Can contourplot plot based on the peaks of the function?

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Hi,

I need to plot a contourplot based on the peaks of my function. I know the maximum of my function but the contourplot maximum is different than that. I need to plot something like the below fig. (the data in 2 figures are not the same)

but I get this one in which, the real peak and the contour peak are not the same. Is there any command for edditing this?

Thank.

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Cris LaPierre 2019-4-3

Share your code, or at a minimum the data.

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Answer 1

Cris LaPierre 2019-4-3

编辑：Cris LaPierre 2019-4-3

在 MATLAB Online 中打开

0 个投票

I suspect there is an error in your code. Here's a simple example using the peaks funciton in MATLAB

[x,y,z] = peaks(100);
surf(x,y,z)
view(2)
figure
contour(x,y,z,20)

Here is what the peak function would look like from the top down (view(2))

And here is what the corresponding contour plot looks like. Notice that the peaks align.

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ASH 2019-4-3

在 MATLAB Online 中打开

Thanks for your reply,

My variables and function are more complex than simple vectors and I'm not sure how to use peaks for them. Moreover, I think because one of my variables is a large number I got error of "maximum array size" when using meshgrid. Also I need to have three of these contour plots (for another code) in one figure, but "hold on" doesn't work.

Because it's a long code I preferred not to share it but if you need it I attached that below. Hopefully, you can find out what's going on.

clear
clc
global T
global K
global pbs
global V_bpi
T=[1600;1600;1600;1600;1600;2000;2000;2000;2000;2000];
K=[2.72925e9;1.43158e9;2.41354e9;2.17249e9;3.89502e8;1.05276e10;1.03348e10;1.08849e10;1.07148e10;1.07487e10];
pbs=[6.19204e12;14354];
psd=[6.03563e12;514.848];
n=length(T);
p=2;
%best estimates
fun = @(x)sseval(x,T,K);
x0 = [1.2e13,14000];
bestx = fminsearch(fun,x0);
c=[bestx(1) bestx(2)]; 
K_p=c(1)*exp(-c(2)./T);
err=sum((K_p-K).^2)/(n-p);
f=exp(-c(2)./T);
g=-(c(1)./T).*exp(-c(2)./T);
F=[f g];
Ft=F';
FtF=Ft*F;
%covariance matrix
V_br=inv(FtF)*err;
cor=V_br(1,2)/(V_br(1,1)*V_br(2,2))^.5;
V_bp=[psd(1),cor*(psd(1)*psd(2))^.5;cor*(psd(1)*psd(2))^.5,psd(2)];
V_bpi=inv(V_bp);
det=det(V_bp);
%Correlated
P_ab=@Calculate_P_ab;
Mat=@Matrix;
D=@(te1,te2)exp(-.5*P_ab(te1,te2)).*exp(-Mat(te1,te2)/(2*err))/((2*pi*err)^(n/2)*((2*pi)^2*det)^.5);
% Y_p=integral2(D,1.1e13,1.2e13,14000,14100);
P_y=@(te1,te2)D(te1,te2);
x0=[7e12 13900];
[xmax,fval]=fmincon(@(x)-P_y(x(1),x(2)),x0);
% figure
% surf(J,H,D(J,H), 'EdgeColor','none', 'FaceAlpha',0.5)
[J,H] = meshgrid(1.1e13:1.2e13,14000:14200);
figure
contour(J,H,P_y(J,H),10)
hold on
plot3(xmax(1),xmax(2), P_y(xmax(1),xmax(2)), 'gp', 'MarkerSize', 10, 'MarkerFaceColor','g')
hold off
xlabel('slope')
ylabel('intercept')
function val_bs = sseval(x,T,K)
  A=x(1);
  ER=x(2);
    val_bs=sum(K-(A*exp(-ER./T))).^2;  
end
function Val=Matrix(te1,te2)
    Val=0;
    global T
    global K
    global n
    for k=1:n 
        var1=T(k);
        var2=K(k);
        Val=Val+(var2-(te1.*exp(-te2./var1))).^2;
    end
end
function val_ab = Calculate_P_ab(te1,te2)
    global V_bpi
    global pbs
    val_ab=V_bpi(1,1)*(te1-pbs(1)).^2+2*V_bpi(1,2)*(te1-pbs(1)).*(te2-pbs(2))+V_bpi(2,2)*(te2-pbs(2)).^2;
end

Thanks.

Cris LaPierre 2019-4-4

在 MATLAB Online 中打开

I got the values by doing this:

[J,H] = meshgrid(linspace(1.1e13,1.2e13,1000),linspace(14000,14200));
a=P_y(J,H);

I then opened a in the variable editor (double click on it in the workspace).

ASH 2019-4-4

Thanks a lot!

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