From: Jan Strube <jstrube@sl...> - 2005-07-28 08:31:31
first of all thanks a lot for this amazing release. I am absolutely
excited that PAIDA works now in Jython !!!
I am trying to plot an IFunction. Your 2d example works fine in jython,
but I will have to create my own subclass of IFunction.
Now, for some reason the Plotter kicks me out with
... paida_core/IPlotterRegion.py", line 697, in plot
The problem is that my class doesn't get recognized as a subclass of
IFunction in the IPlotterRegion class. The funny thing is that I
explicitly do check if the class is an instance of IFunction.
In other words:
myFunc = MyFunction()
if isinstance(myFunc, IFunction):
print success # <-- This step passes !
region.plot(myFunc) # <-- In IPlotterRegion the test elif
(isinstance(data1, IFunction)) and (data2 == None) and (data3 == None):
I do not understand how this test can pass once and fail the second
Could you give me an example of plotting a class that derives from
IFunction, please ?
Jan Fridolf Strube University of Oregon
Stanford Linear Accelerator Center
bldg 48, rm 244
phone: (650) 926-2913
From: Koji KISHIMOTO <korry@us...> - 2005-07-28 12:09:06
I'm sorry. It's my fault.
For Jython2.1 which does not have some features of Python2.3, I
re-defined isinstance() in paida.paida_core.PAbsorber but incorrectly.
I'll release a bug fix version 3.2.1_2.8.1 soon.
Thanks for your pointing out !