Re: [matplotlib-devel] AttributeError: 'module' object has no attribute 'CallLater'

[Nils W. <nw...@ia...>]

On Tue, 29 Jul 2008 16:43:42 -0500
  "John Hunter" <jd...@gm...> wrote:
> On Tue, Jul 29, 2008 at 4:38 PM, Paul Kienzle 
><pki...@ni...> wrote:
> 
>> Okay, how about
>>    if not hasattr(self,'IsVisible'):
>>        self.IsVisible = lambda self: True
>> in __init__
>>
>> That way it will always think it is visible.
> 
> OK, Nils, I committed a minor variant of that in r5926. 
> Test again...
> 
> JDH

Hi John,

Sorry for the delay and thank you for your patience.
Here is the output of the "never-ending story"

Traceback (most recent call last):
   File 
"/usr/lib/python2.4/site-packages/matplotlib/backends/backend_wx.py", 
line 1099, in _onPaint
     self.draw(drawDC=drawDC)
   File 
"/usr/lib/python2.4/site-packages/matplotlib/backends/backend_wxagg.py", 
line 64, in draw
     self.gui_repaint(drawDC=drawDC)
   File 
"/usr/lib/python2.4/site-packages/matplotlib/backends/backend_wx.py", 
line 1000, in gui_repaint
     if self.IsShownOnScreen():
TypeError: <lambda>() takes exactly 1 argument (0 given)
Traceback (most recent call last):
   File 
"/usr/lib/python2.4/site-packages/matplotlib/backends/backend_wx.py", 
line 1101, in _onPaint
     self.gui_repaint(drawDC=drawDC)
   File 
"/usr/lib/python2.4/site-packages/matplotlib/backends/backend_wx.py", 
line 1000, in gui_repaint
     if self.IsShownOnScreen():
TypeError: <lambda>() takes exactly 1 argument (0 given)
  
BTW, the entries are identical for both revisions.
Is that intended ?

r5923 | jdh2358 | 2008-07-29 20:01:16 +0200 (Di, 29 Jul 
2008) | 1 line

special case contains test for degenerate rectangle
------------------------------------------------------------------------
r5922 | jdh2358 | 2008-07-29 19:47:50 +0200 (Di, 29 Jul 
2008) | 1 line

special case contains test for degenerate rectangle