[Matplotlib-users] xticks and yticks on central axis
I would like the xticks and yticks on the central axis to look like the attached png. Right now, I am just using axvline and axhline. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo <>- This SF.Net email is sponsored by the Moblin Your Move Developer's challenge Build the coolest Linux based applications with Moblin SDK & win great prizes Grand prize is a trip for two to an Open Source event anywhere in the world http://moblin-contest.org/redirect.php?banner_id=100&url=/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] colorbar()
Is there a way to get the colorbar to work with an axes instance. ax2 = axes([0.2, 0.1, 0.6, 0.8], axisbg='w') ax2.fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') ax2.pcolormesh(X, Y, newa, shading='flat', cmap=cm.YlOrRd)#gray_r) ax2.axvline(x=0, color='gray', linestyle='--') ax2.axhline(y=0, color='gray', linestyle='--') ax2.plot([offaxisX], [offaxisY], 'r+', mew=1) ax2.colorbar() AttributeError: 'Axes' object has no attribute 'colorbar' colorbar() , line 499, in colorbar() File "C:\Python25\Lib\site-packages\matplotlib\pyplot.py", line 1129, in colorbar ret = gcf().colorbar(mappable, cax = cax, ax=ax, **kw) File "C:\Python25\Lib\site-packages\matplotlib\figure.py", line 956, in colorbar cb = cbar.Colorbar(cax, mappable, **kw) File "C:\Python25\Lib\site-packages\matplotlib\colorbar.py", line 558, in __init__ mappable.autoscale_None() # Ensure mappable.norm.vmin, vmax AttributeError: 'NoneType' object has no attribute 'autoscale_None' I am not sure how the mappable, ax, and cax options work. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo "Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius - and a lot of courage - to move in the opposite direction. " -Albert Einstein - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Rotating an axes instance
I have an axes instance that I would like to rotate. I see that there is a rotation keyword for text and would like to do something like that with a plot. Is this possible. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo "Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius - and a lot of courage - to move in the opposite direction. " -Albert Einstein - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Rotating an axes instance
I have an axes instance that I would like to rotate. I see that there is a rotation keyword for text and would like to do something like that with a plot. - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Posting to Numpy and Scipy
Has anyone had a problem posting to either of these mailing lists. I am a member and have sent a few posts to each of them over the last couple months, but none of them show up in the list. I always receive a 'awaiting moderator approval' email. I have sent an email to the owner about this, but have not received a response. Bryan - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] missing lines on graph with upgrade to 0.98.0
I have copied the new patches.py and axes.py. It fixed the fill, but I still have an axes instance that is not closed. x1, x2, y1, y2 = -4, 4, -4, 4 ax1 = axes([0.0, 0.0, 1.0, 1.0], axisbg='0.95') ax2 = axes([0.2, 0.1, 0.6, 0.8], axisbg='w') ax2.fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') xlim(x1-2,x2+2) ylim(y1-2,y2+2) Do I need more than the files that I have copied? On Thu, Jun 5, 2008 at 4:19 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan, > > Thanks for pointing this out. Mike D. has made a change in the svn trunk > to restore the automatic closing of polygons made with the patches.Polygon > constructor, which is used by the fill command. > > Eric > > Bryan Fodness wrote: > >> It seems like it does not connect the last point to the first point. This >> also happens with the matplotlib.patches Polygon. >> from pylab import fill, xlim, ylim, savefig >> x1, x2, y1, y2 = -4, 4, -4, 4 >> fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') >> xlim(-5,5) >> ylim(-5,5) >> savefig('edge_test') >> >> >> On Thu, Jun 5, 2008 at 1:18 AM, Eric Firing <[EMAIL PROTECTED] > [EMAIL PROTECTED]>> wrote: >> >>Bryan Fodness wrote: >> >>I just upgraded to 0.98.0 and recreated a few graphs. I am >>missing parts of the edges of a fill and polygon. Any suggestions? >> >> >>Please post an illustrative script, as simple as possible. >> >>Eric >> >> >> >> >> -- >> "The game of science can accurately be described as a never-ending insult >> to human intelligence." - João Magueijo >> > > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] fill function
Is there a way to get the underlying array that the fill function uses to graph a polygon? This is assuming that it uses an array. I would like to be able to multiple an array by an array that describes the polygon (one if inside the polygon and zero if outside) -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] missing lines on graph with upgrade to 0.98.0
It seems like it does not connect the last point to the first point. This also happens with the matplotlib.patches Polygon. from pylab import fill, xlim, ylim, savefig x1, x2, y1, y2 = -4, 4, -4, 4 fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') xlim(-5,5) ylim(-5,5) savefig('edge_test') On Thu, Jun 5, 2008 at 1:18 AM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan Fodness wrote: > >> I just upgraded to 0.98.0 and recreated a few graphs. I am missing parts >> of the edges of a fill and polygon. Any suggestions? >> >> > Please post an illustrative script, as simple as possible. > > Eric > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] missing lines on graph with upgrade to 0.98.0
I just upgraded to 0.98.0 and recreated a few graphs. I am missing parts of the edges of a fill and polygon. Any suggestions? -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo <>- Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] trying to change the number of elements in array while preserving data
I tried posting this to numpy, but my posts never show up. So, I was hoping someone here might be able to help me. I have two arrays that are different sizes and i would like to be able to add them for plotting. If I have an array a and b, [[1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9] [1 2 3 4 5 6 7 8 9]] [[0 0 0 0 0] [0 3 3 3 0] [0 3 3 3 0] [0 3 3 3 0] [0 0 0 0 0]] but I would like to change b to look like this, [[0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 3 3 3 3 3 0 0] [0 0 3 3 3 3 3 0 0] [0 0 3 3 3 3 3 0 0] [0 0 3 3 3 3 3 0 0] [0 0 3 3 3 3 3 0 0] [0 0 3 3 3 3 3 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0]] so I can get the sum of a and b. My data will not be regular like these. I have a 400x60 array with irregular data that I would like as a 400x400 array. Does anybody know of an easy way to accomplish this? Bryan -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://sourceforge.net/services/buy/index.php___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] using facecolor='None' for fill on Windows and Fedora
Thanks for your help! I was forcing it to look in an old location with, #!/usr/bin/python as my first line. On Fri, May 16, 2008 at 1:44 PM, Bryan Fodness <[EMAIL PROTECTED]> wrote: > i didn't have 0.92.1, i have 0.91.2 > > [bryan@ ~]$ python > Python 2.5.2 (r252:60911, May 13 2008, 22:14:05) > [GCC 4.1.2 20070502 (Red Hat 4.1.2-12)] on linux2 > Type "help", "copyright", "credits" or "license" for more information. > >>> import matplotlib > >>> matplotlib.__version__ > '0.91.2' > >>> import numpy > >>> numpy.__version__ > '1.0.4' > >>> > > [bryan@ ~]$ locate pylab.py > /usr/lib/python2.5/site-packages/pylab.py > /usr/lib/python2.5/site-packages/pylab.pyc > /usr/lib/python2.5/site-packages/pylab.pyo > /usr/lib/python2.5/site-packages/matplotlib/pylab.py > /usr/lib/python2.5/site-packages/matplotlib/pylab.pyc > /usr/lib/python2.5/site-packages/matplotlib/pylab.pyo > /usr/local/lib/python2.5/site-packages/pylab.py > /usr/local/lib/python2.5/site-packages/pylab.pyc > /usr/local/lib/python2.5/site-packages/matplotlib/pylab.py > /usr/local/lib/python2.5/site-packages/matplotlib/pylab.pyc > [EMAIL PROTECTED] > > On Fri, May 16, 2008 at 1:35 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > >> Bryan, >> >> It appears that you are not using the mpl version you think you are. When >> I look at colors.py for 0.92.1, it doesn't match your error message. >> >> What do you get when you do, from a python prompt: >> >> import matplotlib >> print matplotlib.__version__ >> >> You might try using "locate pylab.py" to find out where the various >> versions are. >> >> Eric >> >> Bryan Fodness wrote: >> >>> I have updated to matplotlib 0.92.1 and numpy 1.0.4 and still receive an >>> error. >>> /space/work/Bryan$ ./Check_0.1.py <http://check_0.1.py/> < >>> http://Check_0.1.py <http://check_0.1.py/>> >>> Traceback (most recent call last): >>> File "./Check_0.1.py <http://check_0.1.py/> >>> <http://Check_0.1.py<http://check_0.1.py/>>", >>> line 344, in >>>savefig(outfile) >>> File "/usr/lib/python2.5/site-packages/matplotlib/pylab.py", line 796, >>> in savefig >>>return fig.savefig(*args, **kwargs) >>> File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 727, >>> in savefig >>>self.canvas.print_figure(*args, **kwargs) >>> File >>> "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line >>> 456, in print_figure >>>self.draw() >>> File >>> "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line >>> 392, in draw >>>self.figure.draw(renderer) >>> File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 569, >>> in draw >>>for a in self.axes: a.draw(renderer) >>> File "/usr/lib/python2.5/site-packages/matplotlib/axes.py", line 1155, >>> in draw >>>a.draw(renderer) >>> File "/usr/lib/python2.5/site-packages/matplotlib/patches.py", line 209, >>> in draw >>>else: rgbFace = colorConverter.to_rgb(self._facecolor) >>> File "/usr/lib/python2.5/site-packages/matplotlib/colors.py", line 429, >>> in to_rgb >>>raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % (str(arg), exc)) >>> ValueError: to_rgb: Invalid rgb arg "None" >>> invalid literal for float(): None >>> Does anyone have an idea? It works in Windows. >>> On Fri, May 9, 2008 at 5:18 PM, Eric Firing <[EMAIL PROTECTED]>> [EMAIL PROTECTED]>> wrote: >>> >>>Bryan Fodness wrote: >>> >>>i have used this command on windows vista with no problem. >>> >>>fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') >>> >>>but when i run on fedora 8, i receive the following. >>> >>>[...] >>> >>>File >>>"/usr/lib/python2.5/site-packages/matplotlib/colors.py", line >>>429, in tob >>> raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % >>>(str(arg), exc)) >>> ValueError: to_rgb: Invalid rgb arg "None" >>> invalid literal for float(): None >>>could someone tell me if i am missing something? >>> >>> >>>Sounds like you have an older mpl version on the Fedora machine than >>>on the Win box. >>> >>>Eric >>> >>> >>> >>> >>> -- >>> "The game of science can accurately be described as a never-ending insult >>> to human intelligence." - João Magueijo >>> >> >> > > > -- > "The game of science can accurately be described as a never-ending insult > to human intelligence." - João Magueijo > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] using facecolor='None' for fill on Windows and Fedora
i didn't have 0.92.1, i have 0.91.2 [bryan@ ~]$ python Python 2.5.2 (r252:60911, May 13 2008, 22:14:05) [GCC 4.1.2 20070502 (Red Hat 4.1.2-12)] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> import matplotlib >>> matplotlib.__version__ '0.91.2' >>> import numpy >>> numpy.__version__ '1.0.4' >>> [bryan@ ~]$ locate pylab.py /usr/lib/python2.5/site-packages/pylab.py /usr/lib/python2.5/site-packages/pylab.pyc /usr/lib/python2.5/site-packages/pylab.pyo /usr/lib/python2.5/site-packages/matplotlib/pylab.py /usr/lib/python2.5/site-packages/matplotlib/pylab.pyc /usr/lib/python2.5/site-packages/matplotlib/pylab.pyo /usr/local/lib/python2.5/site-packages/pylab.py /usr/local/lib/python2.5/site-packages/pylab.pyc /usr/local/lib/python2.5/site-packages/matplotlib/pylab.py /usr/local/lib/python2.5/site-packages/matplotlib/pylab.pyc [EMAIL PROTECTED] On Fri, May 16, 2008 at 1:35 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan, > > It appears that you are not using the mpl version you think you are. When I > look at colors.py for 0.92.1, it doesn't match your error message. > > What do you get when you do, from a python prompt: > > import matplotlib > print matplotlib.__version__ > > You might try using "locate pylab.py" to find out where the various > versions are. > > Eric > > Bryan Fodness wrote: > >> I have updated to matplotlib 0.92.1 and numpy 1.0.4 and still receive an >> error. >> /space/work/Bryan$ ./Check_0.1.py <http://check_0.1.py/> < >> http://Check_0.1.py <http://check_0.1.py/>> >> Traceback (most recent call last): >> File "./Check_0.1.py <http://check_0.1.py/> >> <http://Check_0.1.py<http://check_0.1.py/>>", >> line 344, in >>savefig(outfile) >> File "/usr/lib/python2.5/site-packages/matplotlib/pylab.py", line 796, in >> savefig >>return fig.savefig(*args, **kwargs) >> File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 727, >> in savefig >>self.canvas.print_figure(*args, **kwargs) >> File >> "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line >> 456, in print_figure >>self.draw() >> File >> "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line >> 392, in draw >>self.figure.draw(renderer) >> File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 569, >> in draw >>for a in self.axes: a.draw(renderer) >> File "/usr/lib/python2.5/site-packages/matplotlib/axes.py", line 1155, in >> draw >>a.draw(renderer) >> File "/usr/lib/python2.5/site-packages/matplotlib/patches.py", line 209, >> in draw >>else: rgbFace = colorConverter.to_rgb(self._facecolor) >> File "/usr/lib/python2.5/site-packages/matplotlib/colors.py", line 429, >> in to_rgb >>raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % (str(arg), exc)) >> ValueError: to_rgb: Invalid rgb arg "None" >> invalid literal for float(): None >> Does anyone have an idea? It works in Windows. >> On Fri, May 9, 2008 at 5:18 PM, Eric Firing <[EMAIL PROTECTED] > [EMAIL PROTECTED]>> wrote: >> >>Bryan Fodness wrote: >> >>i have used this command on windows vista with no problem. >> >>fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') >> >>but when i run on fedora 8, i receive the following. >> >>[...] >> >>File >>"/usr/lib/python2.5/site-packages/matplotlib/colors.py", line >>429, in tob >> raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % >>(str(arg), exc)) >> ValueError: to_rgb: Invalid rgb arg "None" >> invalid literal for float(): None >>could someone tell me if i am missing something? >> >> >>Sounds like you have an older mpl version on the Fedora machine than >>on the Win box. >> >>Eric >> >> >> >> >> -- >> "The game of science can accurately be described as a never-ending insult >> to human intelligence." - João Magueijo >> > > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] using facecolor='None' for fill on Windows and Fedora
I have updated to matplotlib 0.92.1 and numpy 1.0.4 and still receive an error. /space/work/Bryan$ ./Check_0.1.py Traceback (most recent call last): File "./Check_0.1.py", line 344, in savefig(outfile) File "/usr/lib/python2.5/site-packages/matplotlib/pylab.py", line 796, in savefig return fig.savefig(*args, **kwargs) File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 727, in savefig self.canvas.print_figure(*args, **kwargs) File "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line 456, in print_figure self.draw() File "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", line 392, in draw self.figure.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 569, in draw for a in self.axes: a.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/axes.py", line 1155, in draw a.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/patches.py", line 209, in draw else: rgbFace = colorConverter.to_rgb(self._facecolor) File "/usr/lib/python2.5/site-packages/matplotlib/colors.py", line 429, in to_rgb raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % (str(arg), exc)) ValueError: to_rgb: Invalid rgb arg "None" invalid literal for float(): None Does anyone have an idea? It works in Windows. On Fri, May 9, 2008 at 5:18 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan Fodness wrote: > >> i have used this command on windows vista with no problem. >> >> fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') >> >> but when i run on fedora 8, i receive the following. >> > [...] > >> File "/usr/lib/python2.5/site-packages/matplotlib/colors.py", line >> 429, in tob >> raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % (str(arg), >> exc)) >> ValueError: to_rgb: Invalid rgb arg "None" >> invalid literal for float(): None >> could someone tell me if i am missing something? >> >> > > Sounds like you have an older mpl version on the Fedora machine than on the > Win box. > > Eric > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] using facecolor='None' for fill on Windows and Fedora
i have used this command on windows vista with no problem. fill([x1,x2,x2,x1], [y1,y1,y2,y2], fc='None', ec='r') but when i run on fedora 8, i receive the following. (most recent call last): File "./program.py", line 361, in savefig(outfile) File "/usr/lib/python2.5/site-packages/matplotlib/pylab.py", line 796, in savg return fig.savefig(*args, **kwargs) File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 727, in sag self.canvas.print_figure(*args, **kwargs) File "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", le self.draw() File "/usr/lib/python2.5/site-packages/matplotlib/backends/backend_agg.py", lw self.figure.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/figure.py", line 569, in drw for a in self.axes: a.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/axes.py", line 1155, in draw a.draw(renderer) File "/usr/lib/python2.5/site-packages/matplotlib/patches.py", line 209, in dw else: rgbFace = colorConverter.to_rgb(self._facecolor) File "/usr/lib/python2.5/site-packages/matplotlib/colors.py", line 429, in tob raise ValueError('to_rgb: Invalid rgb arg "%s"\n%s' % (str(arg), exc)) ValueError: to_rgb: Invalid rgb arg "None" invalid literal for float(): None could someone tell me if i am missing something? -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by the 2008 JavaOne(SM) Conference Don't miss this year's exciting event. There's still time to save $100. Use priority code J8TL2D2. http://ad.doubleclick.net/clk;198757673;13503038;p?http://java.sun.com/javaone___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] using fill, but really need a box
i have been using the fill function to highlight a region on my plot, but now i do not want it to be filled. i have tried using alpha=0.1, but that also makes my edgecolor transparent. is there a "box" function that does not fill a region yet still has the outline of the "box". i tried using patch.rectangle, but it was not what i needed. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by the 2008 JavaOne(SM) Conference Don't miss this year's exciting event. There's still time to save $100. Use priority code J8TL2D2. http://ad.doubleclick.net/clk;198757673;13503038;p?http://java.sun.com/javaone___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Rotating an axes
Can someone tell me if there is a way to rotate an axes instance? I have, figure(1) ax1 = axes([0.0, 0.0, 1.0, 1.0], axisbg='0.95') ax2 = axes([0.2, 0.1, 0.6, 0.8], axisbg='white') and I would like to rotate ax2 30 degrees. It would look something like the attached file. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo <>- This SF.net email is sponsored by the 2008 JavaOne(SM) Conference Don't miss this year's exciting event. There's still time to save $100. Use priority code J8TL2D2. http://ad.doubleclick.net/clk;198757673;13503038;p?http://java.sun.com/javaone___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Drawing shapes outside the plot area
I would like to be able to draw a triangle on the graph outside the axes and plot area. I have used fill before, but that was in the plot area. Can someone push me in the right direction? -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by the 2008 JavaOne(SM) Conference Don't miss this year's exciting event. There's still time to save $100. Use priority code J8TL2D2. http://ad.doubleclick.net/clk;198757673;13503038;p?http://java.sun.com/javaone___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Fwd: speed of clearing and creating a new array
I am starting with a zero array, populate it according to some condition, save the graph, and then reset the array to zero. Can someone tell me if there is a better (faster) way to do this? Or am I just doing that much, and that is how long it will take? I have a couple hundred graphs that I am creating, that will ultimately be combined for animation. ay=4000 ax=60 #create empty array a = zeros((ay,ax), int) while something: pcolormesh(X, Y, a) figname = 'Frame%03d' %int(field) print figname savefig(figname) # reset values a = zeros((ay,ax), int) if something: if something: a[y,x] = a[y,x]+1 - This SF.net email is sponsored by the 2008 JavaOne(SM) Conference Register now and save $200. Hurry, offer ends at 11:59 p.m., Monday, April 7! Use priority code J8TLD2. http://ad.doubleclick.net/clk;198757673;13503038;p?http://java.sun.com/javaone___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Rotating data in graph
Is there an easy way to rotate the view of the data in a graph window without changing the axis and titles. I only need the option for 90, 180, and 270 degrees. The window will always be symmetric. Bryan -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] plotting a pcolormesh and fill
I have a pcolormest and a fill that are plotted on the same graph (they have the same scale). Is there an easy way to only plot the values of pcolormesh that lie in the fill? pcolormesh(oX, oY, test, shading='flat', cmap=cm.gray_r) fill([-x1,x2,x2,-x1], [-y1,-y1,y2,y2], 'b', alpha=0.2, edgecolor='r') -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] plotting a pcolormesh and a fill
I have a pcolormest and a fill that are plotted on the same graph (they have the same scale). Is there an easy way to only plot the values of pcolormesh that lie in the fill? pcolormesh(oX, oY, test, shading='flat', cmap=cm.gray_r) fill([-x1,x2,x2,-x1], [-y1,-y1,y2,y2], 'b', alpha=0.2, edgecolor='r') -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] drawing a box
I do want a rectangle. And, I have tried, axvline(x=x1, ymin=y1, ymax=y2) axvline(x=x2, ymin=y1, ymax=y2) axhline(y=y1, xmin=x1, xmax=x2) axhline(y=y2, xmin=x1, xmax=x2) On Dec 18, 2007 10:40 PM, John Hunter <[EMAIL PROTECTED]> wrote: > On Dec 18, 2007 9:00 PM, Bryan Fodness <[EMAIL PROTECTED]> wrote: > > I would like to draw a polygon using a x1, x2, y1, and y2. > > At a minimum, x1, x2, y1, and y2 define a line segment, or at most a > rectangle. You say a "polygon". What exactly do you mean, and what > have you tried (code please)? > > JDH > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - SF.Net email is sponsored by: Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://ad.doubleclick.net/clk;164216239;13503038;w?http://sf.net/marketplace___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] drawing a box
I would like to draw a polygon using a x1, x2, y1, and y2. I tried to use axhline and axvline with the min and max values but it does give the desired result. It changes the axis limits and does not draw a line at all. Any help would be appreciated. Bryan -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - SF.Net email is sponsored by: Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://ad.doubleclick.net/clk;164216239;13503038;w?http://sf.net/marketplace___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] viewing my data correctly
Thanks, this is what I wanted. I overlooked the cumsum() on the end of my array, so the output of the array was not what I expected or needed. Could you help me figure out how to get an axis label and numbers on the right side of my graph. It is the same data and scale as the left side except it will read Leaf B. On Dec 14, 2007 9:09 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan Fodness wrote: > > I would also like to get the area of the mesh element when I fill the > > corresponding array element. > > > > if a[1,0] > > area = 1.0 * 0.01 > > > > if a[30,0] > > area = 0.5 * 0.01 > > > > Is this possible? > > I'm sorry, but I don't understand what you are asking. Are you asking > how to calculate an array of areas corresponding to the grid? You know > what the delta-Y values are: > > dy = numpy.array([1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]) > > Are you looking for > > area = dy * 0.01 > > ? > > Eric > > > On Nov 26, 2007 7:02 PM, Eric Firing <[EMAIL PROTECTED] > > <mailto:[EMAIL PROTECTED]>> wrote: > > > > Bryan Fodness wrote: > > > Could someone give me an idea how to get started with this so it > > > coincides with my array of values. I took a look at the > > quadmesh_demo > > > in the examples and do not see a straightforward way to do this > > > > Maybe the docstrings make it sound more complicated than it really > is. > > In your case you have an array of rectangles, not general > > quadrilaterals. All you need are two 1-D arrays, one each for the x > > and > > y grid boundaries. Something like this: > > > > Z = numpy.random.rand(60,4000) > > X = numpy.arange(4001) > > Y = numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4 > ]).cumsum() > > pcolor(X, Y, Z) > > > > pcolormesh should work the same, but when I try it now with svn it > > doesn't; I don't know what is going on with it. If you are using a > > release version of mpl, I expect it will work. > > > > Eric > > > > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <[EMAIL PROTECTED] > > <mailto:[EMAIL PROTECTED]>> wrote: > > >> You can provide mesh coordinates to the pcolor command: > > >> > > >> X and Y, if given, specify the (x,y) coordinates of the > colored > > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > > >> should be one greater than those of C; if the dimensions are > the > > >> same, then the last row and column of C will be ignored. > > >> > > >> Actually generating the mesh is up to you (wink), but hopefully > that > > >> allows for what you need to do. > > >> > > >> Cheers, > > >> Mike > > >> > > >> > > >> Bryan Fodness wrote: > > >>> I am wondering if there is a way to view my data with respect > > to the > > >>> physical size of what my array element is suppose to be. > > >>> > > >>> I have an array that is 60 x 4000 where, > > >>> the first row has a height of 1.4 > > >>> the next nine has a height of 1 > > >>> the next forty has a height of 0.5 > > >>> the next nine has a height of 1 > > >>> and the last one has a height of 1.4 > > >>> > > >>> When viewing this with contourf or pcolor, the image is more > narrow > > >>> than it should be. Is there an easy way to view this properly. > > >>> > > >>> Bryan > > >>> > > >> -- > > >> Michael Droettboom > > >> Science Software Branch > > >> Operations and Engineering Division > > >> Space Telescope Science Institute > > >> Operated by AURA for NASA > > >> > > > > > > > > > > > > > > > > > > > -- > > "The game of science can accurately be described as a never-ending > > insult to human intelligence." - João Magueijo > > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo <>- SF.Net email is sponsored by: Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://ad.doubleclick.net/clk;164216239;13503038;w?http://sf.net/marketplace___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] viewing my data correctly
I would also like to get the area of the mesh element when I fill the corresponding array element. if a[1,0] area = 1.0 * 0.01 if a[30,0] area = 0.5 * 0.01 Is this possible? On Nov 26, 2007 7:02 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan Fodness wrote: > > Could someone give me an idea how to get started with this so it > > coincides with my array of values. I took a look at the quadmesh_demo > > in the examples and do not see a straightforward way to do this > > Maybe the docstrings make it sound more complicated than it really is. > In your case you have an array of rectangles, not general > quadrilaterals. All you need are two 1-D arrays, one each for the x and > y grid boundaries. Something like this: > > Z = numpy.random.rand(60,4000) > X = numpy.arange(4001) > Y = numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() > pcolor(X, Y, Z) > > pcolormesh should work the same, but when I try it now with svn it > doesn't; I don't know what is going on with it. If you are using a > release version of mpl, I expect it will work. > > Eric > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <[EMAIL PROTECTED]> wrote: > >> You can provide mesh coordinates to the pcolor command: > >> > >> X and Y, if given, specify the (x,y) coordinates of the colored > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > >> should be one greater than those of C; if the dimensions are the > >> same, then the last row and column of C will be ignored. > >> > >> Actually generating the mesh is up to you (wink), but hopefully that > >> allows for what you need to do. > >> > >> Cheers, > >> Mike > >> > >> > >> Bryan Fodness wrote: > >>> I am wondering if there is a way to view my data with respect to the > >>> physical size of what my array element is suppose to be. > >>> > >>> I have an array that is 60 x 4000 where, > >>> the first row has a height of 1.4 > >>> the next nine has a height of 1 > >>> the next forty has a height of 0.5 > >>> the next nine has a height of 1 > >>> and the last one has a height of 1.4 > >>> > >>> When viewing this with contourf or pcolor, the image is more narrow > >>> than it should be. Is there an easy way to view this properly. > >>> > >>> Bryan > >>> > >> -- > >> Michael Droettboom > >> Science Software Branch > >> Operations and Engineering Division > >> Space Telescope Science Institute > >> Operated by AURA for NASA > >> > > > > > > > > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - SF.Net email is sponsored by: Check out the new SourceForge.net Marketplace. It's the best place to buy or sell services for just about anything Open Source. http://ad.doubleclick.net/clk;164216239;13503038;w?http://sf.net/marketplace___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] viewing my data correctly
Thank You! It works great. On Nov 26, 2007 7:02 PM, Eric Firing <[EMAIL PROTECTED]> wrote: > Bryan Fodness wrote: > > Could someone give me an idea how to get started with this so it > > coincides with my array of values. I took a look at the quadmesh_demo > > in the examples and do not see a straightforward way to do this > > Maybe the docstrings make it sound more complicated than it really is. > In your case you have an array of rectangles, not general > quadrilaterals. All you need are two 1-D arrays, one each for the x and > y grid boundaries. Something like this: > > Z = numpy.random.rand(60,4000) > X = numpy.arange(4001) > Y = numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() > pcolor(X, Y, Z) > > pcolormesh should work the same, but when I try it now with svn it > doesn't; I don't know what is going on with it. If you are using a > release version of mpl, I expect it will work. > > Eric > > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <[EMAIL PROTECTED]> wrote: > >> You can provide mesh coordinates to the pcolor command: > >> > >> X and Y, if given, specify the (x,y) coordinates of the colored > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > >> should be one greater than those of C; if the dimensions are the > >> same, then the last row and column of C will be ignored. > >> > >> Actually generating the mesh is up to you (wink), but hopefully that > >> allows for what you need to do. > >> > >> Cheers, > >> Mike > >> > >> > >> Bryan Fodness wrote: > >>> I am wondering if there is a way to view my data with respect to the > >>> physical size of what my array element is suppose to be. > >>> > >>> I have an array that is 60 x 4000 where, > >>> the first row has a height of 1.4 > >>> the next nine has a height of 1 > >>> the next forty has a height of 0.5 > >>> the next nine has a height of 1 > >>> and the last one has a height of 1.4 > >>> > >>> When viewing this with contourf or pcolor, the image is more narrow > >>> than it should be. Is there an easy way to view this properly. > >>> > >>> Bryan > >>> > >> -- > >> Michael Droettboom > >> Science Software Branch > >> Operations and Engineering Division > >> Space Telescope Science Institute > >> Operated by AURA for NASA > >> > > > > > > > > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] viewing my data correctly
Could someone give me an idea how to get started with this so it coincides with my array of values. I took a look at the quadmesh_demo in the examples and do not see a straightforward way to do this On Nov 26, 2007 7:52 AM, Michael Droettboom <[EMAIL PROTECTED]> wrote: > You can provide mesh coordinates to the pcolor command: > > X and Y, if given, specify the (x,y) coordinates of the colored > quadrilaterals; the quadrilateral for C[i,j] has corners at > (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > should be one greater than those of C; if the dimensions are the > same, then the last row and column of C will be ignored. > > Actually generating the mesh is up to you (wink), but hopefully that > allows for what you need to do. > > Cheers, > Mike > > > Bryan Fodness wrote: > > I am wondering if there is a way to view my data with respect to the > > physical size of what my array element is suppose to be. > > > > I have an array that is 60 x 4000 where, > > the first row has a height of 1.4 > > the next nine has a height of 1 > > the next forty has a height of 0.5 > > the next nine has a height of 1 > > and the last one has a height of 1.4 > > > > When viewing this with contourf or pcolor, the image is more narrow > > than it should be. Is there an easy way to view this properly. > > > > Bryan > > > > -- > Michael Droettboom > Science Software Branch > Operations and Engineering Division > Space Telescope Science Institute > Operated by AURA for NASA > -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] colorbar and secondary axis label
I would like to have my colorbar range from 0 to 1 and add a label (Leaf A) exactly like the Leaf B label on the other side of the y-axis. I have attached my .png -- from pylab import * #pcolor(n, shading='flat', cmap=cm.gray_r) contourf(n, cmap=cm.gray_r) title('Intensity Pattern', size=28) xlabel('Distance (0.01 cm)', size=18) ylabel('Leaf B', size=18) axvline(x=ax/2, color='k', linestyle='--') colorbar() savefig('eIntensity') -- Thank you, Bryan -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] viewing my data correctly
I am wondering if there is a way to view my data with respect to the physical size of what my array element is suppose to be. I have an array that is 60 x 4000 where, the first row has a height of 1.4 the next nine has a height of 1 the next forty has a height of 0.5 the next nine has a height of 1 and the last one has a height of 1.4 When viewing this with contourf or pcolor, the image is more narrow than it should be. Is there an easy way to view this properly. Bryan -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] secondary axis same as primary with different label
I would like to have a secondary y-axis that has the same scale, but a different label. Also, I would like to draw a dashed line down the center of the plot. I have used, ax1 = pcolor(a, shading='flat', cmap=cm.gray_r) title('Intensity Pattern') xlabel('Distance') ylabel('Leaf B') ax2=twinx() ylabel('Leaf A') ax2.yaxis.tick_right() colorbar() and the scale is different for the second y-axis as well as prints on top of the colorbar. -Bryan "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] pcolor
I would like to use pcolor with the colors going from white (zero values) to black (largest value). I am using, pcolor(a, shading = 'flat') colorbar() I do not see how to do this. -- "The game of science can accurately be described as a never-ending insult to human intelligence." - João Magueijo - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users