Graphs -how to change the color of a segment joining 2 dots multiple times in the same plot

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Francesco
Francesco 2011-1-27
Hi all,
I am pretty new to MATLAB programming and I have a problem regarding graphing data. My data are contained in a [3xn] matrix, and I want to plot the first two lines in a xy graph, together with a connecting line. That's easy using the plot function. Things become more complicated when I want to change the color of the segment connecting two consecutive points as a function of the values contained in the third line. For example, if the value in a cell of the third line is <0.5 I want to draw the correspondent segment in black, and if it is >0.5 I want to draw it in red. Could anybody suggest how to do it? Thanks a lot!

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回答（3 个）

the cyclist
the cyclist 2011-1-27
This is pretty ugly, with the looping, but it gives you the gist of one way of doing it. Basically, you assign a handle to each line segment, and use that handle to color the object depending on the third variable. (I split out the variables into three column vectors, to try to be a little clearer about the x-y plotting.)
function [] = dynamicPlotLineColor()
nSegments = 20;
x = rand(2*nSegments,1);
y = rand(2*nSegments,1);
c = rand(nSegments,1);
figure
hold on
for ns = 1:nSegments
hs = plot(x(2*ns-1:2*ns),y(2*ns-1:2*ns));
if c(ns)<0.5,
set(hs,'Color','k')
else
set(hs,'Color','r')
end
end
end
Vieniava
Vieniava 2011-1-27
% lets prepair input matrix
MAT=zeros(3,20);
MAT(3,:)=randi(1,20);
MAT(1,:)=1:20;
MAT(2,:)=MAT(1,:).^2;
% lets go
[k n]=size(MAT);
for i=2:n
if MAT(3,i)<0.5
C='r';
else
C='k';
end
line(MAT(1,[i-1 i]), MAT(2,[i-1 i]), 'Color', C);
end
Walter Roberson
Walter Roberson 2011-1-27
plot creates lineseries objects, which can also be manually created with line() . Any one line object is restricted to being a single colour. That is why the posters above break everything into individual line segments between the points rather than plotting the entire line at one go: it is the only way to control the color at that fine of a level.

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