How to make a colour gradient with 60 lines on a graph

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AJ99
AJ99 el 20 de Mayo de 2022
Comentada: Voss el 20 de Mayo de 2022
I have a program that puts 60 lines onto a singular graph. The 60 lines are the same equation being plotted with one variable changing for each iteration using a simple for loop. I was wondering if there was a way to make the lines change colours in a gradient so that the graph is easier to read by implementing a key. For example, as the value increases the colour would change gradually from yellow to orange to, finally, red. Here is the program for reference:
format long
%DECLARE VARIABLES
M = 10000002;
w = zeros(M,1);
e = zeros(M,1);
dw = zeros(M-1,1);
de = zeros(M-1,1);
T = [0:10000000];
e(1) = 0.0549;
w(1) = 0.3844e9;
for g = 12.0:0.1:18.0
for m = 0:10000000
de(m+1) = -(4e41 + (2.e40)/g) * ((w(m+1))^(-13/2)) * e(m+1);
dw(m+1) = -((8.1e41)*(e(m+1)^2) + ((1.2e40)/g)) * (w(m+1))^(-11/2);
e(m+2) = e(m+1) + (de(m+1) * 3.154e7);
w(m+2) = w(m+1) + (dw(m+1) * 3.154e7);
end
e = e(1:end-1);
w = w(1:end-1);
plot (T, w)
hold on
end
hold off

Respuesta aceptada

Voss
Voss el 20 de Mayo de 2022
You can set the ColorOrder property of the axes to control the colors of the plotted lines, without having to specify each line's color when it's plotted.
Something like this (here I've reduced the number of for loop iterations by a few orders of magnitude):
format long
%DECLARE VARIABLES
M = 1002;%0002;
w = zeros(M,1);
e = zeros(M,1);
dw = zeros(M-1,1);
de = zeros(M-1,1);
T = [0:1000];%0000];
e(1) = 0.0549;
w(1) = 0.3844e9;
% set the axes ColorOrder:
cmap = flip(autumn(61),1); % yellow -> red, with 61 colors (for 61 lines)
set(gca(),'ColorOrder',cmap)
hold on
for g = 12.0:0.1:18.0
for m = 0:1000%0000
de(m+1) = -(4e41 + (2.e40)/g) * ((w(m+1))^(-13/2)) * e(m+1);
dw(m+1) = -((8.1e41)*(e(m+1)^2) + ((1.2e40)/g)) * (w(m+1))^(-11/2);
e(m+2) = e(m+1) + (de(m+1) * 3.154e7);
w(m+2) = w(m+1) + (dw(m+1) * 3.154e7);
end
e = e(1:end-1);
w = w(1:end-1);
plot (T, w)
% hold on
end
hold off
  4 comentarios
AJ99
AJ99 el 20 de Mayo de 2022
This is incredibly useful thanks for the help again!
Voss
Voss el 20 de Mayo de 2022
You're welcome!

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