matplotlib-users Mailing List for matplotlib

On 2012年09月14日 10:15 AM, Benjamin Root wrote:
>
>
> On Fri, Sep 14, 2012 at 3:50 PM, Eric Firing <ef...@ha...
> <mailto:ef...@ha...>> wrote:
>
> On 2012年09月14日 9:00 AM, Benjamin Root wrote:
> > tricontourf() might be more what you are looking for. Another
> > possibility is pcolor() (note that for irregularly spaced grids,
> > pcolormesh() would not work).
>
> Huh? I don't think there is anything pcolor can handle that pcolormesh
> can't handle faster. In both cases, the grids must be quadrilateral,
> but that's all.
>
> Eric
>
>
> Clarification: pcolormesh() must have a grid of coordinates (not
> necessarially equally spaced).
>
> As for pcolormesh() being able to handle anything that pcolor() can
> handle, I have run into situations where that was not the case. I don't
> remember the details, though. I think pcolorfast() operates like that
> (falling back to pcolor() as a last resort).
No, pcolorfast never falls back to pcolor. In order of fastest to 
slowest, it tries to use image rendering, then a variant of nonuniform 
image rendering, and then a quadmesh. It is a bit fussier about inputs 
than pcolor and pcolormesh, and does not draw lines. pcolor and 
pcolormesh differ in the mechanism they use (pcolor uses a 
PolyCollection) and in the way masked data are handled (pcolor draws 
nothing in masked regions).
Eric
>
> Ben Root
>

On Fri, Sep 14, 2012 at 3:50 PM, Eric Firing <ef...@ha...> wrote:
> On 2012年09月14日 9:00 AM, Benjamin Root wrote:
> > tricontourf() might be more what you are looking for. Another
> > possibility is pcolor() (note that for irregularly spaced grids,
> > pcolormesh() would not work).
>
> Huh? I don't think there is anything pcolor can handle that pcolormesh
> can't handle faster. In both cases, the grids must be quadrilateral,
> but that's all.
>
> Eric
>
>
Clarification: pcolormesh() must have a grid of coordinates (not
necessarially equally spaced).
As for pcolormesh() being able to handle anything that pcolor() can handle,
I have run into situations where that was not the case. I don't remember
the details, though. I think pcolorfast() operates like that (falling back
to pcolor() as a last resort).
Ben Root

Wonderful...pcolor is doing the job without processing, it takes exactly what
I already have...the n+1 values for x and y coordinates defining the
boundaries of the cells and the nxn matix itself. 
pcolormesh and pcolorfast also work.
Thank you very much.
--
View this message in context: http://matplotlib.1069221.n5.nabble.com/Need-to-plot-z-at-given-x-y-contour-or-something-tp38926p38930.html
Sent from the matplotlib - users mailing list archive at Nabble.com.

On 2012年09月14日 9:00 AM, Benjamin Root wrote:
> tricontourf() might be more what you are looking for. Another
> possibility is pcolor() (note that for irregularly spaced grids,
> pcolormesh() would not work).
Huh? I don't think there is anything pcolor can handle that pcolormesh 
can't handle faster. In both cases, the grids must be quadrilateral, 
but that's all.
Eric

Hi folks,
you may have already seen this, but in case you haven't, I'm thrilled
to share that the Python Software Foundation has just created its
newest and highest distinction, the Distinguished Service Award, and
has chosen John as its first recipient:
http://pyfound.blogspot.com/2012/09/announcing-2012-distinctive-service.html
This is a fitting tribute to his many contributions.
Cheers,
f

On Fri, Sep 14, 2012 at 2:51 PM, gsal <sal...@gm...> wrote:
> Hi, everybody:
>
> I don't have experience with images or contours and need some help plotting
> a 'z' quantity for given x,y coordinates.
>
> What are the choices?
>
> Here is a small sample of the data:
>
> The first row has the i-th x-coordinate at which the field starts to have
> the value [i,j].
> The first column has the j-th y-coordinate at which the field starts to
> have the value [i,j].
>
> The coordinate steps are not constant, nor the same for both dimensions;
> they can be anything because they come from some odd finite difference
> program.
>
> My first shot at this is getting combersome, I am hoping for a better way.
>
> So far, because the second decimal place in the x,y coordinate is alwasy
> zero, I simply turned those coordinates into integers by multiplying by ten
> and truncating; then by subtracting the first value from the rest, they
> look
> very much like matrix indeces (except for the missing ones):
>
> Then, because I don't know any better, I broadcast the values onto another
> matrix, to fill in the in-between values:
>
> Now, I have a matrix where every i,j has its own z-value and I am supposed
> to be able to plot it with ax.contourf(mymatrix)...which I can, up until
> about mymatrix[:,:1900] or so, afte that, I get the following error:
>
> Traceback (most recent call last):
> File "C:\findiff\t1.py", line 73, in <module>
> ax.contourf(full[:,:2000])
> File "C:\Python26\lib\site-packages\matplotlib\axes.py", line 7322, in
> contourf
> return mcontour.QuadContourSet(self, *args, **kwargs)
> File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1106, in
> __init__
> ContourSet.__init__(self, ax, *args, **kwargs)
> File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 700, in
> __init__
> self._process_args(*args, **kwargs)
> File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1130, in
> _process_args
> C = _cntr.Cntr(x, y, z.filled(), _mask)
> ValueError: Arguments x, y, z, mask (if present) must be 2D arrays.
> x, y, z must be castable to double.
>
>
>
> My current matrix is about 12000x5000.
>
> Any asistance would be greatly appreciated.
>
> Thanks,
>
> Germán
>
>
>
tricontourf() might be more what you are looking for. Another possibility
is pcolor() (note that for irregularly spaced grids, pcolormesh() would not
work).
Cheers!
Ben Root

Hi, everybody:
I don't have experience with images or contours and need some help plotting
a 'z' quantity for given x,y coordinates. 
What are the choices?
Here is a small sample of the data:
 The first row has the i-th x-coordinate at which the field starts to have
the value [i,j].
 The first column has the j-th y-coordinate at which the field starts to
have the value [i,j].
 
 The coordinate steps are not constant, nor the same for both dimensions;
they can be anything because they come from some odd finite difference
program. 
 
 My first shot at this is getting combersome, I am hoping for a better way.
 
 So far, because the second decimal place in the x,y coordinate is alwasy
zero, I simply turned those coordinates into integers by multiplying by ten
and truncating; then by subtracting the first value from the rest, they look
very much like matrix indeces (except for the missing ones):
 Then, because I don't know any better, I broadcast the values onto another
matrix, to fill in the in-between values:
Now, I have a matrix where every i,j has its own z-value and I am supposed
to be able to plot it with ax.contourf(mymatrix)...which I can, up until
about mymatrix[:,:1900] or so, afte that, I get the following error:
Traceback (most recent call last):
 File "C:\findiff\t1.py", line 73, in <module>
 ax.contourf(full[:,:2000])
 File "C:\Python26\lib\site-packages\matplotlib\axes.py", line 7322, in
contourf
 return mcontour.QuadContourSet(self, *args, **kwargs)
 File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1106, in
__init__
 ContourSet.__init__(self, ax, *args, **kwargs)
 File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 700, in
__init__
 self._process_args(*args, **kwargs)
 File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1130, in
_process_args
 C = _cntr.Cntr(x, y, z.filled(), _mask)
ValueError: Arguments x, y, z, mask (if present) must be 2D arrays.
x, y, z must be castable to double.
My current matrix is about 12000x5000.
Any asistance would be greatly appreciated.
Thanks,
Germán 
--
View this message in context: http://matplotlib.1069221.n5.nabble.com/Need-to-plot-z-at-given-x-y-contour-or-something-tp38926.html
Sent from the matplotlib - users mailing list archive at Nabble.com.

All,
John Hunter, lead author of matplotlib, has been awarded PSF's
Distinguished Service Award.
For details, see
http://pyfound.blogspot.co.uk/2012/09/announcing-2012-distinctive-service.html
Some of you may have already seen this. For those who haven't, it serves as
a poignant reminder to the immense effort of John and the matplotlib
developers over the past decade.
Best,
Damon
-- 
Damon McDougall
http://www.damon.is-a-geek.com
B2.39
Mathematics Institute
University of Warwick
Coventry
West Midlands
CV4 7AL
United Kingdom

Thanks for raising this. I have simplified and opened an issue for the bug (
https://github.com/matplotlib/matplotlib/issues/1246) and will be looking
at this asap.
All the best,
Phil

________________________________
 From: Jeff Whitaker <js...@fa...>
To: mat...@li... 
Cc: raw...@ya... 
Sent: Thursday, September 13, 2012 9:44 PM
Subject: Re: [Matplotlib-users] error installing basemap
 
Michael: The NetCDFFile function was deprecated a few releases back, and recently removed. If you have netcdf4-python installed you can do
from netCDF4 import Dataset as NetCDFFile
and the script should work as before.
Regarding the second error, you must open the file for write access
 (mode='w') if you want to add attributes to the data variables.
-Jeff
From: Michael Rawlins <raw...@ya...>
To: Jeff Whitaker <js...@fa...>; 
"mat...@li..." 
<mat...@li...> 
Sent: Thursday, September 13, 2012 10:28 PM
Subject: Re: [Matplotlib-users] error installing basemap
Making some progress. Following here:
http://code.google.com/p/netcdf4-python/wiki/UbuntuInstall
I've installed HDF5 after installing build-essential package. With just gcc installed I got an error.
The ./configure in netcdf4 directory failed, but completed with --disable-netcdf-4. I went ahead anyway hoping I don't need netCDF-4 formats or the additional netCDF-4 functions. Besides that issue, what's also not clear is where is setup.py file for the last step. It is not in the netcdf directory after the make install finished. I assume netcdf-python package will not conflict with the python-netcdf and netcdf versions installed through package manager. 
Mike
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I get an error when trying to use axvline in a gridspec subplot when the Y axis is set to log scale in matplotlib 1.3.x (and I think recent 1.2.x versions) from github under python 2.7.3 and OS X 10.8.1. This worked with matplotlib versions from github earlier this year. Here is a simple code snippet to reproduce the problem, and the error that is generated. The vertical line is drawn in the vsw sublot if I comment out the dens.axvline line. The problem seems to be caused by the log scale in the dens subplot.
Any help is appreciated,
Scott
---------------------------------------------------------------------------
import matplotlib as mpl
import matplotlib.dates
import matplotlib.pyplot as plt
import numpy as np
import datetime as dt
starttime = dt.datetime(2012,9,12,20,15)
stoptime = starttime + dt.timedelta(hours=1)
fig = plt.figure(1,figsize=(480.0/72.0, 620.0/72.0), dpi=72)
gs = mpl.gridspec.GridSpec(4, 1, height_ratios=[1.,1.,1.,.67], top=.95, bottom=0.10, left=0.12, hspace=0.12)
vsw = plt.subplot(gs[0], label="vsw", autoscale_on=True, xlim=[starttime, stoptime])
vsw.plot_date(mpl.dates.drange(starttime, stoptime, dt.timedelta(minutes=10)), np.random.random(6))
dens = plt.subplot(gs[1], label="dens", autoscale_on=True, xlim=[starttime, stoptime], yscale='log')
dens.plot_date(mpl.dates.drange(starttime, stoptime, dt.timedelta(minutes=10)), np.random.random(6))
vsw.axvline(dt.datetime(2012,9,12,20,30))
dens.axvline(dt.datetime(2012,9,12,20,30))
plt.show()
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-36-781867aaf919> in <module>()
 15 dens.plot_date(mpl.dates.drange(starttime, stoptime, dt.timedelta(minutes=10)), np.random.random(6))
 16 vsw.axvline(dt.datetime(2012,9,12,20,30))
---> 17 dens.axvline(dt.datetime(2012,9,12,20,30))
 18 plt.show()
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib-1.3.x-py2.7-macosx-10.7-intel.egg/matplotlib/axes.pyc in axvline(self, x, ymin, ymax, **kwargs)
 3571 self.transData, self.transAxes)
 3572 l = mlines.Line2D([x,x], [ymin,ymax] , transform=trans, **kwargs)
-> 3573 self.add_line(l)
 3574 self.autoscale_view(scalex=scalex, scaley=False)
 3575 return l
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib-1.3.x-py2.7-macosx-10.7-intel.egg/matplotlib/axes.pyc in add_line(self, line)
 1504 line.set_clip_path(self.patch)
 1505 
-> 1506 self._update_line_limits(line)
 1507 if not line.get_label():
 1508 line.set_label('_line%d' % len(self.lines))
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib-1.3.x-py2.7-macosx-10.7-intel.egg/matplotlib/axes.pyc in _update_line_limits(self, line)
 1525 # identify the transform to go from line's coordinates
 1526 # to data coordinates
-> 1527 trans_to_data = line_trans - self.transData
 1528 
 1529 # if transData is affine we can use the cached non-affine component
/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib-1.3.x-py2.7-macosx-10.7-intel.egg/matplotlib/transforms.pyc in __sub__(self, other)
 1204 return self + other.inverted()
 1205 else:
-> 1206 raise ValueError('It is not possible to compute transA - transB '
 1207 'since transB cannot be inverted and there is no '
 1208 'shortcut possible.')
ValueError: It is not possible to compute transA - transB since transB cannot be inverted and there is no shortcut possible.

________________________________
 From: Jeff Whitaker <js...@fa...>
To: mat...@li... 
Cc: raw...@ya... 
Sent: Thursday, September 13, 2012 9:44 PM
Subject: Re: [Matplotlib-users] error installing basemap
 
On 9/13/12 2:34 PM, Michael Rawlins wrote:
>
>
>
>
>________________________________
> From: Michael Droettboom <md...@st...>
>To: mat...@li... 
>Sent: Thursday, September 13, 2012 2:09 PM
>Subject: Re: [Matplotlib-users] error installing basemap
>
>
>You need to also install the python development package (python-dev), which contains the headers. 
>
>Mike
>
>
>
>
>
>
>
>________________________________
> From: Michael Rawlins <raw...@ya...>
>To: Michael Droettboom <md...@st...>; "mat...@li..." <mat...@li...> 
>Sent: Thursday, September 13, 2012 3:11 PM
>Subject: Re: [Matplotlib-users] error installing basemap
>ailing list Mat...@li... https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>
>OK basemap installed. Thanks. But I'm
 getting an error running a script that
 worked with previous installation(s) of
 python, matplotlib, and basemap. The error:
>
>user@comsys:~>python map2_TempDiff_4panels.py
>Traceback (most recent call last):
> File "map2_TempDiff_4panels.py", line 27,
 in <module>
>  from mpl_toolkits.basemap import 
 NetCDFFile
>ImportError: cannot import name NetCDFFile
>
>
>I installed python-mpltoolkits.basemap from
 package manager, before testing my script.
>
>MR
>
>
>An update: My test script, which works with
 previously, now gets past the header
 initializations. Here they are:
>
>import sys,getopt
>from mpl_toolkits.basemap import Basemap,
 shiftgrid, cm 
>#from mpl_toolkits.basemap import 
 NetCDFFile
>from Scientific.IO.NetCDF import NetCDFFile
>from pylab import *
>import matplotlib.pyplot as plt
>
>
>Notr clear why the first import NetCDFFile
 statement does not work. Farther down the
 script, the code stops on this statement:
>
>data.missing_value=-9.99
>
>There error to standard output:
>
>Traceback (most recent call last):
> File "map2_TempDiff_4panels.py", line 266,
 in <module>
>  data.missing_value=-9.99
>IOError: netcdf: write access to read-only
 file
>
>
Michael: The NetCDFFile function was deprecated a few releases
 back, and recently removed. If you have netcdf4-python installed
 you can do
from netCDF4 import Dataset as NetCDFFile
and the script should work as before.
Regarding the second error, you must open the file for write access
 (mode='w') if you want to add attributes to the data variables.
-Jeff
Jeff,
No I don't have netCDF4 installed. It's not in the package manager. Wasn't sure if there would be a conflict with python-netcdf that's installed. Guess that's the reverse interface. Now I'm having some trouble compiling netcdf4-python from sources following:
http://code.google.com/p/netcdf4-python/wiki/UbuntuInstall
The configure ended in error:
checking whether the C compiler works... no
configure: error: in `/home/rawlins/Downloads/netcdf-4.2.1.1':
configure: error: C compiler cannot create executables
 
Mike

On 9/13/12 2:34 PM, Michael Rawlins wrote:
>
>
> ------------------------------------------------------------------------
> *From:* Michael Droettboom <md...@st...>
> *To:* mat...@li...
> *Sent:* Thursday, September 13, 2012 2:09 PM
> *Subject:* Re: [Matplotlib-users] error installing basemap
>
> You need to also install the python development package (python-dev), 
> which contains the headers.
>
> Mike
>
>
>
>
> ------------------------------------------------------------------------
> *From:* Michael Rawlins <raw...@ya...>
> *To:* Michael Droettboom <md...@st...>; 
> "mat...@li..." 
> <mat...@li...>
> *Sent:* Thursday, September 13, 2012 3:11 PM
> *Subject:* Re: [Matplotlib-users] error installing basemap
> ailing list Mat...@li... 
> <mailto:Mat...@li...> 
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>
> OK basemap installed. Thanks. But I'm getting an error running a 
> script that worked with previous installation(s) of python, 
> matplotlib, and basemap. The error:
>
> user@comsys:~>python map2_TempDiff_4panels.py 
> <http://map2_tempdiff_4panels.py/>
> Traceback (most recent call last):
> File "map2_TempDiff_4panels.py", line 27, in <module>
> from mpl_toolkits.basemap import NetCDFFile
> ImportError: cannot import name NetCDFFile
>
>
> I installed python-mpltoolkits.basemap from package manager, before 
> testing my script.
>
> MR
>
>
> An update: My test script, which works with previously, now gets past 
> the header initializations. Here they are:
>
> import sys,getopt
> from mpl_toolkits.basemap import Basemap, shiftgrid, cm
> #from mpl_toolkits.basemap import NetCDFFile
> from Scientific.IO.NetCDF import NetCDFFile
> from pylab import *
> import matplotlib.pyplot as plt
>
>
> Notr clear why the first import NetCDFFile statement does not work. 
> Farther down the script, the code stops on this statement:
>
> data.missing_value=-9.99
>
> There error to standard output:
>
> Traceback (most recent call last):
> File "map2_TempDiff_4panels.py", line 266, in <module>
> data.missing_value=-9.99
> IOError: netcdf: write access to read-only file
>
Michael: The NetCDFFile function was deprecated a few releases back, 
and recently removed. If you have netcdf4-python installed you can do
from netCDF4 import Dataset as NetCDFFile
and the script should work as before.
Regarding the second error, you must open the file for write access 
(mode='w') if you want to add attributes to the data variables.
-Jeff