Interpreter doesn't work in legend when legend has multiple outputs (2016b)?
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Dan Gianotti
am 31 Jan. 2017
Kommentiert: Vince Badagnani
am 3 Apr. 2017
Hi, I think I may have found a weird bug in 2016b. The following works as expected:
p = plot(1:2,2:3);
l = legend({'My Line'},'Interpreter','latex');
But if I request an additional output the interpreter stops working:
p = plot(1:2,2:3);
[l,a] = legend({'My Line'},'Interpreter','latex');
This is even true if instead you write:
[l,~] = legend({'My Line'},'Interpreter','latex');
This isn't just me, right?
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Walter Roberson
am 31 Jan. 2017
When you use multiple outputs of legend() in R2014b or later, it constructs the graphics differently, creating line and text objects instead of handling the objects internally. You can extract the text objects out of what it returns and set their Interpreter property.
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Star Strider
am 31 Jan. 2017
From the documentation:
- [lgd,icons,plots,txt] = legend(_) additionally returns the objects used to create the legend icons, the objects plotted in the graph, and an array of the label text. Note: This syntax is not recommended. It creates a legend that does not support all graphics features. Instead, use the lgd = legend() syntax to return the legend object and set Legend Properties. (Emphasis mine)
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Vince Badagnani
am 3 Apr. 2017
Yes, I'm running into this as well. My code worked in R2015b as follows:
[FFT_legend, ObjH, ~, ~] = legend(ltext, 'Location', 'best', 'Interpreter', 'none');
FFT_legend.FontSize = 20;
The variable ltext is a Mx1 cell array. But now I'm finding in R2016b that I have to write my code as follows:
[FFT_legend, ObjH, ~, ~] = legend(ltext);
txt = findobj(ObjH, 'Type', 'Text');
arrayfun(@(x) set(x, 'FontSize', 20, 'Interpreter', 'none'), txt);
FFT_legend.Location = 'best';
But it gives me a warning when attempting to set the FFT_legend location.
Warning: Error updating Legend.
Struct contents reference from a non-struct array object.
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