390-how-to-call-a-function-with-vector-input-in-the-fit-type-function

编辑： Catalytic 2022-6-2

I am trying to fit an equation to a model and I need to call a function "kkrebook2" in my fit type funtion (For "kkrebook2", the inputs are two vectors and its output is a vector). This is the snippet of my fit type function code where I call "kkrebook2" function. But it doesn't work when I call my fit type function "SCR" in my fit code. Could you please let me know what is the issue of my code?

I think it is related to the way I am calling "kkrebook2" in my fit type function "SCR" but I don't figure out how to fix it. I have attached the "kkrebook2" function as well.

Thank you in advance!

function p = SCR(nu,numGaussians,a,center,sigma)
for j = 1:length(nu)
    for i = 1:length(nu)
        for k = 1 : numGaussians
            if k==1   
                b = Start;
            else 
                b = Start + (center * (k-1));
            end
            thisGaussian(i) = a.*exp(-((nu(i)-b).^2)/(2.*(sigma.^2)));
            % Add into accumulator array:
            gaussEqn1(i) = gaussEqn1(i) + thisGaussian(i);
            z(i)= gaussEqn1(i);   
        end
    end
    Rn(:) = kkrebook2(nu,z,0);   
    hi2(j) = (4*pi*n.*nu(j)*L)/c;
    hi1(j)=(Rn(j));
    f(j)=(-r1+r2*exp(-(z(j))).*exp(-1i*hi1(j)).*exp(-1i*hi2(j)));
    g(j)=(1-r1*r2*exp(-(z(j))).*exp(-1i*hi1(j)).*exp(-1i*hi2(j)));
    h(j)= f(j)./g(j);
    m(j)= abs(h(j));
    p(j)= (m(j).^2);       
end
end

7 个评论
显示 5更早的评论隐藏 5更早的评论

William Rose 2022-6-1

@Shaily_T,

When ever I write a non-trivial function in Matlab, I also write a relatively simple script to test it. I encourage you to do the same. I am attaching examples of what I mean. You can modify the test scripts to pass parameters that are more realistic for your situation. I do not know what typical values are for the arguments to these functions.

The test script for kkrebook2.m works fine, when omega nd imchi are vector of length 2 or vectors of length 10. The script kkrebook2test.m fails if omega and imchi are scalars. This is to be expected, because if omega is a scalar then deltaomega cannot be computed.

The test script for SCR fails, because SCR fails before it even tries to call kkrebook2(). SCR crashes before that, because it does not know what "Start" is. This was already correctly pointed out by @Catalytic. Fix that first.

Run the simple test scripts, and make adjustments to the functions SCR() and kkrebook2(), until you have the functions returning reasonable results. Then try calling SCR from your more complicated parent program.

Good luck.

Matt J 2022-6-1

编辑：Matt J 2022-6-1

在 MATLAB Online 中打开

It is hard to comment without being sure what code you are actually working with. The code you have posted for us does not run:

nu0=1;
SCR(nu0,3,1,1,1)
Not enough input arguments.

Error in solution>SCR (line 10)
                b = Start;

Matt J 2022-6-1

编辑：Matt J 2022-6-1

在 MATLAB Online 中打开

kkrebook2.m

Even when I give your code an input value for the missing "Start" Parameter and fix the missing declaration of gaussEqn1, it still does not produce an output at the given nu0, which I believe is what @Catalytic is talking about in his answer below.

nu0=1;
SCR(nu0,3,1,1,1,0)
Index exceeds the number of array elements. Index must not exceed 1.

Error in kkrebook2 (line 19)
deltaomega=omega(2)-omega(1);

Error in solution>SCR (line 22)
    Rn(:) = kkrebook2(nu,z,0);   
function p = SCR(nu,numGaussians,a,center,sigma,Start)
[gaussEqn1,thisGaussian]=deal(zeros(1,length(nu)));
for j = 1:length(nu)
    for i = 1:length(nu)
        for k = 1 : numGaussians
            if k==1   
                b = Start;
            else 
                b = Start + (center * (k-1));
            end
            thisGaussian(i) = a.*exp(-((nu(i)-b).^2)/(2.*(sigma.^2)));
            % Add into accumulator array:
            gaussEqn1(i) = gaussEqn1(i) + thisGaussian(i);
            z(i)= gaussEqn1(i);   
        end
    end
    Rn(:) = kkrebook2(nu,z,0);   
    hi2(j) = (4*pi*n.*nu(j)*L)/c;
    hi1(j)=(Rn(j));
    f(j)=(-r1+r2*exp(-(z(j))).*exp(-1i*hi1(j)).*exp(-1i*hi2(j)));
    g(j)=(1-r1*r2*exp(-(z(j))).*exp(-1i*hi1(j)).*exp(-1i*hi2(j)));
    h(j)= f(j)./g(j);
    m(j)= abs(h(j));
    p(j)= (m(j).^2);       
end
end

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Catalytic 2022-5-31

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1730390-how-to-call-a-function-with-vector-input-in-the-fit-type-function#answer_974805

编辑：Catalytic 2022-5-31

SCR has to be a 1D function of the independent variable nu, meaning it has to give valid output when nu is a scalar. That is not the case for your function.

Essentially, you appear to be fitting some N-dimensional surface function

, given only a single sample,

.

5 个评论
显示 3更早的评论隐藏 3更早的评论

Matt J 2022-6-1

编辑：Matt J 2022-6-1

在 MATLAB Online 中打开

kkrebook2.m

Further to what @Catalytic points out, I've done a small test below using a version of your code from previous threads. Notice that at nu0=3.778651600000000e+14, a different value of SCR(nu0) is obtained depending on what other values of nu accompany nu0 as input. That should not happen for a 1D function. A curve sample f(x0) should be completely defined by the input value x0, not any other x.

format long
a =0.228738094823143;
center = 2.428378513756761e+07;
sigma = 3.833690154350237e+06;
d0 =0.047400255501257;
fun=@(nu) SCR(nu,9,a,center,sigma,d0);
nuA=linspace(3.778651600000000e14 ,  3.778653400000000e14,100);
nuB=linspace(3.778651600000000e14 ,  3.778653400000000e14,200);
scrA=fun(nuA); 
scrB=fun(nuB);
nuA(1)
ans = 
     3.778651600000000e+14
scrA(1)
ans = 
   0.296831567474829
nuB(1)
ans = 
     3.778651600000000e+14
scrB(1)
ans = 
   0.330814936483596
function p = SCR(nu,numGaussians,a,center,sigma,d0)
r1= 0.667746135604857;
r2= 0.998500100870347;
L= 0.00400002911721321;
c= 2.99792458E8;
n = 1.79997756058945;
StartComb = 377865161049434;
p = zeros(size(nu));
d = zeros(size(nu));
gaussEqn1 = zeros(1,length(nu));
thisGaussian = zeros(1,length(nu));
% alphad = zeros(1,length(nu));
% T = zeros(1,length(nu)); 
% Imag_n = zeros(1,length(nu)); 
% Real_n = zeros(1,length(nu)); 
phi = zeros(1,length(nu));
f = zeros(1,length(nu));
g = zeros(1,length(nu));
h = zeros(1,length(nu));
m = zeros(1,length(nu));
for i = 1:length(nu)
    
    for k = 1 : numGaussians
        
        if k==1 
            
        b = StartComb;
        
        else 
    
        b = StartComb + (center * (k-1));
        
        end
   
        thisGaussian(i) = a.*exp(-((nu(i)-b).^2)/(2.*(sigma.^2)));
        
        % Add into accumulator array:
        gaussEqn1(i) = gaussEqn1(i) + thisGaussian(i);
        d(i)= gaussEqn1(i);
        
    end
 end
       
        alphad = d/L;
        T = exp(-alphad);
        Imag_n =(-log(T)*c)./(nu*(4*pi));  %imaginary index of reffraction
        Real_n=kkrebook2(nu,Imag_n,0) + n; 
for i = 1:length(nu)
        phi(i)=(4*pi*Real_n(i).*nu(i)*L)/c;
        
        Finesse=center/(sigma*sqrt(8*log(2)));
        f(i)=(-r1+r2*exp(-d0)*exp(-(d(i)/Finesse)).*exp(-1i*phi(i)));
        g(i)=(1-r1*r2*exp(-d0)*exp(-(d(i)/Finesse)).*exp(-1i*phi(i)));
        h(i)= f(i)./g(i);
        m(i)= abs(h(i));
        p(i)= (m(i).^2);
   
        
end
 p=abs(p);
end

Shaily_T 2022-6-2

My "SCR" function inputs are a vector of "nu" which is the independent variable and some scalar parameters like numGaussians, a center, and sigma, and the output of "SCR" is a vector. I think for fitting a function to data we have a vector of the dependent variable and a vector of the independent variable for data. And so the function to be fitted should also get a vector of independent variables and give back a vector of dependent variables. The way I have written it seems to do the same but when I call "SCR" it doesn't work.

Catalytic 2022-6-2

编辑：Catalytic 2022-6-2

"I think for fitting a function to data we have a vector of the dependent variable and a vector of the independent variable for data."

That's true in general, but the Curve Fitting Toolbox doesn't support general N-dimensional fitting. The Curve Fitting Toolbox only supports the fitting of functions with a 1-dimensional or 2-dimensional domain, whereas your function has an N-dimensional domain.

To put it more formally, if your code implements a mapping y=f(x):

, then for f() to be considered a 1D curve, each y(i) can depend only on the corresponding x(i). However, in your SCR function each y(i) depends on all of the x(j) simultaneously.

For more general function fitting problems, you could look at lsqnonlin.

请先登录，再进行评论。

How to call a function with vector input in the fit type function?

7 个评论
显示 5更早的评论隐藏 5更早的评论

回答（1 个）

5 个评论
显示 3更早的评论隐藏 3更早的评论

另请参阅

类别

标签

Community Treasure Hunt

How to call a function with vector input in the fit type function?

7 个评论 显示 5更早的评论隐藏 5更早的评论

回答（1 个）

5 个评论 显示 3更早的评论隐藏 3更早的评论

另请参阅

类别

标签

Community Treasure Hunt

7 个评论
显示 5更早的评论隐藏 5更早的评论

5 个评论
显示 3更早的评论隐藏 3更早的评论