matplotlib-users Mailing List for matplotlib

Hi Jae-Joon,
Thanks for the quick fix! Just looked in the svn browser, and noticed 
you changed line 5290 of axes.py to
'o' : (0,0,3),
Should this not be
'o' : (0,3,0),
?
Thanks,
Tom
On Jun 21, 2009, at 2:59 PM, Jae-Joon Lee wrote:
> Thanks for the report.
> And, this turned out to be a bug. The symbol style code was simply
> ignored when its value is 3.
>
> While the bug should now be fixed (both in the trunk and the maint.
> branch), you may use marker style like (20,0,0) (or increase the first
> number when symbol is large) for a workaround.
>
> Regards,
>
> -JJ
>
>
> On Sun, Jun 21, 2009 at 1:09 PM, Thomas
> Robitaille<tho...@gm...> wrote:
>>
>> Hi,
>>
>> I'm trying to use the scatter method, making use of the option to 
>> specify
>> the marker as a tuple. From the documentation, it would seem that 
>> specifying
>> marker=(0,3,0) should draw a circle. However, this is not the case. 
>> If you
>> consider the following code:
>>
>> import matplotlib
>> matplotlib.use('Agg')
>> import matplotlib.pyplot as mpl
>>
>> fig = mpl.figure()
>> ax = fig.add_subplot(111)
>> #ax.scatter([x],[y],c='red',marker=(0,3,0.))
>> ax.scatter([10.],[10.],c='red',marker=(3,3,0.))
>> ax.scatter([11.],[10.],c='red',marker=(6,3,0.))
>> ax.set_xlim(5.,15.)
>> fig.savefig('scatter.png')
>>
>> The first ax.scatter causes an error, the second plots a triangle, 
>> and the
>> third a hexagon. However, the documentation states that (a) setting 
>> the
>> second element to '3' should plot a circle, and (b) the other 
>> arguments
>> should be ignored, so the first ax.scatter should not cause an error.
>>
>> Is this a bug, or am I misunderstanding the documentation?
>>
>> Thanks,
>>
>> Thomas
>> --
>> View this message in context: http://www.nabble.com/possible-bug-with-scatter-tp24136534p24136534.html
>> Sent from the matplotlib - users mailing list archive at Nabble.com.
>>
>>
>> ------------------------------------------------------------------------------
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>>

Thanks for the report.
And, this turned out to be a bug. The symbol style code was simply
ignored when its value is 3.
While the bug should now be fixed (both in the trunk and the maint.
branch), you may use marker style like (20,0,0) (or increase the first
number when symbol is large) for a workaround.
Regards,
-JJ
On Sun, Jun 21, 2009 at 1:09 PM, Thomas
Robitaille<tho...@gm...> wrote:
>
> Hi,
>
> I'm trying to use the scatter method, making use of the option to specify
> the marker as a tuple. From the documentation, it would seem that specifying
> marker=(0,3,0) should draw a circle. However, this is not the case. If you
> consider the following code:
>
> import matplotlib
> matplotlib.use('Agg')
> import matplotlib.pyplot as mpl
>
> fig = mpl.figure()
> ax = fig.add_subplot(111)
> #ax.scatter([x],[y],c='red',marker=(0,3,0.))
> ax.scatter([10.],[10.],c='red',marker=(3,3,0.))
> ax.scatter([11.],[10.],c='red',marker=(6,3,0.))
> ax.set_xlim(5.,15.)
> fig.savefig('scatter.png')
>
> The first ax.scatter causes an error, the second plots a triangle, and the
> third a hexagon. However, the documentation states that (a) setting the
> second element to '3' should plot a circle, and (b) the other arguments
> should be ignored, so the first ax.scatter should not cause an error.
>
> Is this a bug, or am I misunderstanding the documentation?
>
> Thanks,
>
> Thomas
> --
> View this message in context: http://www.nabble.com/possible-bug-with-scatter-tp24136534p24136534.html
> Sent from the matplotlib - users mailing list archive at Nabble.com.
>
>
> ------------------------------------------------------------------------------
> Are you an open source citizen? Join us for the Open Source Bridge conference!
> Portland, OR, June 17-19. Two days of sessions, one day of unconference: 250ドル.
> Need another reason to go? 24-hour hacker lounge. Register today!
> http://ad.doubleclick.net/clk;215844324;13503038;v?http://opensourcebridge.org
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>

Hi,
I'm trying to use the scatter method, making use of the option to specify
the marker as a tuple. From the documentation, it would seem that specifying
marker=(0,3,0) should draw a circle. However, this is not the case. If you
consider the following code:
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as mpl
fig = mpl.figure()
ax = fig.add_subplot(111)
#ax.scatter([x],[y],c='red',marker=(0,3,0.))
ax.scatter([10.],[10.],c='red',marker=(3,3,0.))
ax.scatter([11.],[10.],c='red',marker=(6,3,0.))
ax.set_xlim(5.,15.)
fig.savefig('scatter.png')
The first ax.scatter causes an error, the second plots a triangle, and the
third a hexagon. However, the documentation states that (a) setting the
second element to '3' should plot a circle, and (b) the other arguments
should be ignored, so the first ax.scatter should not cause an error.
Is this a bug, or am I misunderstanding the documentation?
Thanks,
Thomas
-- 
View this message in context: http://www.nabble.com/possible-bug-with-scatter-tp24136534p24136534.html
Sent from the matplotlib - users mailing list archive at Nabble.com.

Hello,
I'm using blit to animate my plot, and I'd like to add a grid on it
but I failed at it.
What I do is:
1. prepare Figure, Axes, etc without plotting anything
2. save teh background with
 self.bg = self.canvas.copy_from_bbox(self.ax.bbox)
(on graph update)
3. restore background with
 self.canvas.restore_region(self.bg)
4. <line>.set_ydata(...)
5. self.ax.draw_artist(self.<line>)
6. self.canvas.blit(self.ax.bbox)
Now, I tried to place a "self.ax.grid(True)" in almost every possible
position, but never a grid came out :(
What is the right approach to solve it? Thanks in advance
Regards,
-- 
Sandro Tosi (aka morph, morpheus, matrixhasu)
My website: http://matrixhasu.altervista.org/
Me at Debian: http://wiki.debian.org/SandroTosi