Legend and Colors for Iteration function

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A.J.M 2020-9-14

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评论： A.J.M 2020-9-15

Good Day;

I need the steps that I can add Legend to my iteration function, where there are four curves, therefore, for each curve there must have specific color and legend that point to it.

Many Thanks

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Answer 1

Adam Danz 2020-9-14

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编辑：Adam Danz 2020-9-14

在 MATLAB Online 中打开

Use the displayname property to assign the legend string and use the "color" property to set the line color.

% Define colors 
n = 12; 
colors = jet(n); 
% Plot within loop
hold on % Important!
for i = 1:n
    plot(1:5, rand(1,5), '-', 'Color', colors(i,:),...
        'DisplayName', sprintf('Line #%d',i));
end
legend()

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A.J.M 2020-9-15

在 MATLAB Online 中打开

Good day Mr. Adam Danz;

This is my code, I plotted without loop, the new problem is the legend colors don't match the curves.

Z=100:250:5000;                                          %Distance Between The Transmitter and Reciever in m
pointing_Err_angle=1*10^-5;                        %Pointing Error Angle in mrad        
lambda=1550e-9;                                          %Wavelength in nm
wo=5*10^-2;                                                  %Beam Waist  at Z=0 in m
 Cn=sqrt(5e-15);                                          %Refractive Index Structure Parameter
% --------------------Calculation Section------------------------------
k=2*pi/lambda;                                                              %The number of optical wave
po=(0.55*(Cn)^2*k^2*Z).^(-3/5);                                    %The Coherence Length
E=(1+2*wo^2./po.^2);
wz=wo*(1+E.*((lambda.*Z)./(pi*wo^2)).^2).^(1/2);           %The Beam Waist 
r=Z*(pointing_Err_angle);                                                %The Radial Displacement B
%----------------------------------------Calculate Hp-------------------
v=(sqrt(pi)*a)./(sqrt(2)*wz);
a0=(erf(v)).^2;                       %Maximal Fraction of Collected Power at r=0
Wzeq=sqrt(wz.^2*((sqrt(pi).*erf(v))/(2*v.*exp(-v.^2))));    
Hp=a0.*exp((-2*r.^2)./((Wzeq).^2));                       
%--------------------------Intensity Distribution---------------
Rytov_var=[0.1,0.2,0.5,0.8];            %Log irradiance variance (Roytov variance)
for i=1:length(Rytov_var)
    for j=1:5:length(Hp)
A1(i,j)=1./(Hp(j).*sqrt(2*pi*Rytov_var(i)));
A2(i,j)=((log(Hp(j))+0.5.*Rytov_var(i)).^2)./2.*Rytov_var(i).^2;
PDF_Hp(i,j)=A1(i,j).*exp(-(A2(i,j)));
    end
end
%--------------------------Plot Section---------------------------------
z=1250:1250:5000;               %Distance for Plot Purpose
figure(1)
semilogy(z,PDF_Hp,'LineWidth',4);
xlabel('The Distance Z')
ylabel('The Probability Density Function ')
title('The PDF of Hp Vs The Distance')
legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
grid on
figure(2)
semilogy(Rytov_var,PDF_Hp,'LineWidth',3);
xlabel('Roytov variance')
ylabel('The Probability Density Function ')
title('The PDF of Hp Vs Rytove Variance')
legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
grid on

Adam Danz 2020-9-15

在 MATLAB Online 中打开

I see the problem. Look at the values you're plotting.

semilogy(z,PDF_Hp,'LineWidth',4);

PDF_Hp =
  Columns 1 through 12
       59.204            0            0            0            0       74.572            0            0            0            0       130.59            0
       33.684            0            0            0            0       41.243            0            0            0            0        66.92            0
       5.2821            0            0            0            0       5.3479            0            0            0            0       5.1881            0
      0.42448            0            0            0            0      0.30818            0            0            0            0      0.12088            0
  Columns 13 through 16
            0            0            0       242.89
            0            0            0       112.93
            0            0            0       4.5326
            0            0            0     0.032869

A line is created for each column so your plot actually has 16 lines, not 4.

The reason they don't appear is because in a log plot, 0 is not defined. However, those line objects are still produced - they are just not shown. As evidence of that, check out the 16 object handles.

h = semilogy(z,PDF_Hp,'LineWidth',4)
%     h = 
%       16×1 Line array:
%     
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line
%       Line

The reason the "wrong" colors are defined is because you're only asking for the first 4 objects to appear in the legend. But objects 2-3-4 are all non-visible lines.

legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')

Solution

Get rid of the columns with 0s in PDF_Hp

A.J.M 2020-9-15

Thanks a lot, Mr. Adam Danz, the problem was solved. I appreciate your hard efforts

Accept my respect.

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Legend and Colors for Iteration function

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