How do you change the color of a selected area under a curve
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I want to change the color of the selected area to say black. Ultimately, I want to have several areas filled with different colors. I know Matlab has specific built-in functions for normal curves, but I want to use my own for learning. Here is some code:
function [P] = N(u,sd,L1,L2)
y1=NC(u,sd,L1);
y2=NC(u,sd,L2);
x=[u-(sd*5):.01:u+(sd*5)];
xL=(L1:.1:L2);
Px = (1/(sqrt(2*pi)*sd)).*exp((-(x-u).^2)./(2*sd.^2));
PxL = (1/(sqrt(2*pi)*sd)).*exp((-(xL-u).^2)./(2*sd.^2));
hold on;
plot(x,Px,'k','LineWidth',4);
area(xL,PxL); %**** THIS IS THE COLOR I WANT TO CHANGE *****
title('Normal Plot - P(x) vs. x');
xlabel('x'); ylabel('P(x)');
axis([u-(sd*5),u+(sd*5),0,max(Px)]);
line([L1,L1],[0,y1],'color','r','LineWidth',2);
line([L2,L2],[0,y2],'color','r','LineWidth',2);
hold off;
function [y]=NC(u, sd, L)
y=(1/(sqrt(2*pi)*sd)).*exp((-(L-u).^2)./(2*sd.^2));
Thanks
1 Commento
Jonathan Epperl
il 11 Dic 2012
What is wrong with area? See http://www.mathworks.com/help/matlab/ref/area.html and in particular the example on the very bottom.
Alternatively you could use the lower-level function patch.
Risposte (1)
Babak
il 11 Dic 2012
Here it is: Note that I only changed your plot() function to fill() and I added the end point and the first point of your data to the data again. With this, the curve will be closed! and fill would fill color int he closed curve... hope it helps you and answers your question.
function [P] = N(u,sd,L1,L2)
y1=NC(u,sd,L1);
y2=NC(u,sd,L2);
x=[u-(sd*5):.01:u+(sd*5)];
xL=(L1:.1:L2);
Px = (1/(sqrt(2*pi)*sd)).*exp((-(x-u).^2)./(2*sd.^2));
PxL = (1/(sqrt(2*pi)*sd)).*exp((-(xL-u).^2)./(2*sd.^2));
hold on;
fill([x , x(end), x(1)],[Px, Px(end),Px(1)],'k'); %,'LineWidth',4);
%area(xL,PxL); %**** THIS IS THE COLOR I WANT TO CHANGE *****
title('Normal Plot - P(x) vs. x');
xlabel('x'); ylabel('P(x)');
axis([u-(sd*5),u+(sd*5),0,max(Px)]);
line([L1,L1],[0,y1],'color','r','LineWidth',2);
line([L2,L2],[0,y2],'color','r','LineWidth',2);
hold off;
function [y]=NC(u, sd, L)
y=(1/(sqrt(2*pi)*sd)).*exp((-(L-u).^2)./(2*sd.^2));
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