Dear Jonathan,
a late reply: just to say that what you outline below works well - thank you.
Best wishes,
Hans
On Fri, 27 May 2011, Jonathan Guyer wrote:
On May 27, 2011, at 10:15 AM, Fangohr H. wrote:
we are currently using the fipy.Viewer class to visualise a scalar field as a
function of time (as was demonstrated in the diffusion example provided by
fipy); example code attached below.
In addition, I would like to mark a particular x,y position on that plot which
changes from iteration to iteration.
If I was not using fipy, I could use matplotlib to update for every iteration a
plot of using
plot( [x], [y], 'o')
to plot a circle at position (x,y), say.
Sure, that's doable.
I couldn't make this work together with fipy -- presumably because fipy wraps
up some matplotlib functionality to make things more convenient.
It does, but we've recently added the ability to more easily access the
matplotlib internals for exactly this sort of use. You can see how to tell FiPy
to use a particular set of matplotlib axes at:
http://www.ctcms.nist.gov/fipy/fipy/generated/viewers.matplotlibViewer.html#module-fipy.viewers.matplotlibViewer
You need to specify that you want a MatplotlibViewer instead of a general
Viewer, and pass it the axes you want it to use. You are then free to
manipulate those axes in any way you wish. In your script, you could do:
from matplotlib import pyplot
fig = pyplot.figure()
ax = fig.add_axes((0.1, 0.1, 0.8, 0.8))
viewer = fipy.MatplotlibViewer(vars=phi, datamin=-phi_outside*1.,
datamax=phi_outside*1, axes=ax)
and then
ll = []
for step in range(steps):
.. eq.solve(var=phi,
.. boundaryConditions=BCs,
.. dt=timeStepDuration)
.. for l in ll:
.. l.remove()
.. ll = ax.plot([step*0.2], [20 - step*0.15], 'ro')
.. viewer.plot()
Thanks in advance, and thank you for providing fipy and the user support.
our pleasure
--
Hans Fangohr
School of Engineering Sciences
University of Southampton
Phone: +44 (0) 238059 8345
Email: [email protected]
http://www.soton.ac.uk/~fangohr
