matplotlib-users Mailing List for matplotlib (Page 10)

Jerzy Karczmarczuk
<jer...@un...> writes:
> Could you provide a /working/ example with the geometry you really want?
> I believe I thought more or less about it as Tony Yu did. If it is 
> wrong, be more precise, please.

I have a data set that looks like this:

mydata = numpy.copy([

# lambda,  data

# First data row
 [[5002., 0.5],
  [5200., 0.34],
  [5251., -1.2],
#  ...
  [8997., 2.4]],

# second data row
 [[5002., 0.72],
  [5251., 0.9],
#  ... 
  [8997.,  0.1]],

# other data rows to follow
#  ...
  ])

where I want to put the first column (lambda) on the Y axis, which each
data row as one colorbar (like in your code), and the data as the color
of that data point -- interpolated vertically.

Best regards

Ole

...Earn money right from your computer.  http://ponto.kelly-systems.com.br/job.link.php?gid_friend=78u5

            ------------------------------
            Sent using Verizon.net Mobile

Could you provide a /working/ example with the geometry you really want?
I believe I thought more or less about it as Tony Yu did. If it is 
wrong, be more precise, please.

Jerzy Karczmarczuk

On Tue, Feb 14, 2012 at 10:33 AM, Olе Streicher <ole...@gm...>wrote:

> Tony Yu <ts...@gm...> writes:
> > Does the following do what you need?
> >
> > #----
> > import numpy as np
> > import matplotlib.pyplot as plt
> >
> > width = 0.5
> > height = 10
> >
> > for x in np.arange(11):
> >     strip = np.random.random(size=(50, 1))
> >     plt.imshow(strip, extent=(x-width/2, x+width/2, 0, height))
> >
> > plt.xlim(-width, x+width)
> >
> > plt.show()
> > #----
>
> Not completely since it requires an equally spaced "strip". I have two
> vectors (one for the vertical axis, one for the data).
>
> Best regards
>
> Ole
>
>
I'm not sure what you mean by equally spaced strip. If you need to scale
the height, you can just calculate it as from the length of the strip. If
you mean the placement of the strips, you can substitue the x values above
with your data instead of the `arange` call---you'd have to appropriately
scale `width` so that matches your x-spacing.

-Tony

Andrea, I believe that if you find ONE good visual aspect ratio, 
according to your figure size, that should do. Yo know your
figsize, and if you know all in your axes([bot,lft,wid, height]), that 
this visual ratio should be easy to fix.

BTW, I did not understand why do you play with your "angle" = 
arctan(...), then compute sin and cos, etc. Two issues.
1. You are wasting your time.
2. If your data is a functional sequence, with x growing, its OK, but 
for ANY data you may get zero in the denominator, and your arctan will 
locally explode, producing holes in the plotted data. Use arctan2. Or 
rather, don't compute angles at all.

Concerning 1.
If you compute dx = xhigh-xlow; dy=yhigh-ylow;
then the parameters of a normal offset are (nx,ny) = (-dy,dx), 
appropriately normalized, and that's all.

All the best.

Jerzy

Hi Jerzy,

On 14 February 2012 16:03, Jerzy Karczmarczuk wrote:
> Andrea Gavana :
>> after some great help from the Numpy users list, I have managed to
>> create "parallel curves".
>>
>>
>> But I can't really do that with the set of data I have, as the X and Y
>> variables have different order of magnitude and I need a single
>> subplot on the figure to have rectangular axes (not square).
> (...)
>> So, my question would be: how do I scale the
>> X and Y vectors so that the parallels look parallel to the main curve
>> even if the axes are not square and the X and Y variables have
>> different data-ranges/magnitudes?
> Andrea, you have TWO problems.
>
> The first is to scale your offset according to your axes range. This can
> be done using ax.get_data_ratio(). In your case you will get 5, and this
> factor should enhance your vertical offset wrt. the horizontal.
> (Or, use .get_xlim() and .get_ylim() and do the computations yourself).

Thank you for your answer, I have implemented this and it looks a bit
better (on the real X/Y pairs I have).

> The second problem is that your FIGURE scales your plot visually,
> independently of your axes, so without special scaling it will have
> different aspects according to your manipulation. An arbitrary affine
> transform will keep straight lines parallel, but no chance with
> arbitrary curves.  You may play with fig.get_figwidth(), etc., but here
> my digging stops.

Will this argument still stand if I am only interested in a single
figure size (maximized window on my screen, plus set_size_inches(20,
12)) and fixed axes positions (set by figure.subplots_adjust)? If not,
should the further scaling simply be the ratio between the x-axis
extent and y-axis extent (in pixels)? Or am I missing something
(again)?

Thank you in advance for your suggestions.


Andrea.

"Imagination Is The Only Weapon In The War Against Reality."
http://xoomer.alice.it/infinity77/

Tony Yu <ts...@gm...> writes:
> Does the following do what you need?
>
> #----
> import numpy as np
> import matplotlib.pyplot as plt
>
> width = 0.5
> height = 10
>
> for x in np.arange(11):
>     strip = np.random.random(size=(50, 1))
>     plt.imshow(strip, extent=(x-width/2, x+width/2, 0, height))
>
> plt.xlim(-width, x+width)
>
> plt.show()
> #----

Not completely since it requires an equally spaced "strip". I have two
vectors (one for the vertical axis, one for the data).

Best regards

Ole

On Tue, Feb 14, 2012 at 9:48 AM, Olе Streicher <ole...@gm...>wrote:

> Hi list,
>
> I have a list of 48 individual data sets (lambda,y) which I want to plot in
> the following diagram:
>
> lambda
> ^
> |
> |
> |
> |
> +--------------->
>  1 2 3 4 5 6 .. Dataset #
>
> The "y" values should be color coded here, in a similar fashion as it is
> done in a color bar (f.e. that I can interpolate between data points).
>
> Practically, I want to have 48 vertical color bars here, all with the
> same scaling. From the documentation, I could not find out how to do
> this; and also I could not find a good example for this.
>
> Could anyone help me here?
>
> Best
>
> Ole
>
>
>
Does the following do what you need?

#----
import numpy as np
import matplotlib.pyplot as plt

width = 0.5
height = 10

for x in np.arange(11):
    strip = np.random.random(size=(50, 1))
    plt.imshow(strip, extent=(x-width/2, x+width/2, 0, height))

plt.xlim(-width, x+width)

plt.show()
#----

-Tony

Andrea Gavana :
> after some great help from the Numpy users list, I have managed to
> create "parallel curves".
>
>
> But I can't really do that with the set of data I have, as the X and Y
> variables have different order of magnitude and I need a single
> subplot on the figure to have rectangular axes (not square).
(...)
> So, my question would be: how do I scale the
> X and Y vectors so that the parallels look parallel to the main curve
> even if the axes are not square and the X and Y variables have
> different data-ranges/magnitudes?
Andrea, you have TWO problems.

The first is to scale your offset according to your axes range. This can 
be done using ax.get_data_ratio(). In your case you will get 5, and this 
factor should enhance your vertical offset wrt. the horizontal.
(Or, use .get_xlim() and .get_ylim() and do the computations yourself).

The second problem is that your FIGURE scales your plot visually, 
independently of your axes, so without special scaling it will have 
different aspects according to your manipulation. An arbitrary affine 
transform will keep straight lines parallel, but no chance with 
arbitrary curves.  You may play with fig.get_figwidth(), etc., but here 
my digging stops.

Good luck.

Jerzy Karczmarczuk

On 14/02/2012 13:52, Debashish Saha wrote:
> import numpy
>
> from enthought.mayavi import mlab
>
> #def test_mesh():
> #"""A very pretty picture of spherical harmonics translated from
>
> #the octaviz example."""
> for r in range (1,5):
>     print r
>
>
>     pi = numpy.pi
>
>     cos = numpy.cos
>
>     sin = numpy.sin
>
>     dphi, dtheta, dz = pi/250.0, pi/250.0, 0.01
>
>     #[phi,theta] = numpy.mgrid[0:pi+dphi*1.5:dphi,0:2*pi+dtheta*1.5:dtheta]
>     [phi,z] = numpy.mgrid[0:2*pi+dphi*1.5:dphi,0:2+dz*1.5:dz]
>
>     m0 = 4; m1 = 3; m2 = 2; m3 = 3; m4 = 6; m5 = 2; m6 = 6; m7 = 4;
>
>    # r = sin(m0*phi)**m1 + cos(m2*phi)**m3 + 5*sin(m4*theta)**m5 +
> cos(m6*theta)**m7
>
>     #x = 1*sin(phi)*cos(theta)
>
>     #y = 1*sin(phi)*sin(theta)
>
>     #z = 1*cos(phi);
>     x=r*cos(phi)
>     y=r*sin(phi)
>     z=z
>     i=['Reds','greens','autumn','purples']
>     print i[r-1]
>     e=i[r-1]
>
>     mlab.mesh(x, y, z,colormap='e')
>     #print i[r-1]
>
> Error:
> TypeError                                 Traceback (most recent call last)
> C:\Python27\lib\site-packages\IPython\utils\py3compat.pyc in
> execfile(fname, glob, loc)
>     166             else:
>     167                 filename = fname
> -->  168             exec compile(scripttext, filename, 'exec') in glob, loc
>     169     else:
>     170         def execfile(fname, *where):
>
> C:\Users\as\jhgf.py in<module>()
>      24     print i[r-1]
>      25     e=i[r-1]
> --->  26     mlab.mesh(x, y, z,'e')
>      27     #print i[r-1]
>
>      28
>
> C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
> the_function(*args, **kwargs)
>      32 def document_pipeline(pipeline):
>      33     def the_function(*args, **kwargs):
> --->  34         return pipeline(*args, **kwargs)
>      35
>      36     if hasattr(pipeline, 'doc'):
>
> C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
> __call__(self, *args, **kwargs)
>      77             scene.disable_render = True
>      78         # Then call the real logic
>
> --->  79         output = self.__call_internal__(*args, **kwargs)
>      80         # And re-enable the rendering, if needed.
>
>      81         if scene is not None:
>
> C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
> __call_internal__(self, *args, **kwargs)
>     830         filters.
>     831         """
> -->  832         self.source = self._source_function(*args, **kwargs)
>     833         kwargs.pop('name', None)
>     834         self.store_kwargs(kwargs)
>
> TypeError: grid_source() takes exactly 3 arguments (4 given)
>
> ------------------------------------------------------------------------------
> Keep Your Developer Skills Current with LearnDevNow!
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Didn't this get answered on the python tutor mailing list within the 
last couple of hours?  What's with it with you?

-- 
Cheers.

Mark Lawrence.

Hi list,

I have a list of 48 individual data sets (lambda,y) which I want to plot in
the following diagram:

lambda
^ 
|
|
|
|
+--------------->
 1 2 3 4 5 6 .. Dataset #

The "y" values should be color coded here, in a similar fashion as it is
done in a color bar (f.e. that I can interpolate between data points).

Practically, I want to have 48 vertical color bars here, all with the
same scaling. From the documentation, I could not find out how to do
this; and also I could not find a good example for this.

Could anyone help me here?

Best

Ole

import numpy

from enthought.mayavi import mlab

#def test_mesh():
#"""A very pretty picture of spherical harmonics translated from

#the octaviz example."""
for r in range (1,5):
   print r


   pi = numpy.pi

   cos = numpy.cos

   sin = numpy.sin

   dphi, dtheta, dz = pi/250.0, pi/250.0, 0.01

   #[phi,theta] = numpy.mgrid[0:pi+dphi*1.5:dphi,0:2*pi+dtheta*1.5:dtheta]
   [phi,z] = numpy.mgrid[0:2*pi+dphi*1.5:dphi,0:2+dz*1.5:dz]

   m0 = 4; m1 = 3; m2 = 2; m3 = 3; m4 = 6; m5 = 2; m6 = 6; m7 = 4;

  # r = sin(m0*phi)**m1 + cos(m2*phi)**m3 + 5*sin(m4*theta)**m5 +
cos(m6*theta)**m7

   #x = 1*sin(phi)*cos(theta)

   #y = 1*sin(phi)*sin(theta)

   #z = 1*cos(phi);
   x=r*cos(phi)
   y=r*sin(phi)
   z=z
   i=['Reds','greens','autumn','purples']
   print i[r-1]
   e=i[r-1]

   mlab.mesh(x, y, z,colormap='e')
   #print i[r-1]

Error:
TypeError                                 Traceback (most recent call last)
C:\Python27\lib\site-packages\IPython\utils\py3compat.pyc in
execfile(fname, glob, loc)
   166             else:
   167                 filename = fname
--> 168             exec compile(scripttext, filename, 'exec') in glob, loc
   169     else:
   170         def execfile(fname, *where):

C:\Users\as\jhgf.py in <module>()
    24     print i[r-1]
    25     e=i[r-1]
---> 26     mlab.mesh(x, y, z,'e')
    27     #print i[r-1]

    28

C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
the_function(*args, **kwargs)
    32 def document_pipeline(pipeline):
    33     def the_function(*args, **kwargs):
---> 34         return pipeline(*args, **kwargs)
    35
    36     if hasattr(pipeline, 'doc'):

C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
__call__(self, *args, **kwargs)
    77             scene.disable_render = True
    78         # Then call the real logic

---> 79         output = self.__call_internal__(*args, **kwargs)
    80         # And re-enable the rendering, if needed.

    81         if scene is not None:

C:\Python27\lib\site-packages\mayavi\tools\helper_functions.pyc in
__call_internal__(self, *args, **kwargs)
   830         filters.
   831         """
--> 832         self.source = self._source_function(*args, **kwargs)
   833         kwargs.pop('name', None)
   834         self.store_kwargs(kwargs)

TypeError: grid_source() takes exactly 3 arguments (4 given)

On 14 February 2012 14:23,  <a....@ya...> wrote:
> I have already installed scipy , numpy & python 2.6.6. Also , I installed
> matplotlip from its website. I strongly need basemap toolkits but when I try
> to install it I cannot do that always I got error massage or command cannot
> found. So could you please help me to get it. NB: I use mac OS X 10.6.8

I'm replying on list (I accidentally replied directly to you).

You'll need to be far more specific about the steps you are taking to
try and install Basemap and what the error message says.

> ‏من جهاز الـ iPhone الخاص بي
>
> في 14/02/2012، الساعة 2:48 AM، كتب Scott Sinclair
> <sco...@gm...>:
>
> Hi Fadhah,
>
> If you'd like someone to help. You'll need to explain in more detail, what
> steps you're trying and what's going wrong.
>
> Cheers,
> Scott
>
> On Feb 13, 2012 5:44 PM, "love ali" <a....@ya...> wrote:
>>
>> Dear all,
>>
>> I use the mac OS X 10.6.8 and I try to install the matplotlib but I cannot
>> run it in my computer so could you please help me?
>>
>>
>> Thanks,
>>
>> Fadhah

This is relating wxmpl (wxPython+matplotlib). version 2.0dev of wxmpl,
matplotlib ver 1.1.0, and wx version is 2.8.12.1.

Since wxmpl is a very "thin" layer above mpl, I believe this is an issue
with mpl.

I'm using wxmpl and trying to get mouse points and selection. This
fails when there is more than one axes in the same place, which I do
since I'm using twinx() in my original code.

I believe the issue is the overlapping of both axes on the same area.
The following code shows that overlapping two axes makes
the mouse events in the overlapped area fail. For example, the cursor
doesn't change in them and the crosshairs disappear (if enabled).
Removing the ax2 code, or just ensuring ax1 and ax2 do not overlap
allows all the events to reach the handler.

Any idea on how to receive mouse events from the overlapped area?

Thanks,
Oren

----------------------------------------------------

import wxmpl

if __name__ == '__main__':
   class MainWindow(wx.Frame):
       def __init__(self):
           wx.Frame.__init__(self, parent=None, title='Graph Test',
size=(700,700))

           self.plot = wxmpl.PlotPanel(self, -1, dpi=70)

           fig = self.plot.get_figure()
           ax1 = fig.add_axes((0.1, 0.1, .6, .6))
           ax1.plot([0,3], [-1,1])
           ax2 = fig.add_axes( (0.3, 0.3, .6, .6) )
           ax2.plot([2,4], [-10,10])

           wxmpl.EVT_POINT(self.plot, -1, self.handle_event)

           self.Show()

       def handle_event(self, event):
           print event, event.GetId()

   app = wx.App()
   mw = MainWindow()
   app.MainLoop()

Hi All,

    after some great help from the Numpy users list, I have managed to
create "parallel curves".

Basically I have a set of X, Y data (around 1,000 elements each) and I
want to create 2 parallel "curves" (offset curves) to the original
one; "parallel" means curves which are displaced from the base curve
by a constant offset, either positive or negative, in the direction of
the curve's normal. I attach a sample demonstrating what I am doing
(and if you have the Shapely package installed, the two additional
subplots look even nicer).

Now, the generated parallels are really parallels (see right subplot),
but visually it doesn't seem so (see the left subplot) because the X
and Y scales are different and the axes area is not square. I know I
could force the axes to be equals via:

ax.set_aspect('equal', 'datalim')

But I can't really do that with the set of data I have, as the X and Y
variables have different order of magnitude and I need a single
subplot on the figure to have rectangular axes (not square). In my
situation, unfortunately it wouldn't make sense to set the axes
square/equal as the plot will lose its meaning and visual usefulness.

I have been told to scale the X, Y variables normalizing them by the
display units of the plot, but I must be dumber than usual as I can't
get it to work properly. So, my question would be: how do I scale the
X and Y vectors so that the parallels look parallel to the main curve
even if the axes are not square and the X and Y variables have
different data-ranges/magnitudes?

Thank you in advance for your suggestions.

Andrea.

"Imagination Is The Only Weapon In The War Against Reality."
http://xoomer.alice.it/infinity77/

Hello,

Mac OSX 10.6.8
matplotlib 1.1.0


>From the matplotlib website:
=========
Installing OSX binaries
If you want to install matplotlib from one of the binary installers we build, you have two choices: a mpkg installer, which is a typical Installer.app, or a binary OSX egg, which you can install via setuptools’ easy_install.

The mkpg installer will have a “zip” extension, and will have a name like matplotlib-0.99.0.rc1-py2.5-macosx10.5_mpkg.zip. 
==========

Where are those files?  I can't find them anywhere.

Also from the matplotlib website:
=======
You can also use the eggs we build for OSX (see the installation instructions for easy_install if you do not have it on your system already). You can try:

> easy_install matplotlib
which should grab the latest egg from the sourceforge site, but sometimes the naming conventions for OSX eggs can be broken (see below). Therefore, there is no guarantee the right egg will be found. We recommend you download the latest egg from our download site directly to your harddrive, and manually install it, eg:

> easy_install --install-dir=~/dev/lib/python2.5/site-packages/  matplotlib-0.99.0.rc1-py2.5-macosx-10.5-i386.egg
=======

Where are those eggs?  I can't find them anywhere.

On the home page of the matplotlib website, on the right hand side, is a link "downloads", which took me here:

http://sourceforge.net/projects/matplotlib/files/matplotlib/matplotlib-1.1.0/

There isn't anything later than osx 10.3 there, and there are no egg files at all.  Yet the install directions mention osx10.5. 

I installed python 2.7, which automatically installed an interpreter named python2.7, because the Numpy installer I found was for python 2.7 and mac osx 10.6. I installed this for matplotlib:

matplotlib-1.1.0-py2.7-python.org-macosx10.3.dmg

because that is the only thing even close.

This works:

import matplotlib
print(matplotlib.__version__)
print(matplotlib.__file__)

$ python2.7 my_prog.py
1.1.0
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/__init__.pyc
/Users/Me/.matplotlib


But this does not work:

import matplotlib.pyplot as plot
xs = [2, 3, 5, 7, 11]
ys = [4, 9, 5, 9, 1]
plot.plot(xs, ys)
plot.savefig("squaremod10.png")


$ python2.7 my_prog.py
  File "my_prog.py", line 1, in <module>
    import matplotlib.pyplot as plot
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/pyplot.py", line 23, in <module>
    from matplotlib.figure import Figure, figaspect
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/figure.py", line 16, in <module>
    import artist
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/artist.py", line 6, in <module>
    from transforms import Bbox, IdentityTransform, TransformedBbox, TransformedPath
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/transforms.py", line 34, in <module>
    from matplotlib._path import affine_transform
ImportError: dlopen(/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/_path.so, 2): no suitable image found.  Did find:
	/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/_path.so: no matching architecture in universal wrapper

Hi,

1.) For terminal just go to your finder and type in "terminal" in the
search bar, terminal should pop up
2.) I don't think you do, you might need to change the path variable to
point to the python you have. There are instructions for that in the
tutorial I linked to.
3.) I'm not sure what you're asking
4.) What do you mean you can't find them on your computer?
Start a terminal window and type
'python'
This should bring you to the interactive python environment.
Then type,
'import numpy'
Does it successfully import or do you get an error message?

(Also, click "Reply All" to keep this conversation to the list"
- Alexa

On Mon, Feb 13, 2012 at 10:11 AM, <a....@ya...> wrote:

> Hi Alex , thank you so much for your help. I have some questions:
>
> 1- where can I find the terminal in my computer
> 2- I have already had python 2.6.6 so do I need to delete it and start
> again.
> 3- my mac is  os x 10.6.8 so do I need to clean it while it is already do
> not have much download.
> 4- I download the scipy ,numpy and i have these modules successfully
> installed but I cannot find them in my computer.
>
>
> So could you help me ?
>
> Many thanks,
>
> Fadh
>
> ‏من جهاز الـ iPhone الخاص بي
>
> في 14/02/2012، الساعة 5:04 AM، كتب Alexa Villaume <ale...@gm...>:
>
> Hi Fadhah,
>
> This is the tutorial I used to install matplotlib on my computer and it
> was really easy to understand and follow the steps -
>
>
> <http://www.thisisthegreenroom.com/2011/installing-python-numpy-scipy-matplotlib-and-ipython-on-lion/>
> http://www.thisisthegreenroom.com/2011/installing-python-numpy-scipy-matplotlib-and-ipython-on-lion/
>
> If that doesn't help with your problems describe what they are more
> specifically and you'll be able to get better help
>
> Cheers,
> Alexa
>
>
> On Sun, Feb 12, 2012 at 5:39 PM, love ali < <a....@ya...>
> a....@ya...> wrote:
>
>> Dear all,
>>
>> I use the mac OS X 10.6.8 and I try to install the matplotlib but I
>> cannot run it in my computer so could you please help me?
>>
>>
>> Thanks,
>>
>> Fadhah
>>
>> ------------------------------------------------------------------------------
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>> Metro Style Apps, more. Free future releases when you subscribe now!
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Hi folks,

[ I'm broadcasting this widely for maximum reach, but I'd appreciate
it if replies can be kept to the *numpy* list, which is sort of the
'base' list for scientific/numerical work.  It will make it much
easier to organize a coherent set of notes later on.  Apology if
you're subscribed to all and get it 10 times. ]

As part of the PyData workshop (http://pydataworkshop.eventbrite.com)
to be held March 2 and 3 at the Mountain View Google offices, we have
scheduled a session for an open discussion with Guido van Rossum and
hopefully as many core python-dev members who can make it.  We wanted
to seize the combined opportunity of the PyData workshop bringing a
number of 'scipy people' to Google with the timeline for Python 3.3,
the first release after the Python language moratorium, being within
sight: http://www.python.org/dev/peps/pep-0398.

While a number of scientific Python packages are already available for
Python 3 (either in released form or in their master git branches),
it's fair to say that there hasn't been a major transition of the
scientific community to Python3.  Since there is no more development
being done on the Python2 series, eventually we will all want to find
ways to make this transition, and we think that this is an excellent
time to engage the core python development team and consider ideas
that would make Python3 generally a more appealing language for
scientific work.  Guido has made it clear that he doesn't speak for
the day-to-day development of Python anymore, so we all should be
aware that any ideas that come out of this panel will still need to be
discussed with python-dev itself via standard mechanisms before
anything is implemented.  Nonetheless, the opportunity for a solid
face-to-face dialog for brainstorming was too good to pass up.

The purpose of this email is then to solicit, from all of our
community, ideas for this discussion.  In a week or so we'll need to
summarize the main points brought up here and make a more concrete
agenda out of it; I will also post a summary of the meeting afterwards
here.

Anything is a valid topic, some points just to get the conversation started:

- Extra operators/PEP 225.  Here's a summary from the last time we
went over this, years ago at Scipy 2008:
http://mail.scipy.org/pipermail/numpy-discussion/2008-October/038234.html,
and the current status of the document we wrote about it is here:
file:///home/fperez/www/site/_build/html/py4science/numpy-pep225/numpy-pep225.html.

- Improved syntax/support for rationals or decimal literals?  While
Python now has both decimals
(http://docs.python.org/library/decimal.html) and rationals
(http://docs.python.org/library/fractions.html), they're quite clunky
to use because they require full constructor calls.  Guido has
mentioned in previous discussions toying with ideas about support for
different kinds of numeric literals...

- Using the numpy docstring standard python-wide, and thus having
python improve the pathetic state of the stdlib's docstrings?  This is
an area where our community is light years ahead of the standard
library, but we'd all benefit from Python itself improving on this
front.  I'm toying with the idea of giving a lighting talk at PyConn
about this, comparing the great, robust culture and tools of good
docstrings across the Scipy ecosystem with the sad, sad state of
docstrings in the stdlib.  It might spur some movement on that front
from the stdlib authors, esp. if the core python-dev team realizes the
value and benefit it can bring (at relatively low cost, given how most
of the information does exist, it's just in the wrong places).  But
more importantly for us, if there was truly a universal standard for
high-quality docstrings across Python projects, building good
documentation/help machinery would be a lot easier, as we'd know what
to expect and search for (such as rendering them nicely in the ipython
notebook, providing high-quality cross-project help search, etc).

- Literal syntax for arrays?  Sage has been floating a discussion
about a literal matrix syntax
(https://groups.google.com/forum/#!topic/sage-devel/mzwepqZBHnA).  For
something like this to go into python in any meaningful way there
would have to be core multidimensional arrays in the language, but
perhaps it's time to think about a piece of the numpy array itself
into Python?  This is one of the more 'out there' ideas, but after
all, that's the point of a discussion like this, especially
considering we'll have both Travis and Guido in one room.

- Other syntactic sugar? Sage has "a..b" <=> range(a, b+1), which I
actually think is  both nice and useful... There's also the question
of allowing "a:b:c" notation outside of [], which has come up a few
times in conversation over the last few years. Others?

- The packaging quagmire?  This continues to be a problem, though
python3 does have new improvements to distutils.  I'm not really up to
speed on the situation, to be frank.  If we want to bring this up,
someone will have to provide a solid reference or volunteer to do it
in person.

- etc...

I'm putting the above just to *start* the discussion, but the real
point is for the rest of the community to contribute ideas, so don't
be shy.

Final note: while I am here commiting to organizing and presenting
this at the discussion with Guido (as well as contacting python-dev),
I would greatly appreciate help with the task of summarizing this
prior to the meeting as I'm pretty badly swamped in the run-in to
pydata/pycon.  So if anyone is willing to help draft the summary as
the date draws closer (we can put it up on a github wiki, gist,
whatever), I will be very grateful.  I'm sure it will be better than
what I'll otherwise do the last night at 2am :)

Cheers,

f

ps - to the obvious question about webcasting the discussion live for
remote participation: yes, we looked into it already; no,
unfortunately it appears it won't be possible.  We'll try to at least
have the audio recorded (and possibly video) for posting later on.

pps- if you are close to Mountain View and are interested in attending
this panel in person, drop me a line at fer...@be....
We have a few spots available *for this discussion only* on top of the
pydata regular attendance (which is long closed, I'm afraid).  But
we'll need to provide Google with a list of those attendees in
advance.  Please indicate if you are a core python committer in your
email, as we'll give priority for this overflow pool to core python
developers (but will otherwise accommodate as many people as Google
lets us).

On Mon, Feb 13, 2012 at 2:59 PM, Jerzy Karczmarczuk <
jer...@un...> wrote:

>
>  reckoner:
>
>
>>
>> >>> th= array([ 4.65542641,  5.32920696,  2.20928291])
>> >>> p=patch.Polygon(array([cos(th),sin(th)]).T)
>> >>> print p.contains_point((0,0))
>> 1
>> >>> print matplotlib.nxutils.pnpoly(0,0,p.get_path().vertices)
>> 0
>>
>> Thanks!
>>
>>
>>
> I can confirm this for both the current development branch and v1.1.x.
> Strange....
>
> Ben Root
>
>   HYPOTHESIS (I have no time to check now everything..., I did my best.)
>
> If radius=None, it is converted into radius=1 in contains_point (in Patch).
> (This is the linewidth !)
> I am not sure whether it is ok, i.e. in pixels. If this value is
> considered to be relative to axes, it is enormous!
>
> p.contains_point((0,0),radius=0.01)
>
> gives 0 as it should.
>
> Jerzy Karczmarczuk
>
> P.S. I suggest anyway that reckoner submits a complete program with
> imports. What is "patch"??
>
>
>
Jerzy is right.  contains() and contains_point() both takes a radius
argument as this function is intended for use when mouse-clicking.  If no
radius is given, it uses the object's linewidth and passes it to the path's
contains_point() function with its transform.  I *suspect* that the problem
is that the transforms haven't been fully defined yet and might be
incorrectly testing this.

This needs a bit more investigation to see if this is the intended behavior
or if something needs to be fixed.

As a side note, "patch" is an imported module via "from matplotlib import
patches as patch".  Not standard among mpl devs, but not out of the
ordinary.

Ben Root

> reckoner:
>
>
>
>     >>> th= array([ 4.65542641,  5.32920696,  2.20928291])
>     >>> p=patch.Polygon(array([cos(th),sin(th)]).T)
>     >>> print p.contains_point((0,0))
>     1
>     >>> print matplotlib.nxutils.pnpoly(0,0,p.get_path().vertices)
>     0
>
>     Thanks!
>
>
>
> I can confirm this for both the current development branch and 
> v1.1.x.  Strange....
>
> Ben Root
>
HYPOTHESIS (I have no time to check now everything..., I did my best.)

If radius=None, it is converted into radius=1 in contains_point (in Patch).
(This is the linewidth !)
I am not sure whether it is ok, i.e. in pixels. If this value is 
considered to be relative to axes, it is enormous!

p.contains_point((0,0),radius=0.01)

gives 0 as it should.

Jerzy Karczmarczuk

P.S. I suggest anyway that reckoner submits a complete program with 
imports. What is "patch"??

Hmm, on further review, it looks like this is what triplot does!

I'll give it a try...

On 2/13/12 3:30 PM, Howard wrote:
> Hi all
>
> I'm using tricontourf with a triangulation object to draw a surface.  
> For debugging purposes, I'd like to draw just the unfilled triangles.  
> Is there a simple way to do this?
>
> Thanks
> Howard
> -- 
> Howard Lander <mailto:ho...@re...>
> Senior Research Software Developer
> Renaissance Computing Institute (RENCI) <http://www.renci.org>
> The University of North Carolina at Chapel Hill
> Duke University
> North Carolina State University
> 100 Europa Drive
> Suite 540
> Chapel Hill, NC 27517
> 919-445-9651


-- 
Howard Lander <mailto:ho...@re...>
Senior Research Software Developer
Renaissance Computing Institute (RENCI) <http://www.renci.org>
The University of North Carolina at Chapel Hill
Duke University
North Carolina State University
100 Europa Drive
Suite 540
Chapel Hill, NC 27517
919-445-9651

Hi all

I'm using tricontourf with a triangulation object to draw a surface.  
For debugging purposes, I'd like to draw just the unfilled triangles.  
Is there a simple way to do this?

Thanks
Howard
-- 
Howard Lander <mailto:ho...@re...>
Senior Research Software Developer
Renaissance Computing Institute (RENCI) <http://www.renci.org>
The University of North Carolina at Chapel Hill
Duke University
North Carolina State University
100 Europa Drive
Suite 540
Chapel Hill, NC 27517
919-445-9651

On Mon, Feb 13, 2012 at 1:33 PM, reckoner <rec...@gm...> wrote:

> Here's a better example:
>
> >>> th= array([ 4.65542641,  5.32920696,  2.20928291])
> >>> p=patch.Polygon(array([cos(th),sin(th)]).T)
> >>> print p.contains_point((0,0))
> 1
> >>> print matplotlib.nxutils.pnpoly(0,0,p.get_path().vertices)
> 0
>
> Thanks!
>
>
>
I can confirm this for both the current development branch and v1.1.x.
Strange....

Ben Root

Here's a better example:

>>> th= array([ 4.65542641,  5.32920696,  2.20928291])
>>> p=patch.Polygon(array([cos(th),sin(th)]).T)
>>> print p.contains_point((0,0))
1
>>> print matplotlib.nxutils.pnpoly(0,0,p.get_path().vertices)
0

Thanks!




On 2/13/2012 8:47 AM, reckoner wrote:
> I'm trying to test whether or not the origin is contained inside the
> following triangle
> inscribed inside a circle:
>
>>>> th = array([ 2.3913423,  5.3133123,  1.8516171])
>>>> p=patch.Polygon(array([cos(th),sin(th)]).T)
>
>>>> p.contains_point((0,0))
>
> returns 0
>
> but,
>
>>>> matplotlib.nxutils.points_inside_poly([[0,0]],p.get_path().vertices)
>
> returns True
>
> What am I missing here?
>
> Thanks!
>
>
>

Le 13/02/2012 17:47, reckoner a écrit :

I'm trying to test whether or not the origin is contained inside the
following triangle
inscribed inside a circle:

> th = array([ 2.3913423,  5.3133123,  1.8516171])
> p=patch.Polygon(array([cos(th),sin(th)]).T)

What do you mean by "patch" here?

I tested

p=Polygon(array([cos(th),sin(th)]).T)
p.contains_point((0,0))

and the answer is 1.

Jerzy Karczmarczuk