matplotlib-users Mailing List for matplotlib (Page 16)

On Mon, Aug 2, 2010 at 10:15 AM, Richard Lincoln <r.w...@gm...> wrote:
> I am having trouble creating a plot using TeX which has a $ sign in
> one of the axis labels.  If I run:
>
> import matplotlib
>
> matplotlib.rc('font', **{'family': 'serif', 'serif': ['Computer Modern Roman']})
> matplotlib.rc('text', usetex=True)
>
> import pylab
>
> pylab.figure()
> pylab.plot([0.0, 1.0])
> pylab.ylabel("Profit (\$)")
> pylab.savefig('/tmp/fig.pdf')
>
>
> I get the error:

Try:

pylab.ylabel(r"Profit (\$)")

I am having trouble creating a plot using TeX which has a $ sign in
one of the axis labels.  If I run:

import matplotlib

matplotlib.rc('font', **{'family': 'serif', 'serif': ['Computer Modern Roman']})
matplotlib.rc('text', usetex=True)

import pylab

pylab.figure()
pylab.plot([0.0, 1.0])
pylab.ylabel("Profit (\$)")
pylab.savefig('/tmp/fig.pdf')


I get the error:

Traceback (most recent call last):
  File "/home/rwl/python/aes/Pylon/examples/pyreto/thesis/plot_tex.py",
line 12, in <module>
    pylab.savefig('/tmp/fig.pdf')
  File "/usr/lib/pymodules/python2.5/matplotlib/pyplot.py", line 356, in savefig
    return fig.savefig(*args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/figure.py", line 1032,
in savefig
    self.canvas.print_figure(*args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/backend_bases.py",
line 1476, in print_figure
    **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/backend_bases.py",
line 1334, in print_pdf
    return pdf.print_pdf(*args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/backends/backend_pdf.py",
line 2031, in print_pdf
    self.figure.draw(renderer)
  File "/usr/lib/pymodules/python2.5/matplotlib/artist.py", line 46,
in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/figure.py", line 773, in draw
    for a in self.axes: a.draw(renderer)
  File "/usr/lib/pymodules/python2.5/matplotlib/artist.py", line 46,
in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/axes.py", line 1735, in draw
    a.draw(renderer)
  File "/usr/lib/pymodules/python2.5/matplotlib/artist.py", line 46,
in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.5/matplotlib/axis.py", line 757, in draw
    self.label.draw(renderer)
  File "/usr/lib/pymodules/python2.5/matplotlib/text.py", line 556, in draw
    self._fontproperties, angle)
  File "/usr/lib/pymodules/python2.5/matplotlib/backends/backend_pdf.py",
line 1437, in draw_tex
    psfont = self.tex_font_mapping(dvifont.texname)
  File "/usr/lib/pymodules/python2.5/matplotlib/backends/backend_pdf.py",
line 1269, in tex_font_mapping
    return self.tex_font_map[texfont]
  File "/usr/lib/pymodules/python2.5/matplotlib/dviread.py", line 673,
in __getitem__
    result = self._font[texname]
KeyError: 'tcrm1200'


I am using Debian/unstable and have the latest texlive installed.  Can
anyone tell me what the problem might be?

Richard

On Sun, Aug 1, 2010 at 10:51 PM, David Goldsmith <d.l...@gm...>wrote:

> Hi!  Is there a way to see data values when imaging an array, say, e.g.,
> when holding the cursor over a point?
>
> DG
>

I found the data_browser.py example - how do I use it if I'm not using
pylab.show, i.e., I'm using oo?

DG

Hi!  Is there a way to see data values when imaging an array, say, e.g.,
when holding the cursor over a point?

DG

On Sun, Aug 1, 2010 at 4:10 PM, Benjamin Root <ben...@ou...> wrote:

> On Sun, Aug 1, 2010 at 3:48 PM, Tom Arens <ta...@gm...> wrote:
>
>> Hello,
>>
>> I thought that it has to be like Ben wrote:
>>
>> -    def contour(self, X, Y, Z, levels=10, **kwargs):
>> +    def contour(self, X, Y, Z, *args, **kwargs):
>>
>>
>> Your suggestion Eric ( contour(X, Y, Z, [0] ) doesn't work as the levels
>> are still the same.
>>
>> Can you translate this thread for a "normal" user? Is this a bug and
>> will be fixed in a newer version of matplotlib or what do I have to do
>> to get "less levels"?
>>
>> Thank you for your help.
>> Tom
>>
>>
>>
> Tom, it is a bug and it is currently getting patched in the released v1.0
> of matplotlib.  I am double-checking a few things in the tests before fully
> committing it.  If you want, you can either install the revised v1.0 of
> matplotlib (when it is ready), or you can edit the axes3d.py file yourself
> to fix the function header.
>
> Thank you for noticing this error.
>
> Ben Root
>

Ok, this has been committed to v1.0 as r8610 and merged to trunk as r8611.
I should note that the image for contour3d_demo2.py will be changed
(personally, for the better).  I don't know if something needs to be done
for that.

Ben Root

On Sun, Aug 1, 2010 at 3:48 PM, Tom Arens <ta...@gm...> wrote:

> Hello,
>
> I thought that it has to be like Ben wrote:
>
> -    def contour(self, X, Y, Z, levels=10, **kwargs):
> +    def contour(self, X, Y, Z, *args, **kwargs):
>
>
> Your suggestion Eric ( contour(X, Y, Z, [0] ) doesn't work as the levels
> are still the same.
>
> Can you translate this thread for a "normal" user? Is this a bug and
> will be fixed in a newer version of matplotlib or what do I have to do
> to get "less levels"?
>
> Thank you for your help.
> Tom
>
>
>
Tom, it is a bug and it is currently getting patched in the released v1.0 of
matplotlib.  I am double-checking a few things in the tests before fully
committing it.  If you want, you can either install the revised v1.0 of
matplotlib (when it is ready), or you can edit the axes3d.py file yourself
to fix the function header.

Thank you for noticing this error.

Ben Root

Hello,

I thought that it has to be like Ben wrote:

-    def contour(self, X, Y, Z, levels=10, **kwargs):
+    def contour(self, X, Y, Z, *args, **kwargs):


Your suggestion Eric ( contour(X, Y, Z, [0] ) doesn't work as the levels
are still the same.

Can you translate this thread for a "normal" user? Is this a bug and
will be fixed in a newer version of matplotlib or what do I have to do
to get "less levels"?

Thank you for your help.
Tom



Am 01.08.2010 20:12 schrieb Eric Firing:
> On 08/01/2010 07:35 AM, Benjamin Root wrote:
>> On Sun, Aug 1, 2010 at 9:55 AM, Tom Arens <ta...@gm...
>> <mailto:ta...@gm...>> wrote:
>>
>>     Hello everyone,
>>
>>     does anybody know why the contour3D function has a fixed set of levels?
>>
>>     contour3D(X, Y, Z, levels=10, **kwargs)
>>
>>     I want to plot only one line for one level. With "contourf" it works:
>>
>>
>>
>>     from mpl_toolkits.mplot3d import axes3d
>>     import matplotlib.pyplot as plt
>>
>>     fig = plt.figure()
>>     ax = axes3d.Axes3D(fig)
>>     X, Y, Z = axes3d.get_test_data(0.05)
>>     cset = ax.contourf(X, Y, Z, 0)  # doesn't work with contour
>>     ax.clabel(cset, fontsize=9, inline=1)
>>
>>     plt.show()
>>
>>
>>
>>     Many greetings,
>>     Tom
>>
>>
>> Hmm, interesting.  Looking at the contour3d call signature, it appears
>> that 'levels' was put into the call signature to basically remove that
>> keyword argument from the kwargs that get passed down to the 2-d version
>> of contour.  It is never used in the body of contour3d().
>>
>> I would guess that this is might be a remnant of some original code that
>> actually used the levels parameter.  Simply removing levels=0 from the
>> call signature seems to fix it (and passing [0] to levels as well since
>> it expects a sequence).
>>
>> As a matter of consistency, I think the call signature should be changed
>> to better match the call signature for contourf3d() and for the 2-d
>> version of contour().
> 
> Ben,
> 
> Good idea, go ahead.  The contourf3d docstring can also be modified to 
> match your change to the contour3d docstring.
> 
> I would consider all this as bug-fixing, so it can go in branch and trunk.
> 
> Thanks.
> 
> Eric
> 
>>
>> Ben Root
>>
> 
> ------------------------------------------------------------------------------
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> 

On 08/01/2010 07:35 AM, Benjamin Root wrote:
> On Sun, Aug 1, 2010 at 9:55 AM, Tom Arens <ta...@gm...
> <mailto:ta...@gm...>> wrote:
>
>     Hello everyone,
>
>     does anybody know why the contour3D function has a fixed set of levels?
>
>     contour3D(X, Y, Z, levels=10, **kwargs)
>
>     I want to plot only one line for one level. With "contourf" it works:
>
>
>
>     from mpl_toolkits.mplot3d import axes3d
>     import matplotlib.pyplot as plt
>
>     fig = plt.figure()
>     ax = axes3d.Axes3D(fig)
>     X, Y, Z = axes3d.get_test_data(0.05)
>     cset = ax.contourf(X, Y, Z, 0)  # doesn't work with contour
>     ax.clabel(cset, fontsize=9, inline=1)
>
>     plt.show()
>
>
>
>     Many greetings,
>     Tom
>
>
> Hmm, interesting.  Looking at the contour3d call signature, it appears
> that 'levels' was put into the call signature to basically remove that
> keyword argument from the kwargs that get passed down to the 2-d version
> of contour.  It is never used in the body of contour3d().
>
> I would guess that this is might be a remnant of some original code that
> actually used the levels parameter.  Simply removing levels=0 from the
> call signature seems to fix it (and passing [0] to levels as well since
> it expects a sequence).
>
> As a matter of consistency, I think the call signature should be changed
> to better match the call signature for contourf3d() and for the 2-d
> version of contour().

Ben,

Good idea, go ahead.  The contourf3d docstring can also be modified to 
match your change to the contour3d docstring.

I would consider all this as bug-fixing, so it can go in branch and trunk.

Thanks.

Eric

>
> Ben Root
>

On Sun, Aug 1, 2010 at 9:55 AM, Tom Arens <ta...@gm...> wrote:

> Hello everyone,
>
> does anybody know why the contour3D function has a fixed set of levels?
>
> contour3D(X, Y, Z, levels=10, **kwargs)
>
> I want to plot only one line for one level. With "contourf" it works:
>
>
>
> from mpl_toolkits.mplot3d import axes3d
> import matplotlib.pyplot as plt
>
> fig = plt.figure()
> ax = axes3d.Axes3D(fig)
> X, Y, Z = axes3d.get_test_data(0.05)
> cset = ax.contourf(X, Y, Z, 0)  # doesn't work with contour
> ax.clabel(cset, fontsize=9, inline=1)
>
> plt.show()
>
>
>
> Many greetings,
> Tom
>
>
Hmm, interesting.  Looking at the contour3d call signature, it appears that
'levels' was put into the call signature to basically remove that keyword
argument from the kwargs that get passed down to the 2-d version of
contour.  It is never used in the body of contour3d().

I would guess that this is might be a remnant of some original code that
actually used the levels parameter.  Simply removing levels=0 from the call
signature seems to fix it (and passing [0] to levels as well since it
expects a sequence).

As a matter of consistency, I think the call signature should be changed to
better match the call signature for contourf3d() and for the 2-d version of
contour().

Ben Root

On 08/01/2010 04:55 AM, Tom Arens wrote:
> Hello everyone,
>
> does anybody know why the contour3D function has a fixed set of levels?
>
> contour3D(X, Y, Z, levels=10, **kwargs)
>
> I want to plot only one line for one level. With "contourf" it works:
>
>
>
> from mpl_toolkits.mplot3d import axes3d
> import matplotlib.pyplot as plt
>
> fig = plt.figure()
> ax = axes3d.Axes3D(fig)
> X, Y, Z = axes3d.get_test_data(0.05)
> cset = ax.contourf(X, Y, Z, 0)	# doesn't work with contour

This is a fluke. Try

cset = ax.contour(X, Y, Z, [0])

and the same for contour3d.  When the levels arg or kwarg is a scalar, 
it specifies the number of levels to be found via auto-scaling; when it 
is a sequence, it specifies the actual levels.

Eric

> ax.clabel(cset, fontsize=9, inline=1)
>
> plt.show()
>
>
>
> Many greetings,
> Tom
>
> ------------------------------------------------------------------------------
> The Palm PDK Hot Apps Program offers developers who use the
> Plug-In Development Kit to bring their C/C++ apps to Palm for a share
> of $1 Million in cash or HP Products. Visit us here for more details:
> http://p.sf.net/sfu/dev2dev-palm
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users

Hello everyone,

does anybody know why the contour3D function has a fixed set of levels?

contour3D(X, Y, Z, levels=10, **kwargs)

I want to plot only one line for one level. With "contourf" it works:



from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt

fig = plt.figure()
ax = axes3d.Axes3D(fig)
X, Y, Z = axes3d.get_test_data(0.05)
cset = ax.contourf(X, Y, Z, 0)	# doesn't work with contour
ax.clabel(cset, fontsize=9, inline=1)

plt.show()



Many greetings,
Tom