Displaying two different variable legends in the same figure

I am trying to display two legends in the same figure.Here, z1,z2 and z3 all are variables.Please suggest.
Suppose z1=0.4, z2=0.9, & z3=0.25
lgd1=legend(['r_th=' num2str(z1)]);
lgd2=legend(['K=' num2str(z2+z3)]);

3 comentarios

Do you need multiple legends or can you instead use one legend with multiple rows?
Thanks. I am trying to get display like this:
r_th=0.4, K=0.9+0.25=1.15.
It can be in same row or multiple rows.
You still did not answer my question though. Yuvaraj's answer shows you how to manipulate the plot by superimposing two axes with one legend on each axes, which is a way to make 2 legends. I believe what you are really after is a single legend with multiple inputs, which is really basic. See the documentation for the correct legend syntax.

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Respuestas (2)

hAx(1) = axes();
hLine(1) = plot(1:10, 'Parent',hAx(1));
set(hAx(1), 'Box','off')
legend(hLine(1), 'line')
hAx(2) = copyobj(hAx(1),gcf);
delete( get(hAx(2),'Children') ) hLine(2) = plot(sin(1:10), 'Color','r', 'Parent',hAx(2));
set(hAx(2), 'Color','none', 'XTick',[], ... 'YAxisLocation','right', 'Box','off') %# make it transparent legend(hLine(2), {'curve'}, 'Location','NorthWest', 'Color','w')
You can only plot one legend in each axes object.
With that restriction, you can plot them as 1 row and 2 columns, (or 2 columns and 1 row in R2018a and later):
legend({sprintf('r_{th} = %.2f',z1), sprintf('K = %.2f + %.2f = %.2f', z2, z3, z2+z3)}, 'Location','N') % One Column, Two Rows
legend({sprintf('r_{th} = %.2f',z1), sprintf('K = %.2f + %.2f = %.2f', z2, z3, z2+z3)}, 'Location','N', 'NumColumns',2) % Two Columns, One Row (R2018a+)

2 comentarios

Both r_th and K are initial parameters (not axes parameter) and I
define their value at the beginning of code. For example:
z1=0.4; z2=0.9; z3=0.2;
r_th=z1;
K=z2+z3;
What I am trying to do is I want to choose their different value
randomly (at the beginning of code) without changing their value also in
legend every time. Please suggest.
I have nothing to suggest. I have no idea what you want.

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Preguntada:

el 3 de Mayo de 2018

Comentada:

el 4 de Mayo de 2018

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