what does the four vector mean in matlab legend 'Position'?

I use the line:

       leg = legend('Ionisation','Recombination');
       pos = get(leg,'Position')

But get returned the answer:

pos =0.638988095238095,0.209126984126984, O.257142857142857,0.1

Can I ask what this means because this not the positions on the axes because when you look by eye it seems incorrect...

All I want to do is find the location (on the axes of the graph) of the bottom left corner so I can add another text box below with a string in it...

Thanks in advance MA

Answer 1

[Azzi Abdelmalek]

pos(1) and pos(2) are the coordinate of the legend windows
pos(3) and pos(4) are width and height of this windows

[Martin]

Thanks Azzi, what do you mean coordinate of legend windows? And what is the coordinate reference t?

Thanks

[Azzi Abdelmalek]

To understand try:

x0=0;
y0=0;
width=0.25;
height=0.25;
plot(1,1,2,2);
legend({'a','b'},'position',[x0 y0 width height])

change the position x0, and y0 then run again

x0=0.5;
y0=1
legend({'a','b'},'position',[x0 y0 width height])

x0,y0 are the coordinate of the left down corner of the legend windows

[Malek Barhoush]

x0=0;

y0=0;

width=0.25;

height=0.25;

plot(1,1,2,2);

legend({'a','b'},'position',[x0 y0 width height])

Answer 2

[Walter Roberson]

In order to understand the coordinates, you need to get() the Units property. Units can be any of several things, but in my experience the three that are used the most are "normal", "pixel", and "data".

Pixel is a measurement in pixels; fractional pixels are allowed.

Normal is based on fractions of the containing object, 0 being none of the object and 1 being the entire object. For example 'normal' and 0.3 as an axis position would mean 3/10 of the way along (from the left) the containing figure (or uipanel)

Data is data coordinates, based around the current xlim, ylim, or zlim. For example in your call

plot(1,1,2,2)

the 1's and 2's are in data coordinates.

If a graphics object has no Units property but has a Position property, then the graphics object is locked into Data coordinates; those kinds of graphics objects must be children of an axes.

Answer 3

[Jan]

You can learn more about the position by experimenting:

leg = legend('Ionisation','Recombination');
pos = get(leg,'Position');

pause(1);
set(leg, 'Position', [0.64, 0.21, O.25, 0.1]);

pause(1);
set(leg, 'Position', [0.74, 0.21, O.25, 0.1]);

pause(1);
set(leg, 'Position', [0.64, 0.31, O.25, 0.1]);

pause(1);
set(leg, 'Position', [0.64, 0.21, O.35, 0.1]);

pause(1);
set(leg, 'Position', [0.64, 0.21, O.25, 0.2]);

pause(1);
set(leg, 'Position', [0.0, 0.0, 1.0, 1.0]);

pause(1);
set(leg, 'Position', [0.5, 0.5, 0.4, 0.4]);

Answer 4

[Martin]

Thanks for all the information on this guys. I think what I have is the dimensions given in 'normal' units. Basically I have to run this simulation a lot of times, each time the y axis can change by 2 orders of magnitude and therefore I can't hardwire the textbox positions. I'm hoping that I can find some sort of conversion factor between 'normal' and 'data' coords as that's what the textbox positions are given in.

[Jan]

The limits of the axes do not matter, whan you use 'normalized' posiotions. Matlab converts the absolute position on the screen automatically for you. This code writes the string to the upper right corner without knowing the data size:

axes('XLim', cumsum(rand(1, 2)), 'YLim', cumsum(rand(1, 2)));
H = text(0.9, 0.9, 'Hello', 'VerticalAlignment', 'top', ...
     'HorizontalAlignment', 'right', ...
     'Units', 'normalized');
drawnow;
% Obtain the position in data units if wanted:
set(H, 'Units', 'data');
dataPos = get(H, 'Position');

[Martin]

Ah that's a really good way to do it, thanks. I appreciate that a lot, and certainly have learned a good deal.