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How to plot multiple time series cell arrays as a shadowed area

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Hi guys!
I have the following cell data (each column represents a different algorithm for comparison; each row represents a experiment run):
* The data is attached.
Using this code:
figure
hold on
cellfun(@plot,mutation1)
I get this plot:
Question 1: How can I represent each column of data in a specific color in the plot using the code above?
Three columns have constant data (single lines: purple, blue and green). Two columns have data with small variations (the other two multicolored lines on the graph).
Question 2: Would it be possible to represent these two multicolored lines (which are made up of a series of other lines) as a shaded area with a middle line?
Thank you very much!

Accepted Answer

Star Strider
Star Strider on 19 Sep 2023
Edited: Star Strider on 19 Sep 2023
Answer 1: See the colororder call.
Answer 2: Yes, although it takes a bit of exploring to determine what those curves are. I left in my ‘exploration’ steps (commented-out). After that, this is straightforward.
Try this —
LD = load('mutation1.mat');
mutation1 = LD.mutation1
mutation1 = 20×5 cell array
{300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double} {300×1 double}
colororder(turbo(numel(mutation1))) % Use The 'turbo' 'colormap' To Define The Colours
mutationmtx = cell2mat(reshape(mutation1, 1,[]));
% figure
% surf(mutationmtx, 'EdgeColor','none')
% colormap(turbo)
% xlabel('Columns')
% ylabel('Rows')
%
% figure
% plot((1:size(mutationmtx,2)), mutationmtx(1,:))
% hold on
% plot((1:size(mutationmtx,2)), mutationmtx(150,:))
% hold off
% grid
% legend('1','150', 'Location','best')
Lv1 = mutationmtx(1,:) > 0.3;
Lv2 = mutationmtx(150,:) > 0.05;
highest = max(mutationmtx(:,Lv1 & ~Lv2),[],2);
lowest = min(mutationmtx(:,Lv1 & Lv2), [],2);
middleline = median([highest lowest],2);
% figure
% plot(highest, 'g')
% hold on
% plot(lowest, 'r')
% hold off
v = (1:numel(highest)).';
figure
hold on
cellfun(@plot,mutation1)
patch([v; flip(v)], [highest; flip(lowest)], [1 1 1]*0.75, 'FaceAlpha',0.5)
plot(middleline, '-g', 'LineWidth',1.5)
% Ax = gca;
% Ax.XScale = 'log';
See the documentation on patch for those details.
EDIT — (19 Sep 2023 at 13:52)
Forgot about ‘middle line’, Now added.
.
  4 Comments
David Franco
David Franco on 20 Sep 2023
Thank you again @Star Strider! I found a workaround:
Since the first three columns of the cell have the same values (they are deterministic), it is not necessary to plot all of them, so I took the average (I could also take just one sample from each of the three columns)...
For columns 4 and 5 (they are stochastic) I took the maximum and minimum values (because these are variables).
So I didn't need to use the cellfun function and I was able to define the colors and create the correct caption.
load mutation1.mat
n = size(mutation1,1);
mutationmtx = cell2mat(reshape(mutation1,1,[]));
ga1 = mean(mutationmtx(:,1:n),2);
ga2 = mean(mutationmtx(:,n+1:2*n),2);
ga3 = mean(mutationmtx(:,2*n+1:3*n),2);
min_ga4 = min(mutationmtx(:,3*n+1:4*n),[],2);
max_ga4 = max(mutationmtx(:,3*n+1:4*n),[],2);
min_ga5 = min(mutationmtx(:,4*n+1:5*n),[],2);
max_ga5 = max(mutationmtx(:,4*n+1:5*n),[],2);
v = (1:numel(min_ga4)).';
figure
hold on
plot(v,ga1,'Color',[0.0000,0.4470,0.7410]')
plot(v,ga2,'Color',[0.9290,0.6940,0.1250]')
plot(v,ga3,'Color',[0.4940,0.1840,0.5560]')
patch([v; flip(v)], [max_ga4; flip(min_ga4)], 'g', 'FaceAlpha',0.5)
patch([v; flip(v)], [max_ga5; flip(min_ga5)], 'r', 'FaceAlpha',0.5)
legend('GA1','GA2','GA3','GA4','GA5')
The result:
I'm sorry if I hadn't explained it clearly from the beginning. As English is not my native language, it is sometimes difficult to correctly express my ideas. =)
Star Strider
Star Strider on 20 Sep 2023
As always, my pleasure!
No worries — some concepts are difficult to put into words regardless of the language, especially some mathematical concepts. I am happy that I could help you get this sorted.
.

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