Re: [Matplotlib-users] imshow and projection axes

[Michael D. <md...@st...>]

BTW -- I think a possible fix may be to do:

         if not im.get_clip_path():
             im.set_clip_path(self.patch)

Mike

On 07/23/2010 09:16 AM, Michael Droettboom wrote:
> I've traced this back to revision 7867:
>
> http://matplotlib.svn.sourceforge.net/viewvc/matplotlib/branches/v0_99_maint/lib/matplotlib/axes.py?r1=7867&r2=7866&pathrev=7867 
> <http://matplotlib.svn.sourceforge.net/viewvc/matplotlib/branches/v0_99_maint/lib/matplotlib/axes.py?r1=7867&r2=7866&pathrev=7867>
>
> Andrew: I'm sure that change is there for a good reason -- do you 
> remember what it was?  I don't just want to revert it.  Interestingly, 
> get_clip_on() is True by default, and I'm not sure in what case it's 
> set to False in this context.
>
> Mike
>
> On 07/23/2010 03:36 AM, Tobias Winchen wrote:
>> Hi,
>>
>>    
>>> On Thursday 22 July 2010 Michael Droettboom wrote:
>>> Which backend are you using?
>>>      
>> I tried this with GTKAgg and Qt4Agg, its was working with 0.99.0 and 0.99.1 on
>> Debian/Squeeze and 0.99.0 on Scientific Linux 5. I first noticed this with
>> 0.99.3 on Debian/Squeeze, but it is the same with with 1.0.0 on Debian/Squeeze
>> and Scientific Linux 5.
>>
>>    
>>> Can you provide a short script that
>>> reproduces the bug?
>>>      
>> This script reproduces the behaviour:
>>
>>
>> import pylab
>> from math import *
>>
>> d = [[1,2],[3,4]]
>>
>> f = pylab.figure()
>> pylab.subplot(111, projection='hammer')
>>
>> pylab.imshow(d, extent=(-pi,pi,-pi/2,pi/2))
>> f.show()
>>
>>
>> It creates the following plot in version 0.99.0
>> http://web.physik.rwth-aachen.de/~winchen/mpl_0990.png
>>
>> and in version 0.99.3
>> http://web.physik.rwth-aachen.de/~winchen/mpl_0993.png
>>
>> Best regards,
>>
>>
>> Tobias
>>    
>>
>>
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>
>
> -- 
> Michael Droettboom
> Science Software Branch
> Space Telescope Science Institute
> Baltimore, Maryland, USA


-- 
Michael Droettboom
Science Software Branch
Space Telescope Science Institute
Baltimore, Maryland, USA