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Scatter plot does not display point 0,0 when double precision

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chris
chris el 23 de Abr. de 2014
Comentada: AJ von Alt el 23 de Abr. de 2014
I have the following code which works as intended, plotting 3 points.
x = [0 0; 12 0; 6 6];
scatter(x(:,1), x(:,2), 100, 'fill');
However, when I interject the following line
x = 1.01 * x;
which effectively turns everything into double precision, then the point 0,0 does not display
why is that, and how can I make it so that it also displays?
Thanks in advance!

Respuesta aceptada

José-Luis
José-Luis el 23 de Abr. de 2014
Trying to understand the mess that is Matlab's graphics engine is a real pain. I can only recommend a workaround:
x = [0 0; 12 0; 6 6];
figure(1)
scatter(x(:,1), x(:,2), 100, 'fill');
x = 1.01 * x;
figure(2)
scatter(x(:,1), x(:,2), 100, 'fill'); %Now readjust the axes so you can see what you actually plotted:
set(gca,'XLim',x([1 2],1)); %you can use min() max() in order to avoid hard coding the indices.
set(gca,'YLim',x([1 3],2));

Más respuestas (3)

AJ von Alt
AJ von Alt el 23 de Abr. de 2014
This looks like an issue with the Painters renderer. Switching to OpenGL resolves the issue on my end.
x = [0 0; 12 0; 6 6];
figure('renderer','OpenGL')
scatter(x(:,1), x(:,2), 100, 'fill');
y = 1.01 * x;
figure('renderer','OpenGL')
scatter(y(:,1), y(:,2), 100, 'fill');
  2 comentarios
chris
chris el 23 de Abr. de 2014
This also works! Although it goes beyond my ability to understand this, it is a cleaner solution. Thanks!
P.S. I think you need the
figure('renderer','OpenGL')
line only once, otherwise it creates an additional empty plot...
AJ von Alt
AJ von Alt el 23 de Abr. de 2014
The renderer is used to draw the plot in the figure. It is defined independently for each figure so you will need to specify it every time you create one.
You can change the renderer of the current figure to "OpenGL" using the command:
set( gcf , 'renderer' , 'OpenGL' )

Iniciar sesión para comentar.


chris
chris el 23 de Abr. de 2014
Thanks for your answer. Attached image is how my system behaves.

chris
chris el 23 de Abr. de 2014
This works! It is weird though why the original version doesn't behave seamlessly.
Anyways, thanks everybody for the help.

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