# How to plot different parts of one vector with different colors?

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miki90 on 30 Jan 2018
Commented: Walter Roberson on 2 Oct 2019
Hello. I have a time vector and ecg vector on y axis which need to be plotted in the same figure with 3 different colors. There are three different zones (time before), suspicious zone in the middle (which I want to be yellow) and time after. How can it be plotted? I tried to plot like in the code below, but it can't be plotted like that.
if true
time1colored(1) = time1(1:523);
time1colored(2) = time1(524:704);
time1colored(3) = time1(705:end);
rr3colored(1) = rr3(1:523);
rr3colored(2) = rr3(524:704);
rr3colored(3) = rr3(705:end);
colors='gyr';
figure; cla;
for i=1:3
plot(time1colored(1),rr3colored(1), 'Color', colors(i))
end
plot(time1,rr3)
xlabel('time [s]'),ylabel('interval [ms]')
end

James Tursa on 30 Jan 2018
Edited: James Tursa on 30 Jan 2018
Something like this if you go with the segmented approach:
ix = { 1:523, 524:704, 704:numel(time1) };
colors = 'gyr';
figure; cla;
for i=1:3
plot(time1(ix{i}),rr3(ix{i}),'Color',colors(i)); hold on
end
##### 3 CommentsShow 1 older commentHide 1 older comment
Vidz on 2 Oct 2019
Edited: Vidz on 2 Oct 2019
gives error as runs out of colors
Index exceeds the number of array elements (3).
So, I did
plot(time1(ix{i}),rr3(ix{i}),'Color',[rand rand rand]); hold on
Walter Roberson on 2 Oct 2019
The color list is hardcoded by way of
colors = 'gyr';
You can expand that to include any of the 8 color codes bcgkmrwy
If you need more than that, then I suggest using one of the color tables, such as
ctab = copper(23);
[...]
plot(time1(ix{i}), rr3(ix{i}), 'Color', ctab(i,:)); hold on

Walter Roberson on 30 Jan 2018
This is not possible in MATLAB using plot() or related calls. plot() and related calls use line() objects, and line() objects are restricted to a single color.
You can break the vector into segments and draw the segments separately.
If you look in the File Exchange you will find a couple of different contributions for drawing colored lines.
miki90 on 30 Jan 2018
Your idea is also good and worked for me. Thanks!

miki90 on 2 Feb 2018
Sorry for posting on an answered post, but, applying this code to plotting with markers, I haven't got only markers but the points connected with lines too. Can anyone tell me why?
if true
ip = { 1:523, 524:824, 824:numel(rr4) };
colors = ['g*','y*','r*'];
figure(); cla;
for i=1:3
plot(rr4(ip{i}),rr5(ip{i}),colors(i)); hold on
end
end
James Tursa on 2 Feb 2018
You could have used your char array as is, but you would have to change that colors(i) to colors(i,:). I think this makes the code a bit harder to read, and also your construction of colors depends on all of your row strings being the same length. The cell array approach looks neater IMO and also allows you to easily use different length char strings for each color if you wanted to.
Walter Roberson on 2 Feb 2018
Also as well as using colors(i,:) you would have had to use colors = ['g*';'y*';'r*'] and all of the specifications would have had to have been the same length. Your existing colors = ['g*','y*','r*']; is building a single character vector with contents 'g*y*r*' . Using cell arrays is better in this case.
Alternately from R2017a onwards you could stay with your existing code but use colors = ["g*","y*","r*"]; which would not require any change to the indexing you are using.