matplotlib-users Mailing List for matplotlib (Page 4)

Hmmm, it isn't the same, but I wonder if it is related to the oddities
around https://github.com/matplotlib/matplotlib/pull/2925 . What I mean is
that in that case, none of our unit tests caught the problem, even though
it should have. In your case, your unit tests is catching a similar
problem, but there is no obvious reason why there should be a problem.
Now, technically speaking, in your case, there is a bug in the unittest
package (values() is an iterator in py3k rather than a list), but I
wouldn't know why that dictionary would change in the first place.
On Mon, Apr 14, 2014 at 2:54 PM, John Evans <joh...@gm...>wrote:
> Helllo, I'm seeing a strange issue when running unittests on python3.3 and
> python3.4 that somehow involves matplotlib. My code has a somewhat
> complicated setup, but I think I've boiled the issue down to the following
> reproduction steps
>
>
> import unittest
> import warnings
>
> import matplotlib.pyplot
>
> class TestMe(unittest.TestCase):
> def test_warn(self):
> with self.assertWarns(UserWarning):
> warnings.warn("a warning", UserWarning)
>
> if __name__ == "__main__":
> unittest.main()
>
>
> It looks like it should pass, but it errors as follows
>
> E
> ======================================================================
> ERROR: test_warn (__main__.TestMe)
> ----------------------------------------------------------------------
> Traceback (most recent call last):
> File "/homes/5/jevans/Downloads/testit.py", line 8, in test_warn
> with self.assertWarns(UserWarning):
> File
> "/space/getafix/1/users/jevans/anaconda/envs/py3k/lib/python3.3/unittest/case.py",
> line 177, in __enter__
> for v in sys.modules.values():
> RuntimeError: dictionary changed size during iteration
>
> ----------------------------------------------------------------------
> Ran 1 test in 0.002
>
>
> If the matplotlib import is changed to just
>
> import matlotlib
>
> or if the matplotlib import is commented out altogether, it then passes.
> I'm seeing the behavior on Anaconda with python 3.3 and matplotlib 1.3.1
> on both mac and linux, but also with MacPorts with pythons 3.3 and 3.4,
> matplotlib 1.3.1. All seems fine with a Fedora 20 laptop with python 3.3
> and also matplotlib 1.3.1.
>
>
> --
> John Evans
>
>
> ------------------------------------------------------------------------------
> Learn Graph Databases - Download FREE O'Reilly Book
> "Graph Databases" is the definitive new guide to graph databases and their
> applications. Written by three acclaimed leaders in the field,
> this first edition is now available. Download your free book today!
> http://p.sf.net/sfu/NeoTech
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>

Dear Phil,
Thank you. This solves my problem. So the title of my mail is wrong, the
behaviour is reasonable but I am using wrong coordinates.
And also thanks to Jeff.
Cheers,
Chao
On Wed, Apr 16, 2014 at 10:14 AM, Phil Elson <pel...@gm...> wrote:
> Hi Chao,
>
> The warning you are getting:
>
> WARNING: x coordinate not monotonically increasing - contour plot
> may not be what you expect. If it looks odd, your can either
> adjust the map projection region to be consistent with your data, or
> (if your data is on a global lat/lon grid) use the shiftgrid
> function to adjust the data to be consistent with the map projection
> region (see examples/contour_demo.py).
>
> Is important here. It looks like the x coordinate is not in appropriate
> longitudes. Printing the first 5 and last 5 longitudes gives us our first
> clue:
>
> First 5: [-180. -178.99720764 -177.99443054 -176.99163818 -175.98886108]
> Last 5 : [ 175.98886108 176.9916687 177.9944458 178.9972229 180.00003052]
>
> Notice that the last longitude wraps around beyond 180. So if we were to
> clip these numbers to -180 and +180 we will see that the warning disappears
> and the contour is correct. This can be achieved with:
>
> lon = np.clip(lon, -180, 180)
>
> Alternatively, we can just construct the latitudes and longitudes
> ourselves directly with:
>
> lon, lat = np.meshgrid(np.linspace(-180, 180, 360), np.linspace(-90, 90,
> 180))
>
> Incidentally, I tried these numbers with cartopy which has been designed
> to handle dateline wrapping automatically, and the contour worked with the
> unmodified longitudes (http://nbviewer.ipython.org/gist/pelson/10830039).
>
> ---------------------------
>
> @JeffWhitaker - This looks like a bug with float tolerances in the
> makegrid function. It currently does:
>
> def makegrid(self,nx,ny,returnxy=False):
> dx = (self.urcrnrx-self.llcrnrx)/(nx-1)
> dy = (self.urcrnry-self.llcrnry)/(ny-1)
>
> But might be better if it did:
>
> def makegrid(self,nx,ny,returnxy=False):
> x = np.linspace(self.llcrnrx, self.urcrnrx, nx)
> y = np.linspace(self.llcrnry, self.urcrnry, ny)
>
> To avoid the multiplicative floating point drift that is currently being
> seen.
>
> HTH,
>
> Phil
>
-- 
please visit:
http://www.globalcarbonatlas.org/
***********************************************************************************
Chao YUE
Laboratoire des Sciences du Climat et de l'Environnement (LSCE-IPSL)
UMR 1572 CEA-CNRS-UVSQ
Batiment 712 - Pe 119
91191 GIF Sur YVETTE Cedex
Tel: (33) 01 69 08 29 02; Fax:01.69.08.77.16
************************************************************************************

Hi Chao,
The warning you are getting:
WARNING: x coordinate not monotonically increasing - contour plot
may not be what you expect. If it looks odd, your can either
adjust the map projection region to be consistent with your data, or
(if your data is on a global lat/lon grid) use the shiftgrid
function to adjust the data to be consistent with the map projection
region (see examples/contour_demo.py).
Is important here. It looks like the x coordinate is not in appropriate
longitudes. Printing the first 5 and last 5 longitudes gives us our first
clue:
First 5: [-180. -178.99720764 -177.99443054 -176.99163818 -175.98886108]
Last 5 : [ 175.98886108 176.9916687 177.9944458 178.9972229 180.00003052]
Notice that the last longitude wraps around beyond 180. So if we were to
clip these numbers to -180 and +180 we will see that the warning disappears
and the contour is correct. This can be achieved with:
lon = np.clip(lon, -180, 180)
Alternatively, we can just construct the latitudes and longitudes ourselves
directly with:
lon, lat = np.meshgrid(np.linspace(-180, 180, 360), np.linspace(-90, 90,
180))
Incidentally, I tried these numbers with cartopy which has been designed to
handle dateline wrapping automatically, and the contour worked with the
unmodified longitudes (http://nbviewer.ipython.org/gist/pelson/10830039).
---------------------------
@JeffWhitaker - This looks like a bug with float tolerances in the makegrid
function. It currently does:
 def makegrid(self,nx,ny,returnxy=False):
 dx = (self.urcrnrx-self.llcrnrx)/(nx-1)
 dy = (self.urcrnry-self.llcrnry)/(ny-1)
But might be better if it did:
 def makegrid(self,nx,ny,returnxy=False):
 x = np.linspace(self.llcrnrx, self.urcrnrx, nx)
 y = np.linspace(self.llcrnry, self.urcrnry, ny)
To avoid the multiplicative floating point drift that is currently being
seen.
HTH,
Phil

Cool notebook. I took the liberty of giving it a go with cartopy, and you
can see the results here http://nbviewer.ipython.org/gist/pelson/10822698
I'd agree that the issue you linked to does look very similar to the issue
you are seeing, so I think this is very likely a bug.
Cheers,
On 16 April 2014 00:54, Scott Henderson <st...@co...> wrote:
> Hello,
>
> I’m trying to make a plot data on a map with the ‘cyl’ projection with a
> shifted centerline (lon_0=180), but I receive an error when shiftdata() is
> called. Since the plot works when lon_0=0, this seems to be a bug.
>
> I’ve posted the code, error, and plots here:
>
> http://nbviewer.ipython.org/gist/anonymous/cbfe6d0f66ff3a8186c8/shiftdata_issue.ipynb
>
> It might be related to this issue:
> https://github.com/matplotlib/basemap/issues/126
>
> Any insight would be appreciated!
> Scott
>
> ------------------------------------------------------------------------------
> Learn Graph Databases - Download FREE O'Reilly Book
> "Graph Databases" is the definitive new guide to graph databases and their
> applications. Written by three acclaimed leaders in the field,
> this first edition is now available. Download your free book today!
> http://p.sf.net/sfu/NeoTech
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>

Hello,
I’m trying to make a plot data on a map with the ‘cyl’ projection with a shifted centerline (lon_0=180), but I receive an error when shiftdata() is called. Since the plot works when lon_0=0, this seems to be a bug. 
I’ve posted the code, error, and plots here:
http://nbviewer.ipython.org/gist/anonymous/cbfe6d0f66ff3a8186c8/shiftdata_issue.ipynb
It might be related to this issue:
https://github.com/matplotlib/basemap/issues/126
Any insight would be appreciated!
Scott

actually this makes a lot of sense..
thanks for the pointers.. I shall keep on digging..
laszlo
On 14 April 2014 18:36, Benjamin Root <ben...@ou...> wrote:
> I think the closest you are going to get is with using the "shade=True"
> kwarg in plot_surface(). This is the only way that mplot3d utilizes normal
> vectors, and that really only makes one side look "duller" than the other.
>
> Since you mentioned wanting to eventually display self-intersecting
> surfaces, I would probably suggest trying out Mayavi2 or glumpy instead as
> those are more geared towards 3d visualization than mplot3d is. mplot3d has
> significant issues with rendering intersecting polygons because it isn't a
> true 3d plotting system (it just computes projections of whole polygons and
> uses a single z value to represent where in the layering the polygon should
> go).
>
> Cheers!
> Ben Root
>
>
>
> On Mon, Apr 14, 2014 at 9:34 AM, László Oroszlány <oro...@gm...>wrote:
>
>> well I sort of wanted to avoid doing two spheres.. later on I wanted to
>> do more complicated surfaces.. and it can get a bit messy.. It is not
>> straight forward to generate the two parallel surfaces in general.. to be
>> honest the problematic case would be when i want to display
>> selfintersecting but still orientable surfaces (NOT Klein bottles or
>> Moebius strips)
>> cheers anyway for the quick response
>>
>> laszlo
>>
>>
>>
>> On 14 April 2014 15:21, Shahar Shani Kadmiel <ka...@po...>wrote:
>>
>>> Hi, I am not aware of such an option (AFAIK) but my suggestion would be
>>> to make two spheres with very small radii difference, paint the slightly
>>> smaller one (inside) blue and the other one red.
>>> Just a quick fix for the problem at hand. I'm sure the experts here will
>>> have plenty of very in depth solutions.
>>>
>>> Shahar
>>> —
>>> Sent from Mailbox <https://www.dropbox.com/mailbox> for iPhone
>>>
>>>
>>> On Mon, Apr 14, 2014 at 1:48 PM, László Oroszlány <oro...@gm...>wrote:
>>>
>>>> Dear matplotlib users,
>>>>
>>>> I recently started using matplotlib to make a couple of educational
>>>> presentations.
>>>> For most of my problems I found the manual and the examples on the web
>>>> enough,
>>>> however I ran into a bit of an issue regarding plotting some surfaces.
>>>> My main problem has to do with plotting orientable (or two sided
>>>> surfaces).
>>>> Simply put I want to plot a sphere cut in half and make the inside red
>>>> and the outside blue.
>>>> I was wondering if there exist some flag or option in the already
>>>> available plotting functions to do this?
>>>>
>>>> Cheers
>>>>
>>>> laszlo
>>>>
>>>>
>>>
>>
>>
>> ------------------------------------------------------------------------------
>> Learn Graph Databases - Download FREE O'Reilly Book
>> "Graph Databases" is the definitive new guide to graph databases and their
>> applications. Written by three acclaimed leaders in the field,
>> this first edition is now available. Download your free book today!
>> http://p.sf.net/sfu/NeoTech
>> _______________________________________________
>> Matplotlib-users mailing list
>> Mat...@li...
>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>
>>
>

Helllo, I'm seeing a strange issue when running unittests on python3.3 and
python3.4 that somehow involves matplotlib. My code has a somewhat
complicated setup, but I think I've boiled the issue down to the following
reproduction steps
import unittest
import warnings
import matplotlib.pyplot
class TestMe(unittest.TestCase):
 def test_warn(self):
 with self.assertWarns(UserWarning):
 warnings.warn("a warning", UserWarning)
if __name__ == "__main__":
 unittest.main()
It looks like it should pass, but it errors as follows
E
======================================================================
ERROR: test_warn (__main__.TestMe)
----------------------------------------------------------------------
Traceback (most recent call last):
 File "/homes/5/jevans/Downloads/testit.py", line 8, in test_warn
 with self.assertWarns(UserWarning):
 File
"/space/getafix/1/users/jevans/anaconda/envs/py3k/lib/python3.3/unittest/case.py",
line 177, in __enter__
 for v in sys.modules.values():
RuntimeError: dictionary changed size during iteration
----------------------------------------------------------------------
Ran 1 test in 0.002
If the matplotlib import is changed to just
import matlotlib
or if the matplotlib import is commented out altogether, it then passes.
 I'm seeing the behavior on Anaconda with python 3.3 and matplotlib 1.3.1
on both mac and linux, but also with MacPorts with pythons 3.3 and 3.4,
matplotlib 1.3.1. All seems fine with a Fedora 20 laptop with python 3.3
and also matplotlib 1.3.1.
-- 
John Evans

I think the closest you are going to get is with using the "shade=True"
kwarg in plot_surface(). This is the only way that mplot3d utilizes normal
vectors, and that really only makes one side look "duller" than the other.
Since you mentioned wanting to eventually display self-intersecting
surfaces, I would probably suggest trying out Mayavi2 or glumpy instead as
those are more geared towards 3d visualization than mplot3d is. mplot3d has
significant issues with rendering intersecting polygons because it isn't a
true 3d plotting system (it just computes projections of whole polygons and
uses a single z value to represent where in the layering the polygon should
go).
Cheers!
Ben Root
On Mon, Apr 14, 2014 at 9:34 AM, László Oroszlány <oro...@gm...>wrote:
> well I sort of wanted to avoid doing two spheres.. later on I wanted to
> do more complicated surfaces.. and it can get a bit messy.. It is not
> straight forward to generate the two parallel surfaces in general.. to be
> honest the problematic case would be when i want to display
> selfintersecting but still orientable surfaces (NOT Klein bottles or
> Moebius strips)
> cheers anyway for the quick response
>
> laszlo
>
>
>
> On 14 April 2014 15:21, Shahar Shani Kadmiel <ka...@po...>wrote:
>
>> Hi, I am not aware of such an option (AFAIK) but my suggestion would be
>> to make two spheres with very small radii difference, paint the slightly
>> smaller one (inside) blue and the other one red.
>> Just a quick fix for the problem at hand. I'm sure the experts here will
>> have plenty of very in depth solutions.
>>
>> Shahar
>> —
>> Sent from Mailbox <https://www.dropbox.com/mailbox> for iPhone
>>
>>
>> On Mon, Apr 14, 2014 at 1:48 PM, László Oroszlány <oro...@gm...>wrote:
>>
>>> Dear matplotlib users,
>>>
>>> I recently started using matplotlib to make a couple of educational
>>> presentations.
>>> For most of my problems I found the manual and the examples on the web
>>> enough,
>>> however I ran into a bit of an issue regarding plotting some surfaces.
>>> My main problem has to do with plotting orientable (or two sided
>>> surfaces).
>>> Simply put I want to plot a sphere cut in half and make the inside red
>>> and the outside blue.
>>> I was wondering if there exist some flag or option in the already
>>> available plotting functions to do this?
>>>
>>> Cheers
>>>
>>> laszlo
>>>
>>>
>>
>
>
> ------------------------------------------------------------------------------
> Learn Graph Databases - Download FREE O'Reilly Book
> "Graph Databases" is the definitive new guide to graph databases and their
> applications. Written by three acclaimed leaders in the field,
> this first edition is now available. Download your free book today!
> http://p.sf.net/sfu/NeoTech
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>

Hi, I am not aware of such an option (AFAIK) but my suggestion would be to make two spheres with very small radii difference, paint the slightly smaller one (inside) blue and the other one red. Just a quick fix for the problem at hand. I'm sure the experts here will have plenty of very in depth solutions. 
Shahar
—
Sent from Mailbox for iPhone
On Mon, Apr 14, 2014 at 1:48 PM, László Oroszlány <oro...@gm...>
wrote:
> Dear matplotlib users,
> I recently started using matplotlib to make a couple of educational
> presentations.
> For most of my problems I found the manual and the examples on the web
> enough,
> however I ran into a bit of an issue regarding plotting some surfaces.
> My main problem has to do with plotting orientable (or two sided surfaces).
> Simply put I want to plot a sphere cut in half and make the inside red and
> the outside blue.
> I was wondering if there exist some flag or option in the already available
> plotting functions to do this?
> Cheers
> laszlo

well I sort of wanted to avoid doing two spheres.. later on I wanted to do
more complicated surfaces.. and it can get a bit messy.. It is not straight
forward to generate the two parallel surfaces in general.. to be honest
the problematic case would be when i want to display selfintersecting but
still orientable surfaces (NOT Klein bottles or Moebius strips)
cheers anyway for the quick response
laszlo
On 14 April 2014 15:21, Shahar Shani Kadmiel <ka...@po...> wrote:
> Hi, I am not aware of such an option (AFAIK) but my suggestion would be to
> make two spheres with very small radii difference, paint the slightly
> smaller one (inside) blue and the other one red.
> Just a quick fix for the problem at hand. I'm sure the experts here will
> have plenty of very in depth solutions.
>
> Shahar
> —
> Sent from Mailbox <https://www.dropbox.com/mailbox> for iPhone
>
>
> On Mon, Apr 14, 2014 at 1:48 PM, László Oroszlány <oro...@gm...>wrote:
>
>> Dear matplotlib users,
>>
>> I recently started using matplotlib to make a couple of educational
>> presentations.
>> For most of my problems I found the manual and the examples on the web
>> enough,
>> however I ran into a bit of an issue regarding plotting some surfaces.
>> My main problem has to do with plotting orientable (or two sided
>> surfaces).
>> Simply put I want to plot a sphere cut in half and make the inside red
>> and the outside blue.
>> I was wondering if there exist some flag or option in the already
>> available plotting functions to do this?
>>
>> Cheers
>>
>> laszlo
>>
>>
>

Dear matplotlib users,
I recently started using matplotlib to make a couple of educational
presentations.
For most of my problems I found the manual and the examples on the web
enough,
however I ran into a bit of an issue regarding plotting some surfaces.
My main problem has to do with plotting orientable (or two sided surfaces).
Simply put I want to plot a sphere cut in half and make the inside red and
the outside blue.
I was wondering if there exist some flag or option in the already available
 plotting functions to do this?
Cheers
laszlo

Dear all,
I am trying to plot some data (see attached data.txt) on global coverage
with 1-degree resolution on the Robinson projection using Basemap. However
I get some strange band on the high latitude, and imshow function by
matplotlib does not show similar thing. Please refer to the two attached
figures.
Could anyone give me some tips? thanks!!! Below is a working example:
import numy as np
import mpl_toolkits.basemap as bmp
import matplotlib.pyplot as plt
#The example file data.txt could be downloaded from dropbox:
https://www.dropbox.com/s/xma4w540qa83sa6/data.txt
#############
data=np.genfromtxt('data.txt',usemask=True,missing_values='0.000000000000000000e+00')
print np.ma.unique(data)
lev = np.arange(0.5,8.6,1)
print lev
### here we have to first build the equal-distance grid, this is inspired
from here:
###
http://matplotlib.1069221.n5.nabble.com/Basemap-plotting-data-on-projection-td40973.html
cyl_basemap = bmp.Basemap(projection='cyl', llcrnrlat=-90, urcrnrlat=90,
 llcrnrlon=-180, urcrnrlon=180, resolution='l')
lon, lat = cyl_basemap.makegrid(360, 180)
fig,ax=plt.subplots(1,1)
m=bmp.Basemap(projection='robin',lon_0=0,resolution='c',ax=ax)
m.drawcoastlines()
x, y = m(lon, np.flipud(lat))
m.contourf(x,y,data=data,levels=lev)
#############################
Cheers,
Chao
-- 
please visit:
http://www.globalcarbonatlas.org/
***********************************************************************************
Chao YUE
Laboratoire des Sciences du Climat et de l'Environnement (LSCE-IPSL)
UMR 1572 CEA-CNRS-UVSQ
Batiment 712 - Pe 119
91191 GIF Sur YVETTE Cedex
Tel: (33) 01 69 08 29 02; Fax:01.69.08.77.16
************************************************************************************

You're on windows, so that last command is just "activate mpl33"
On Fri, Apr 11, 2014 at 7:05 PM, Paul Hobson <pmh...@gm...> wrote:
> You should be using conda to install matplotlib:
>
> conda create --name mpl33 matplotlib python=3.3 ipython-notebok
> source activate mpl33
>
>
> On Fri, Apr 11, 2014 at 8:11 AM, grivet <gr...@cn...> wrote:
>
>> Under Win7pro, I have tried to install matplotlib by running either
>> matplotlib-1.3.0.win32-py3.3.exe or
>> matplotlib-1.3.1.win32-py3.3.exe. In each case, the installer tells
>> pythion-3.3 is not in the registry (although
>> Anaconda was successfully installed previousle). A pop-up window then
>> opens, asking me for the python
>> and installation directories. It proves impossible to write anything in
>> this window.
>> Does anybody know what's wrong in my set up ?
>> TIA for any help
>> JP Grivet*
>>
>>
>>
>> ------------------------------------------------------------------------------
>> Put Bad Developers to Shame
>> Dominate Development with Jenkins Continuous Integration
>> Continuously Automate Build, Test & Deployment
>> Start a new project now. Try Jenkins in the cloud.
>> http://p.sf.net/sfu/13600_Cloudbees
>> _______________________________________________
>> Matplotlib-users mailing list
>> Mat...@li...
>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>
>
>

You should be using conda to install matplotlib:
conda create --name mpl33 matplotlib python=3.3 ipython-notebok
source activate mpl33
On Fri, Apr 11, 2014 at 8:11 AM, grivet <gr...@cn...> wrote:
> Under Win7pro, I have tried to install matplotlib by running either
> matplotlib-1.3.0.win32-py3.3.exe or
> matplotlib-1.3.1.win32-py3.3.exe. In each case, the installer tells
> pythion-3.3 is not in the registry (although
> Anaconda was successfully installed previousle). A pop-up window then
> opens, asking me for the python
> and installation directories. It proves impossible to write anything in
> this window.
> Does anybody know what's wrong in my set up ?
> TIA for any help
> JP Grivet*
>
>
>
> ------------------------------------------------------------------------------
> Put Bad Developers to Shame
> Dominate Development with Jenkins Continuous Integration
> Continuously Automate Build, Test & Deployment
> Start a new project now. Try Jenkins in the cloud.
> http://p.sf.net/sfu/13600_Cloudbees
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>

Hey Adam,
I wouldn't make a circle, instead use a Line2D artist (accessible from
pyplot) with they same (base) symbology as your scatter plot. It has pretty
much the same call signature as ax.plot
-paul
On Fri, Apr 11, 2014 at 9:35 AM, Sterling Smith <sm...@fu...>wrote:
> Adam,
>
> I agree that the Circle ended up with a rectangle in the legend, which I
> wouldn't think of as the expected response. Would the following work for
> your purposes?
>
> figure()
> p,=plot(0,0,marker='o',ls='',color='red')
> legend([p], ["Red Rectangle"],numpoints=1)
> p.remove()
> draw()
>
> -Sterling
>
> On Apr 11, 2014, at 8:14AM, Adam Hughes wrote:
>
> > Hi Paul,
> >
> > I tried out the legend proxy artist, and it works for rectangles in the
> legend, but I can't seem to get a Circle to appear in the legend, which I
> presume should be:
> >
> > p = Circle((0, 0), fc="r")
> > legend([p], ["Red Rectangle"])
> >
> >
> > On Wed, Apr 9, 2014 at 2:20 PM, Adam Hughes <hug...@gm...>
> wrote:
> > Thanks Paul, I will try it out.
> >
> >
> > On Wed, Apr 9, 2014 at 12:21 PM, Paul Hobson <pmh...@gm...> wrote:
> >
> >
> >
> > On Wed, Apr 9, 2014 at 9:00 AM, Adam Hughes <hug...@gm...>
> wrote:
> > Thanks. That's probably the way I'll go. At first, I thought creating
> separate legend markers and removing them from the plot seemed hacky, but I
> guess there's no way that matplotlib could know which legend size I want.
> I wonder if there'd be any interest in a PR to add a keyword to legend to
> handle this situation?
> >
> > Why not just work the other way around with proxy artists. IOW, make the
> artists but never add them to the plot.
> >
> >
> http://matplotlib.org/users/legend_guide.html?highlight=proxy%20artists#using-proxy-artist
> > (works with Line2D artists)
> >
> > -p
> >
> >
> >
> > On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...>
> wrote:
> > Adam,
> >
> > I haven't investigated, but does the discussion of the legend marker at
> [1] help?
> >
> > -Sterling
> >
> > [1]
> https://www.mail-archive.com/mat...@li.../msg25200.html
> >
> > On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
> >
> > > Hello,
> > >
> > > I've been searching but can't seem to find this topic addressed
> (perhaps wrong search terms)
> > >
> > > Simply put, I have a scatter plot with variable size markers, and I'd
> like to have the markers all be a single size in the legend. Is there a
> standard way to do this?
> > >
> > > Thanks.
> > >
> ------------------------------------------------------------------------------
> > > Put Bad Developers to Shame
> > > Dominate Development with Jenkins Continuous Integration
> > > Continuously Automate Build, Test & Deployment
> > > Start a new project now. Try Jenkins in the cloud.
> > >
> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
> > > Matplotlib-users mailing list
> > > Mat...@li...
> > > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> >
> >
> >
> >
> ------------------------------------------------------------------------------
> > Put Bad Developers to Shame
> > Dominate Development with Jenkins Continuous Integration
> > Continuously Automate Build, Test & Deployment
> > Start a new project now. Try Jenkins in the cloud.
> > http://p.sf.net/sfu/13600_Cloudbees
> > _______________________________________________
> > Matplotlib-users mailing list
> > Mat...@li...
> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> >
> >
> >
> >
>
>

Adam,
I agree that the Circle ended up with a rectangle in the legend, which I wouldn't think of as the expected response. Would the following work for your purposes?
figure()
p,=plot(0,0,marker='o',ls='',color='red')
legend([p], ["Red Rectangle"],numpoints=1)
p.remove()
draw()
-Sterling
On Apr 11, 2014, at 8:14AM, Adam Hughes wrote:
> Hi Paul, 
> 
> I tried out the legend proxy artist, and it works for rectangles in the legend, but I can't seem to get a Circle to appear in the legend, which I presume should be:
> 
> p = Circle((0, 0), fc="r")
> legend([p], ["Red Rectangle"])
> 
> 
> On Wed, Apr 9, 2014 at 2:20 PM, Adam Hughes <hug...@gm...> wrote:
> Thanks Paul, I will try it out.
> 
> 
> On Wed, Apr 9, 2014 at 12:21 PM, Paul Hobson <pmh...@gm...> wrote:
> 
> 
> 
> On Wed, Apr 9, 2014 at 9:00 AM, Adam Hughes <hug...@gm...> wrote:
> Thanks. That's probably the way I'll go. At first, I thought creating separate legend markers and removing them from the plot seemed hacky, but I guess there's no way that matplotlib could know which legend size I want. I wonder if there'd be any interest in a PR to add a keyword to legend to handle this situation?
> 
> Why not just work the other way around with proxy artists. IOW, make the artists but never add them to the plot.
> 
> http://matplotlib.org/users/legend_guide.html?highlight=proxy%20artists#using-proxy-artist
> (works with Line2D artists)
> 
> -p
> 
> 
> 
> On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...> wrote:
> Adam,
> 
> I haven't investigated, but does the discussion of the legend marker at [1] help?
> 
> -Sterling
> 
> [1] https://www.mail-archive.com/mat...@li.../msg25200.html
> 
> On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
> 
> > Hello,
> >
> > I've been searching but can't seem to find this topic addressed (perhaps wrong search terms)
> >
> > Simply put, I have a scatter plot with variable size markers, and I'd like to have the markers all be a single size in the legend. Is there a standard way to do this?
> >
> > Thanks.
> > ------------------------------------------------------------------------------
> > Put Bad Developers to Shame
> > Dominate Development with Jenkins Continuous Integration
> > Continuously Automate Build, Test & Deployment
> > Start a new project now. Try Jenkins in the cloud.
> > http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
> > Matplotlib-users mailing list
> > Mat...@li...
> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> 
> 
> 
> ------------------------------------------------------------------------------
> Put Bad Developers to Shame
> Dominate Development with Jenkins Continuous Integration
> Continuously Automate Build, Test & Deployment
> Start a new project now. Try Jenkins in the cloud.
> http://p.sf.net/sfu/13600_Cloudbees
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
> 
> 
> 
> 

I can confirm the inconsistency between behavior and documentation for 1.3.1. The errorbar line gets the color of the line, not the marker. Probably you should file a bug report on github.
-Sterling
On Apr 11, 2014, at 7:50AM, Oliver wrote:
> I apologize if this has been fixed already, I can only check different versions at home. However, the documentation of mpl 1.3.1. has the same information. So unless the code changed to reflect the documentation, this is still present.
> 
> When using errorbar, the documentation says the color of the errorbar lines will match with the color of the markers if ecolor=None. That’s not what I found. Apparently it takes over the color of the Line2D instance which interconnects the markers.
> 
> Short, Self Contained, Correct Example:
> 
> from pylab import *
> plt.ion() # saves typing show
> 
> x = np.arange(10)
> y = np.random.rand(10)
> xerr, yerr = y/4., y/4.
> 
> # Markers in red, but errorlines assume the color of the "trendline" (default rcparams: blue).
> errorbar(x, y, yerr=yerr, mfc='r', marker='o', ecolor=None) 
> 
> # Errorlines get color green now - documentation not in line with results
> 
> figure(); errorbar(x, y, yerr=xerr, mfc='r', marker='o', ecolor=None, color='g')
> # Errorlines get color blue now, because it can be specified - expected behaviour
> figure(); errorbar(x, y, yerr=xerr, mfc='r', marker='o', ecolor='b', color='g')
> 
> Is this an oversight mistake?
> 
> ------------------------------------------------------------------------------
> Put Bad Developers to Shame
> Dominate Development with Jenkins Continuous Integration
> Continuously Automate Build, Test & Deployment 
> Start a new project now. Try Jenkins in the cloud.
> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users

Under Win7pro, I have tried to install matplotlib by running either 
matplotlib-1.3.0.win32-py3.3.exe or
matplotlib-1.3.1.win32-py3.3.exe. In each case, the installer tells 
pythion-3.3 is not in the registry (although
Anaconda was successfully installed previousle). A pop-up window then 
opens, asking me for the python
and installation directories. It proves impossible to write anything in 
this window.
Does anybody know what's wrong in my set up ?
TIA for any help
JP Grivet*

Hi Paul,
I tried out the legend proxy artist, and it works for rectangles in the
legend, but I can't seem to get a Circle to appear in the legend, which I
presume should be:
p = Circle((0, 0), fc="r")
legend([p], ["Red Rectangle"])
On Wed, Apr 9, 2014 at 2:20 PM, Adam Hughes <hug...@gm...> wrote:
> Thanks Paul, I will try it out.
>
>
> On Wed, Apr 9, 2014 at 12:21 PM, Paul Hobson <pmh...@gm...> wrote:
>
>>
>>
>>
>> On Wed, Apr 9, 2014 at 9:00 AM, Adam Hughes <hug...@gm...>wrote:
>>
>>> Thanks. That's probably the way I'll go. At first, I thought creating
>>> separate legend markers and removing them from the plot seemed hacky, but I
>>> guess there's no way that matplotlib could know which legend size I want.
>>> I wonder if there'd be any interest in a PR to add a keyword to legend to
>>> handle this situation?
>>>
>>
>> Why not just work the other way around with proxy artists. IOW, make the
>> artists but never add them to the plot.
>>
>>
>> http://matplotlib.org/users/legend_guide.html?highlight=proxy%20artists#using-proxy-artist
>> (works with Line2D artists)
>>
>> -p
>>
>>
>>
>>>
>>> On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...>wrote:
>>>
>>>> Adam,
>>>>
>>>> I haven't investigated, but does the discussion of the legend marker at
>>>> [1] help?
>>>>
>>>> -Sterling
>>>>
>>>> [1]
>>>> https://www.mail-archive.com/mat...@li.../msg25200.html
>>>>
>>>> On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
>>>>
>>>> > Hello,
>>>> >
>>>> > I've been searching but can't seem to find this topic addressed
>>>> (perhaps wrong search terms)
>>>> >
>>>> > Simply put, I have a scatter plot with variable size markers, and I'd
>>>> like to have the markers all be a single size in the legend. Is there a
>>>> standard way to do this?
>>>> >
>>>> > Thanks.
>>>> >
>>>> ------------------------------------------------------------------------------
>>>> > Put Bad Developers to Shame
>>>> > Dominate Development with Jenkins Continuous Integration
>>>> > Continuously Automate Build, Test & Deployment
>>>> > Start a new project now. Try Jenkins in the cloud.
>>>> >
>>>> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
>>>> > Matplotlib-users mailing list
>>>> > Mat...@li...
>>>> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>>>
>>>>
>>>
>>>
>>> ------------------------------------------------------------------------------
>>> Put Bad Developers to Shame
>>> Dominate Development with Jenkins Continuous Integration
>>> Continuously Automate Build, Test & Deployment
>>> Start a new project now. Try Jenkins in the cloud.
>>> http://p.sf.net/sfu/13600_Cloudbees
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Mat...@li...
>>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>>
>>>
>>
>

I apologize if this has been fixed already, I can only check different
versions at home. However, the documentation of mpl
1.3.1<http://matplotlib.org/1.3.1/api/pyplot_api.html#matplotlib.pyplot.errorbar>.
has the same information. So unless the code changed to reflect the
documentation, this is still present.
When using errorbar, the documentation says the color of the errorbar lines
will match with the color of the markers if ecolor=None. That’s not what I
found. Apparently it takes over the color of the Line2D instance which
interconnects the markers.
Short, Self Contained, Correct Example:
from pylab import *
plt.ion() # saves typing show
x = np.arange(10)
y = np.random.rand(10)
xerr, yerr = y/4., y/4.
# Markers in red, but errorlines assume the color of the "trendline"
(default rcparams: blue).
errorbar(x, y, yerr=yerr, mfc='r', marker='o', ecolor=None)
# Errorlines get color green now - documentation not in line with results
figure(); errorbar(x, y, yerr=xerr, mfc='r', marker='o', ecolor=None, color='g')
# Errorlines get color blue now, because it can be specified -
expected behaviour
figure(); errorbar(x, y, yerr=xerr, mfc='r', marker='o', ecolor='b', color='g')
Is this an oversight mistake?

Hello everyone,
Just a quick reminder that the EuroScipy call for abstracts closes on the
14th: don't forget to submit your talk proposal! It is in four days only!
In short, EuroScipy is a cross-disciplinary gathering focused on the use
and development of the Python language in scientific research. This event
strives to bring together both users and developers of scientific tools, as
well as academic research and state of the art industry.
EuroSciPy 2014, the Seventh Annual Conference on Python in Science, takes
place in *Cambridge, UK on 27 - 30 August 2014*. The conference features
two days of tutorials followed by two days of scientific talks. The day
after the main conference, developer sprints will be organized on projects
of interest to attendees.
The topics presented at EuroSciPy are very diverse, with a focus on
advanced software engineering and original uses of Python and its
scientific libraries, either in theoretical or experimental research, from
both academia and the industry. The program includes keynotes, contributed
talks and posters.
Submissions for talks and posters are welcome on our website (
http://www.euroscipy.org/2014/). In your abstract, please provide details
on what Python tools are being employed, and how. The deadline for
submission is 14 April 2014.
Also until 14 April 2014, you can apply for a sprint session on 31 August
2014. See https://www.euroscipy.org/2014/calls/sprints/ for details.
Thanks,
N

 I want to plot audio data (samples versus time), and then animate a marker
moving across the plot while the audio is played back. The marker should
stay in sync with the audio data that is being played at that moment. Is
there a way to do this using the animation capabilities of matplotlib? If I
could sync the movement of the market with an accurate timer (like a QTimer
in Qt) it should be good enough.
Mike

Thanks Paul, I will try it out.
On Wed, Apr 9, 2014 at 12:21 PM, Paul Hobson <pmh...@gm...> wrote:
>
>
>
> On Wed, Apr 9, 2014 at 9:00 AM, Adam Hughes <hug...@gm...>wrote:
>
>> Thanks. That's probably the way I'll go. At first, I thought creating
>> separate legend markers and removing them from the plot seemed hacky, but I
>> guess there's no way that matplotlib could know which legend size I want.
>> I wonder if there'd be any interest in a PR to add a keyword to legend to
>> handle this situation?
>>
>
> Why not just work the other way around with proxy artists. IOW, make the
> artists but never add them to the plot.
>
>
> http://matplotlib.org/users/legend_guide.html?highlight=proxy%20artists#using-proxy-artist
> (works with Line2D artists)
>
> -p
>
>
>
>>
>> On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...>wrote:
>>
>>> Adam,
>>>
>>> I haven't investigated, but does the discussion of the legend marker at
>>> [1] help?
>>>
>>> -Sterling
>>>
>>> [1]
>>> https://www.mail-archive.com/mat...@li.../msg25200.html
>>>
>>> On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
>>>
>>> > Hello,
>>> >
>>> > I've been searching but can't seem to find this topic addressed
>>> (perhaps wrong search terms)
>>> >
>>> > Simply put, I have a scatter plot with variable size markers, and I'd
>>> like to have the markers all be a single size in the legend. Is there a
>>> standard way to do this?
>>> >
>>> > Thanks.
>>> >
>>> ------------------------------------------------------------------------------
>>> > Put Bad Developers to Shame
>>> > Dominate Development with Jenkins Continuous Integration
>>> > Continuously Automate Build, Test & Deployment
>>> > Start a new project now. Try Jenkins in the cloud.
>>> >
>>> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
>>> > Matplotlib-users mailing list
>>> > Mat...@li...
>>> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>>
>>>
>>
>>
>> ------------------------------------------------------------------------------
>> Put Bad Developers to Shame
>> Dominate Development with Jenkins Continuous Integration
>> Continuously Automate Build, Test & Deployment
>> Start a new project now. Try Jenkins in the cloud.
>> http://p.sf.net/sfu/13600_Cloudbees
>> _______________________________________________
>> Matplotlib-users mailing list
>> Mat...@li...
>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>
>>
>

On Wed, Apr 9, 2014 at 9:00 AM, Adam Hughes <hug...@gm...> wrote:
> Thanks. That's probably the way I'll go. At first, I thought creating
> separate legend markers and removing them from the plot seemed hacky, but I
> guess there's no way that matplotlib could know which legend size I want.
> I wonder if there'd be any interest in a PR to add a keyword to legend to
> handle this situation?
>
Why not just work the other way around with proxy artists. IOW, make the
artists but never add them to the plot.
http://matplotlib.org/users/legend_guide.html?highlight=proxy%20artists#using-proxy-artist
(works with Line2D artists)
-p
>
> On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...>wrote:
>
>> Adam,
>>
>> I haven't investigated, but does the discussion of the legend marker at
>> [1] help?
>>
>> -Sterling
>>
>> [1]
>> https://www.mail-archive.com/mat...@li.../msg25200.html
>>
>> On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
>>
>> > Hello,
>> >
>> > I've been searching but can't seem to find this topic addressed
>> (perhaps wrong search terms)
>> >
>> > Simply put, I have a scatter plot with variable size markers, and I'd
>> like to have the markers all be a single size in the legend. Is there a
>> standard way to do this?
>> >
>> > Thanks.
>> >
>> ------------------------------------------------------------------------------
>> > Put Bad Developers to Shame
>> > Dominate Development with Jenkins Continuous Integration
>> > Continuously Automate Build, Test & Deployment
>> > Start a new project now. Try Jenkins in the cloud.
>> >
>> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
>> > Matplotlib-users mailing list
>> > Mat...@li...
>> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>
>>
>
>
> ------------------------------------------------------------------------------
> Put Bad Developers to Shame
> Dominate Development with Jenkins Continuous Integration
> Continuously Automate Build, Test & Deployment
> Start a new project now. Try Jenkins in the cloud.
> http://p.sf.net/sfu/13600_Cloudbees
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>

Thanks. That's probably the way I'll go. At first, I thought creating
separate legend markers and removing them from the plot seemed hacky, but I
guess there's no way that matplotlib could know which legend size I want.
 I wonder if there'd be any interest in a PR to add a keyword to legend to
handle this situation?
On Wed, Apr 9, 2014 at 1:44 AM, Sterling Smith <sm...@fu...>wrote:
> Adam,
>
> I haven't investigated, but does the discussion of the legend marker at
> [1] help?
>
> -Sterling
>
> [1]
> https://www.mail-archive.com/mat...@li.../msg25200.html
>
> On Apr 8, 2014, at 3:44PM, Adam Hughes wrote:
>
> > Hello,
> >
> > I've been searching but can't seem to find this topic addressed (perhaps
> wrong search terms)
> >
> > Simply put, I have a scatter plot with variable size markers, and I'd
> like to have the markers all be a single size in the legend. Is there a
> standard way to do this?
> >
> > Thanks.
> >
> ------------------------------------------------------------------------------
> > Put Bad Developers to Shame
> > Dominate Development with Jenkins Continuous Integration
> > Continuously Automate Build, Test & Deployment
> > Start a new project now. Try Jenkins in the cloud.
> >
> http://p.sf.net/sfu/13600_Cloudbees_______________________________________________
> > Matplotlib-users mailing list
> > Mat...@li...
> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>