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Can I have same plot colors between 2 different figures?

2 visualizaciones (últimos 30 días)
Nimrodb
Nimrodb el 4 de Mzo. de 2013
Hi,
I have 2 Figures. Each has a Axes area with 4 plots each.
When I do 'hold all' - Matlab assign colors to the plots automatically.
I want to have the two figures plot colors the same meaning - Line#1 in Axis#1 will have the same color as Line#1 in Axes#2....
Is it possible to do so without setting manually the colors to each line?
Is there a 'hold' sequence where this might work?

Respuestas (1)

Azzi Abdelmalek
Azzi Abdelmalek el 4 de Mzo. de 2013
Example
t=0:0.1:10
y11=cos(t)
y12=2*cos(t)
h1(1)=plot(t,y11)
hold on
h1(2)=plot(t,y12)
set(h1(1),'color','red')
set(h1(2),'color','green')
figure
y21=sin(t)
y22=2*sin(t)
h2(1)=plot(t,y21)
hold on
h2(2)=plot(t,y22)
set(h2(1),'color','red')
set(h2(2),'color','green')
  6 comentarios
Azzi Abdelmalek
Azzi Abdelmalek el 5 de Mzo. de 2013
Editada: Azzi Abdelmalek el 5 de Mzo. de 2013
Then What is the problem? Just use hold all, with the order of plotting you want. The colors are givens in the order
Nimrodb
Nimrodb el 5 de Mzo. de 2013
Editada: Nimrodb el 5 de Mzo. de 2013
My code is as such (If you have suggesting on changing it but getting the same result - I'm happy to hear it):
hold all
%%Plotting x plots:
for i = 1:size(Data(:,1),1)
h.Plots(i,1) = plot(Data(i,:),(TxArray+TxArray*0.1*i));
end
%%Assigning the x plots to to Axes(1): {Matlab sets plots color automatically}
for i = 1:size(Data(:,1),1)
if ~isempty(h.Plots(i))
set(h.Plots(i),...
'Parent',h.Axes(1),...
'LineWidth',[3.0],...
'DisplayName',STAName);
end
end
h.Legend(1) = legend(h.Axes(1),'show');
%%Plotting same number of plots as in Axes(1) but a bit differently:
for i = 1:size(Data(:,1),1)
h.Plots(i,2) = plot(Data(i,:),(TxArray/4+(i-1)/3));
end
%%Assigning the x plots to Axes(2):
for i = 1:size(Data(:,1),1)
if ~isempty(h.Plots(i))
set(h.Plots(i,2),...
'Parent',h.Axes(2),...
'LineWidth',[3.0],...
'color',get(h.Plots(i,1),'color'),...%%This is my current solution
'DisplayName',STAName);
end
end

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