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

Question

David Franco 2023-9-19

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https://ww2.mathworks.cn/matlabcentral/answers/2023082-how-to-plot-multiple-time-series-cell-arrays-as-a-shadowed-area

评论： Star Strider 2023-9-20

采纳的回答： Star Strider

mutation1.mat

在 MATLAB Online 中打开

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!

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Answer 1

Star Strider 2023-9-19

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链接

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https://ww2.mathworks.cn/matlabcentral/answers/2023082-how-to-plot-multiple-time-series-cell-arrays-as-a-shadowed-area#answer_1313342

编辑：Star Strider 2023-9-19

在 MATLAB Online 中打开

mutation1.mat

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.

.

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Star Strider 2023-9-20

在 MATLAB Online 中打开

mutation1.mat

As always, my pleasure!

I was not quite certain what you wanted. I am happy that you could use my code to get your desired result.

There are 100 lines, so the legend would look busy. You might be able to accommodate that by using the 'NumColumns' name-value pair, however it would still be crowded. The best way to be certain everything was assigned values in the legend would be something like this —

load mutation1.mat

mutationmtx = cell2mat(reshape(mutation1,1,[]));

min_ga4 = min(mutationmtx(:,61:80),[],2);

max_ga4 = max(mutationmtx(:,61:80),[],2);

min_ga5 = min(mutationmtx(:,81:100),[],2);

max_ga5 = max(mutationmtx(:,81:100),[],2);

v = (1:numel(min_ga4)).';

lgdc = compose('Line %3d',1:size(mutationmtx,2));

colororder(turbo(size(mutationmtx,2)));

cm = colormap(turbo(size(mutationmtx,2)));

figure

hold on

for k = 1:size(mutationmtx,2)

hcp(k,:) = plot(mutationmtx(:,k), 'DisplayName',lgdc{k}, 'Color',cm(k,:));

end

hp1 = patch([v; flip(v)], [max_ga4; flip(min_ga4)], 'g', 'FaceAlpha',0.5, 'DisplayName','ga4');

hp2 = patch([v; flip(v)], [max_ga5; flip(min_ga5)], 'r', 'FaceAlpha',0.5, 'DisplayName','ga5');

% for k = 1:numel(hcp)

% hcp(k).DisplayName = lgdc{k};

% end

legend([hcp; hp1; hp2], 'Location','southoutside', 'NumColumns',6, 'FontSize',5)

I have no idea why colororder wasn’t working correctly, since it even initially failed with the loop. I changed the code to plot the ‘mutationmtx’ columns and associated colours individually, along with their 'DisplayName' properties, since that made the legend easier to work with.

The legend is going to be a problem with 102 entries. See its documentation for more options.

.

David Franco 2023-9-20

在 MATLAB Online 中打开

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 2023-9-20

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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