Thread Subject:

Inser text in graph

Hi,
I am having one figure which shows there different solutions in three different colors. I want to add a text in the upper right corner of the figure with the same colors. Any help will be appreciated.

On 6/5/2012 9:53 AM, Sanaa wrote:
> Hi,
> I am having one figure which shows there different solutions in three
>different colors. I want to add a text in the upper right corner of
>the figure with the same colors. Any help will be appreciated.

You can use legend(), or if you want to put the text yourself
you can use text()


"text Text annotation.
     text(X,Y,'string') adds the text in the quotes to location (X,Y)
     on the current axes,"


" legend Display legend.
     legend(string1,string2,string3, ...) puts a legend on the current plot
     using the specified strings as labels"

Thanks a lot for your help. But I need the strings appears with the same corresponding colors!!
Thanks a lot in advance

On 6/5/2012 10:13 AM, Sanaa wrote:
> Thanks a lot for your help. But I need the strings appears with
> the same corresponding colors!!
> Thanks a lot in advance

But that what legend does. Have you tried it?

x = 0:pi/100:2*pi;
plot(x,sin(x),'r',x,cos(x),'k');
legend('sin(x)','cos(x)');

screen shot:

http://12000.org/tmp/060512/legend_example.png

So, I have no idea what is the problem you are having.

--Nasser

Thanks a lot for your help. I will explain more. I have to plot the trajectories of a map with three different parameters. So, I am having the code

% define the vector of values that r takes
r_vec = [0.4, 07, 1.5];
% define the vector of colors, to plot the dat aof reach value of r in
% different color
color_vec = ['r', 'b', 'k'];


% do a loop of values of r
for j = 1 : size(r_vec,2)
% define delay
r = r_vec(j);
% define coefficient rho
rho = 0.5;
% define the number of deiscrete times in interval [k*r, (k+1)*r]
n = 50;
% define the nimber of iterations k = 1,.....,Nit
Nit = 10;
% define the initial vector which is a vector of size n
x0 = ones(1,n);
% define vectors x_next and x_previous
x_next = zeros(1,n);
x_previous = zeros(1,n);
time = zeros(1,n);

% initialize x_previous
x_previous = x0;
x_next = x_previous;
time = linspace(0,r,n);
    
  % set the figure details
    figure(1)
    xlabel('t')
    ylabel('x')
    plot(time,x_next,'.')
    hold on
for i = 1: Nit
    
    x_next = 1 - rho*x_previous.^2;
    x_previous = x_next;
    time = linspace(i*r,(i+1)*r,n);
    
    
    plot(time,x_next,'.', 'Color', char(color_vec(j)))
    
end
   
   
end
hold off

As you can see, I am having tow plot commands for just one ''r''. In the example you wrote you have sine and cos....I am confused

On 6/5/2012 5:18 PM, Sanaa wrote:
> Thanks a lot for your help. I will explain more. I have to plot the
> trajectories of a map with three different parameters. So, I am having
> the code
>
  r_vec = [0.4, 07, 1.5];
  color_vec = ['r', 'b', 'k'];
...

>
> As you can see, I am having tow plot commands for just one ''r''. In the
> example you wrote you have sine and cos....I am confused

[lh,oh]=legend({'r=0.25';'r=0.5';'r=1'});
for ix=1:3,set(oh(ix*2),'color',color_vec(ix)),end


By your drawing as you did, there are six object handles, the first of
which is the text the 2nd,4th,6th are the line objects corresponding to
the legend texts (you can find this out by exploring the objects'
properties in the object handle vector via set()--look at 'type' for
starters).

--

On 6/5/2012 5:18 PM, Sanaa wrote:
> Thanks a lot for your help. I will explain more. I have to plot the trajectories of a map with three different parameters. So, I am having the code
>
> % define the vector of values that r takes
> r_vec = [0.4, 07, 1.5];
> % define the vector of colors, to plot the dat aof reach value of r in
> % different color
> color_vec = ['r', 'b', 'k'];
>
>
> % do a loop of values of r
> for j = 1 : size(r_vec,2)
> % define delay
> r = r_vec(j);
> % define coefficient rho
> rho = 0.5;
> % define the number of deiscrete times in interval [k*r, (k+1)*r]
> n = 50;
> % define the nimber of iterations k = 1,.....,Nit
> Nit = 10;
> % define the initial vector which is a vector of size n
> x0 = ones(1,n);
> % define vectors x_next and x_previous
> x_next = zeros(1,n);
> x_previous = zeros(1,n);
> time = zeros(1,n);
>
> % initialize x_previous
> x_previous = x0;
> x_next = x_previous;
> time = linspace(0,r,n);
>
> % set the figure details
> figure(1)
> xlabel('t')
> ylabel('x')
> plot(time,x_next,'.')
> hold on
> for i = 1: Nit
>
> x_next = 1 - rho*x_previous.^2;
> x_previous = x_next;
> time = linspace(i*r,(i+1)*r,n);
>
>
> plot(time,x_next,'.', 'Color', char(color_vec(j)))
>
> end
>
>
> end
> hold off
>
> As you can see, I am having tow plot commands for just one ''r''.
>In the example you wrote you have sine and cos....I am confused

btw, I think your code above is not efficient. Try this, does the same
for less than half the code. You define things you do not use. You
create things not needed. NO need to put things inside a loop
if they do not change. etc...

---------------------------------------
clear all; close all;
r_vec = [0.4, 7, 1.5];
annot = {'-ro','-.b','*k'};
rho = 0.5;
n = 50;
Nit = 10;

figure(1);
hold on

for j = 1 : length(r_vec)
     
     x_previous = ones(1,n);
     x_next = x_previous;
     time = linspace(0,r_vec(j),n);
                
% plot(time,x_next,'.') %why?
     for i = 1: Nit
     
        x_next = 1 - rho*x_previous.^2;
        x_previous = x_next;
        time = linspace(i*r_vec(j),(i+1)*r_vec(j),n);
        plot(time,x_next,char(annot(j)))
     end
end

xlabel('t')
ylabel('x')
----------------------------

I see you got an answer for the legend issue from dpb now,
so I hope all is set now.

--Nasser

Thanks a lot for your modification. However, when I used legend I got one small problem which is:
the values of the parameters are reversed that is, not in the same order as in the plot '' I mean their colors'' and one of them is written completely in that color.
To see what I mean please run this code

clear all; close all;
r_vec = [0.25, 0.5, 1];
annot = {'r','b','k'};
rho = 0.5;
n = 50;
Nit = 10;

figure(1);
hold on

for j = 1 : length(r_vec)
     
     x_previous = ones(1,n);
     x_next = x_previous;
     time = linspace(0,r_vec(j),n);
                
% plot(time,x_next,'.') %why?
     for i = 1: Nit
     
        x_next = 1 - rho*x_previous.^2;
        x_previous = x_next;
        time = linspace(i*r_vec(j),(i+1)*r_vec(j),n);
        plot(time,x_next,char(annot(j)),'linewidth',3)
     end
end
[lh,oh]=legend({'r=0.25';'r=0.5';'r=1'});
for ix=1:3,set(oh(ix*2),'color',annot(ix)),end
xlabel('t')
ylabel('x')

Thanks a lot for your help

On 6/6/2012 2:22 PM, Sanaa wrote:
> Thanks a lot for your modification. However, when I used legend I got
> one small problem which is:
> the values of the parameters are reversed that is, not in the same order
> as in the plot '' I mean their colors'' and one of them is written
> completely in that color. To see what I mean please run this code
...

 >> for ix=1:3,set(oh(ix*2),'color',color_vec(ix)),end

Seems to work here...I get the above annotation w/ a red,blue,black line
in that order from the top...

Apparently there's a modification in the internals of legend() between
versions; what does the above show you for the handles to the legend
objects?

 >> get(oh,'type')
ans =
     'text'
     'line'
     'line'
     'line'
     'line'
     'line'
     'line'
 >> get(oh(2:end),'color');
 >> ans{:}
ans =
      1 0 0
ans =
      1 0 0
ans =
      0 0 1
ans =
      0 0 1
ans =
      0 0 0
ans =
      0 0 0
 >>

These are the color triplets for the lines; the first two are red, next
blue, last two black.

If they are indeed different, then you'll have to look into them and see
which ones you need to change.

--

On 6/6/2012 2:40 PM, dpb wrote:
...

> >> for ix=1:3,set(oh(ix*2),'color',color_vec(ix)),end
>
> Seems to work here...I get the above annotation w/ a red,blue,black line
> in that order from the top...
>
...

That is, the legend is

r=0.25
r=0.5
r=1

in that order w/ the lines respectively r, b, k and the text in black
after executing the above set() loop

--

Thanks a lot for helping me. I have modified the code according to what Me. Nasser wrote. That is I don't have color_vec anymore, I am having annot instead, that's why I am getting the error

??? Error using ==> set
Color value must be a 3 element numeric vector

Error in ==> Trajectories_modified at 27
for ix=1:3,set(oh(ix*2),'color',annot(ix)),end

The code is

clear all; close all;
r_vec = [0.25, 0.5, 1];
annot = {'r','b','k'};
rho = 0.5;
n = 50;
Nit = 10;

figure(1);
hold on

for j = 1 : length(r_vec)
     
     x_previous = ones(1,n);
     x_next = x_previous;
     time = linspace(0,r_vec(j),n);
                
% plot(time,x_next,'.') %why?
     for i = 1: Nit
     
        x_next = 1 - rho*x_previous.^2;
        x_previous = x_next;
        time = linspace(i*r_vec(j),(i+1)*r_vec(j),n);
        plot(time,x_next,char(annot(j)),'linewidth',3)
     end
end
[lh,oh]=legend({'r=0.25';'r=0.5';'r=1'});
for ix=1:3,set(oh(ix*2),'color',annot(ix)),end
%for ix=1:3,set(oh(ix*2),'color',color_vec(ix)),end
xlabel('t')
ylabel('x')

Thanks a lot in advance.

On 6/6/2012 3:47 PM, Sanaa wrote:
> Thanks a lot for helping me. I have modified the code according to what
> Me. Nasser wrote. That is I don't have color_vec anymore, I am having
> annot instead, that's why I am getting the error
>
> ??? Error using ==> set
> Color value must be a 3 element numeric vector
>
> Error in ==> Trajectories_modified at 27
> for ix=1:3,set(oh(ix*2),'color',annot(ix)),end
>
> The code is
>
> clear all; close all;
> r_vec = [0.25, 0.5, 1];
> annot = {'r','b','k'};
...

Well, that's because you're passing a cell-string to set() not a
character. I don't know why you would change, but if you're going to
use a cell string instead of a character string, you'll have to
dereference it w/ the curly brackets instead of passing the cell.

--

On 6/6/2012 4:30 PM, dpb wrote:
...

> ... if you're going to use
> a cell string instead of a character string, you'll have to dereference
> it w/ the curly brackets instead of passing the cell.

Or, that's why there's the char() casting function in the other place
you've used it; but imo writing the {} is simpler altho results in same
thing.

But, raises the question of why? if all you're using it for is to get
the character that is required anyway, why not simply use a character
string instead and eliminate the need for the cast?

--