Common legend obscures tiled layout

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Daniel Rowe
Daniel Rowe el 14 de Jun. de 2022
Comentada: Daniel el 23 de Sept. de 2023
Hello,
I have a tiledlayout(1,3) with square axis plots. When I try to add a common legend, it obscures the plots. Is there any workaround for this?
Note that this only happens with squares axes, not default.
Probably very simple but I cannot find a solution for this. Code and image attached.
I tried to upload a .mat file but the compressed data >> 5 Mb, so if someone could produce some random data and plot in tiles to present the workaround, I'd be very grateful
Cheers,
Daniel
% Preallocation for Taylor microscales
TLS_A = zeros(1,n);
% Plot - stage 2 (fit parabola)
h(5) = figure;
tiledlayout(1,n)
for i = 1:n
[xData, yData] = prepareCurveData( rii_xcorr{i}(1:(find(Rii_xcorr{i}<=yy(i),1,'first'))) , Rii_xcorr{i}(1:(find(rii_xcorr{i}>=xx(i),1,'first')))); % transforms data for curve fitting with the 'fit' function
[fitresult, gof] = fit( xData, yData, ft, opts ); % returns goodness-of-fit statistics in the structure 'gof'
nexttile(i)
plot(rii_xcorr{i},feval(fitresult,rii_xcorr{i}),'r','LineWidth',2)
hold on
plot(rii_xcorr{i},Rii_xcorr{i},'MarkerEdgeColor','k','MarkerSize',6,'Marker','o', 'LineStyle','none','LineWidth',1.5)
ylabel(sprintf(('$R_{%s}$'),V_i{i}))
xlabel('$r$ (m)')
set(gca, 'FontSize',18)
axis square
xline(xx(i),'--r')
xlim([0 round(xx(i)*3,3)])
ylim([0 1.05])
TLS_A(i) = fitresult.a;
title(sprintf(('$\\lambda_{%.1s}$ = %.3f m'),V_i{i},TLS_A(i)))
end
% Common legend
leg = legend('$R_{ii}(r)=1-\frac{r^2}{\lambda^2}$','Orientation','Horizontal');
legend boxoff
leg.Layout.Tile = 'south';
set(legend,'FontSize',14)

Respuestas (1)

Tushar
Tushar el 20 de Sept. de 2023
Editada: Tushar el 20 de Sept. de 2023
Hi Daniel,
I understand that the legend you are trying to add to the plot obstructs the axes labels in a tiled layout. You can display the legend in a separate tile of the layout.
Here is a code snippet for the same:
% Define the tiled layout
t=tiledlayout('flow','TileSpacing','loose'); % 'flow' ensures that the layout can accomodate any number of axes
title(t,"output of the code");
% make a new tile using the nexttile command
nexttile;
p1=plot(rand(5));
xlabel('x-axis');
ylabel('y-axis');
% second plot
nexttile;
p2=plot(rand(5));
xlabel('x-axis');
ylabel('y-axis');
% third plot
nexttile;
p3=plot(rand(5));
xlabel('x-axis');
ylabel('y-axis');
% specify the legend
lgd = legend;
lgd.Layout.Tile=4; % in your case, this would be (n+1)
legend boxoff;
lgd.Layout.TileSpan=[1 3]; % in your case, this would be [1 n]
The above is also explained as an example in the documentation linked below:
For more info on tiled layouts, refer to the documentation here:
Hope this helps,
Tushar Agarwal
  1 comentario
Daniel
Daniel el 23 de Sept. de 2023
Hi thanks for your response.
I am aware of the approach described in the documentation, however, this is not an elegant solution in that the legend takes up a disproportionate amount of the window as it occupies the space of an entire tile. I am looking for a solution where only three tiles are used and the legend remains in the 'south' location

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