Re: [Matplotlib-users] Polar plot - problem with negative values for radius
I am using a workaround now. But that is a hackery solution. Before plotting my data I convert it to dBs and limit it to the lowest value I want to display. Then I plot it using a regular polar plot with a custom formatting function that sets the tick labels with respect to the data offset. Since I use a custom Navigation-Toolbar anyways it was no big deal to add the few necessery lines of codes to handle the offset there, too. The plot now looks exactly as I want it. But: I'd still prefer using a scale that does all the work in the background. -- View this message in context: http://old.nabble.com/Polar-plot---problem-with-negative-values-for-radius-tp30936638p30975519.html Sent from the matplotlib - users mailing list archive at Nabble.com. -- The ultimate all-in-one performance toolkit: Intel(R) Parallel Studio XE: Pinpoint memory and threading errors before they happen. Find and fix more than 250 security defects in the development cycle. Locate bottlenecks in serial and parallel code that limit performance. http://p.sf.net/sfu/intel-dev2devfeb ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Polar plot - problem with negative values for radius
On Thu, Feb 17, 2011 at 4:12 AM, Stephan Markus wrote: > > Small update: > > I tried the very same code with MPL 1.0.1 and Python 2.5.0 on Linux 64 and > Python 2.5.4 on Win32 and it runs w/o throwing any exceptions there! > > But: the behaviour is still not that what I expected. Still these issues > are > remaining: > - the smallest magnitude (center magnitude in other words) is 0.1 and I'd > like to display way smaller values > - the smallest magnitude doesn't even change when zooming > - the grid lines and ticks do not show as I'd expect (see my last message) > when zooming > > > That's why I'd rather stick to a linear scale, doing my own logarithmic > conversion, limit my data at lowest value I need and just use an offset for > the scale. > I see what you mean. This problem is almost identical to another thread going on where we can't seem to correctly do log scale for 3d plots. The tick locators are in the wrong positions (the major ticks should be evenly spaced) and the error message is similar to one I have been encountering in mplot3d. I will look a little further into this and see if I can kill two birds with one stone... Ben Root -- The ultimate all-in-one performance toolkit: Intel(R) Parallel Studio XE: Pinpoint memory and threading errors before they happen. Find and fix more than 250 security defects in the development cycle. Locate bottlenecks in serial and parallel code that limit performance. http://p.sf.net/sfu/intel-dev2devfeb___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Polar plot - problem with negative values for radius
Small update: I tried the very same code with MPL 1.0.1 and Python 2.5.0 on Linux 64 and Python 2.5.4 on Win32 and it runs w/o throwing any exceptions there! But: the behaviour is still not that what I expected. Still these issues are remaining: - the smallest magnitude (center magnitude in other words) is 0.1 and I'd like to display way smaller values - the smallest magnitude doesn't even change when zooming - the grid lines and ticks do not show as I'd expect (see my last message) when zooming That's why I'd rather stick to a linear scale, doing my own logarithmic conversion, limit my data at lowest value I need and just use an offset for the scale. -- View this message in context: http://old.nabble.com/Polar-plot---problem-with-negative-values-for-radius-tp30936638p30948265.html Sent from the matplotlib - users mailing list archive at Nabble.com. -- The ultimate all-in-one performance toolkit: Intel(R) Parallel Studio XE: Pinpoint memory and threading errors before they happen. Find and fix more than 250 security defects in the development cycle. Locate bottlenecks in serial and parallel code that limit performance. http://p.sf.net/sfu/intel-dev2devfeb ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Polar plot - problem with negative values for radius
Ben,
I should have mentioned that I already tried that. When I set the rscale to
'log' the plot crashes when zooming or mpl cannot even create it.
Maybe some example code will help:
from numpy import arange, sin, pi, cos, ones
from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg,
NavigationToolbar2TkAgg
from matplotlib.figure import Figure
import Tkinter as Tk
root = Tk.Tk()
f = Figure(figsize=(5,4), dpi=100)
ax = f.add_subplot(111, projection='polar')
t = arange(0.0,2*pi,0.01)
s1 = ones(len(t))*10
ax.set_rscale('log')
ax.plot(t,s1)
canvas = FigureCanvasTkAgg(f, master=root)
canvas.show()
canvas.get_tk_widget().pack(side=Tk.TOP, fill=Tk.BOTH, expand=1)
toolbar = NavigationToolbar2TkAgg(canvas, root)
toolbar.update()
canvas._tkcanvas.pack(side=Tk.TOP, fill=Tk.BOTH, expand=1)
Tk.mainloop()
Using this code the plot will show up but you can see the 10^-1 and 10^-2
tick labels overlapping in the center. Ok, it shouldn't be a big deal to get
rid of them. The major problem is if you try to zoom out, the tick labels
move away from the center and if you try to zoom into the plot it eventually
throws an exception "ValueError: cannot convert float NaN to integer".
If you try to plot smaller values (e.g. replacing the line 's1 =
ones(len(t))*10' with 's1 = ones(len(t))') mpl also throws an ValueError
exception and does not even create the plot.
Let me know if you need full tracebacks.
Btw: I am using Matplotlib 1.0.1 and Python 2.6.0 on Windows 64.
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Re: [Matplotlib-users] Polar plot - problem with negative values for radius
On Wed, Feb 16, 2011 at 9:16 AM, Stephan Markus wrote:
>
> Hi,
>
> I am trying to display some complex values in a polar plot. Displaying
> linear magnitude vs. angle - of course - works without any issues. But I'd
> rather display the logarithmic magnitute vs. angle. Since the data for the
> radius gets negative then, it'll be wrapped around / rotated by 180deg by
> matplotlib.
>
> How can I display negative values for the radius w/o having them rotated by
> 180deg?
>
> Of course I can add an offset to the data before plotting it but since my
> plot is interactive and I use the xdata/ydata of events I'd like them to
> represent the real data, without an offset. I thought it might be possible
> to achieve with a custom transformation but I don't actually get it to
> work.
>
> -Stephan
> --
> View this message in context:
> http://old.nabble.com/Polar-plot---problem-with-negative-values-for-radius-tp30936638p30936638.html
> Sent from the matplotlib - users mailing list archive at Nabble.com.
>
>
Stephan,
>From the polar axes object, you can call ax.set_rscale('log') and that will
automatically make the radial ticks scale and label the axes properly. You
can then input the original data as you would without worry of negative
values.
Let us know how that works for you!
Ben Root
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[Matplotlib-users] Polar plot - problem with negative values for radius
Hi, I am trying to display some complex values in a polar plot. Displaying linear magnitude vs. angle - of course - works without any issues. But I'd rather display the logarithmic magnitute vs. angle. Since the data for the radius gets negative then, it'll be wrapped around / rotated by 180deg by matplotlib. How can I display negative values for the radius w/o having them rotated by 180deg? Of course I can add an offset to the data before plotting it but since my plot is interactive and I use the xdata/ydata of events I'd like them to represent the real data, without an offset. I thought it might be possible to achieve with a custom transformation but I don't actually get it to work. -Stephan -- View this message in context: http://old.nabble.com/Polar-plot---problem-with-negative-values-for-radius-tp30936638p30936638.html Sent from the matplotlib - users mailing list archive at Nabble.com. -- The ultimate all-in-one performance toolkit: Intel(R) Parallel Studio XE: Pinpoint memory and threading errors before they happen. Find and fix more than 250 security defects in the development cycle. Locate bottlenecks in serial and parallel code that limit performance. http://p.sf.net/sfu/intel-dev2devfeb ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
2010/8/20 Michael Droettboom : >> Yeah, it's my issue, but I'm not happy with fixing it. Currently, >> matplotlib forces the xticks (i.e., the theta ticks) to be at sensible >> values via .set_xticks() and .set_xlabels() (projections/polar.py). >> >> I'm coding a matplotlib extension package which has to clear the axes >> often, but restoring the major locators, the title and stuff after >> clearing. It was agnostic to the specialities of polar axes so far. > > Why and how are you restoring the major locator? It seems like that's the > issue. I don't think preventing the theta locator from being changed is > something we want to do. Polar plots (by default) just set fixed theta > ticks at multiples of pi/4. My package provides support for Layers in matplotlib. And the layers' data can be changed, making a complete redraw of the axes a solution much easier to implement than dealing with the fuzzy API stuff directly. I don't need high-performance. When putting axes.clear(), the locator is being reset. In cartesian coords, it happens that I want a MaxNLocator(nbins=3) or similiar from time to time, and this must be restored then. For cartesian axes, if the user does not specify another locator, I'm setting the AutoLocator(), what is the same what the (cartesian) axes does. Dealing with the case "no locator set" (by "using matplotlib default fallback") is neither nice nor straightforward. (It has to do with complicating the calling conventions. I would have to treat the case 'ignore [x|y]locator' specially.) It would maybe be a good solution to provide some abstract Axes.get_default_locators(), being public. Each SomeClass(Axes) can define what they understand under "default locator". >> I would rather suggest to insert a new Locator class being aware of >> radians. It would suffice to return tick positions dividing 2 pi into >> an integer number of bins. It's not necessary to cover all the >> peculiarities of the old historic division system into 360 parts. > > Perhaps using FixedLocator, rather than explicitly setting the ticks using > set_xticks (as polar plots currently do) would be better. However, the > locator could still be changed, not really addressing your problem. Seems that you misunderstood my problem, if I'm not misunderstanding you :-) I have no problem with a "mutable" locator. (But the user has normally no access to the axes, only the Stack class changes the axes.) But right, I wanted to derive it from the Locator class framework, just specialising the location. > For convenience, however, we could add a locator that given n, divides 2pi > into n parts. Yes, and Ben's idea is quite nice, to make this accessible also to rectangular plots. This implies some simple thoughts on the view lims to take them into account when issuing the tick locations. >> Accompanying would be formatters in radians and degrees with >> adjustable precision (no autodetect necessary). > > Sure. Adding a radian formatter makes sense. If we go into details let's switch to -devel maybe? Friedrich -- Sell apps to millions through the Intel(R) Atom(Tm) Developer Program Be part of this innovative community and reach millions of netbook users worldwide. Take advantage of special opportunities to increase revenue and speed time-to-market. Join now, and jumpstart your future. http://p.sf.net/sfu/intel-atom-d2d ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On Fri, Aug 20, 2010 at 8:33 AM, Michael Droettboom wrote: > On 08/19/2010 05:53 PM, Friedrich Romstedt wrote: > > 2010/8/19 Michael Droettboom: > > > >> On 08/18/2010 06:03 PM, Friedrich Romstedt wrote: > >> > >>> Is the attached issue with a plain polar axes already fixed? I never > >>> encountered this before. 344 degrees happens to be 6.0 rad. I'm on > >>> svn 8626. > >>> > >> How are you creating that graph? By default, polar plots don't do that. > >> > > Yeah, it's my issue, but I'm not happy with fixing it. Currently, > > matplotlib forces the xticks (i.e., the theta ticks) to be at sensible > > values via .set_xticks() and .set_xlabels() (projections/polar.py). > > > > I'm coding a matplotlib extension package which has to clear the axes > > often, but restoring the major locators, the title and stuff after > > clearing. It was agnostic to the specialities of polar axes so far. > > > Why and how are you restoring the major locator? It seems like that's > the issue. I don't think preventing the theta locator from being > changed is something we want to do. Polar plots (by default) just set > fixed theta ticks at multiples of pi/4. > > I would rather suggest to insert a new Locator class being aware of > > radians. It would suffice to return tick positions dividing 2 pi into > > an integer number of bins. It's not necessary to cover all the > > peculiarities of the old historic division system into 360 parts. > > > Perhaps using FixedLocator, rather than explicitly setting the ticks > using set_xticks (as polar plots currently do) would be better. > However, the locator could still be changed, not really addressing your > problem. > > For convenience, however, we could add a locator that given n, divides > 2pi into n parts. > > Accompanying would be formatters in radians and degrees with > > adjustable precision (no autodetect necessary). > > > Sure. Adding a radian formatter makes sense. > > Just curious, this wouldn't have to be just for PolarPlots, right? Could it also be used for regular plots of sinusoids and such. Ben Root -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On 08/19/2010 05:53 PM, Friedrich Romstedt wrote: > 2010/8/19 Michael Droettboom: > >> On 08/18/2010 06:03 PM, Friedrich Romstedt wrote: >> >>> Is the attached issue with a plain polar axes already fixed? I never >>> encountered this before. 344 degrees happens to be 6.0 rad. I'm on >>> svn 8626. >>> >> How are you creating that graph? By default, polar plots don't do that. >> > Yeah, it's my issue, but I'm not happy with fixing it. Currently, > matplotlib forces the xticks (i.e., the theta ticks) to be at sensible > values via .set_xticks() and .set_xlabels() (projections/polar.py). > > I'm coding a matplotlib extension package which has to clear the axes > often, but restoring the major locators, the title and stuff after > clearing. It was agnostic to the specialities of polar axes so far. > Why and how are you restoring the major locator? It seems like that's the issue. I don't think preventing the theta locator from being changed is something we want to do. Polar plots (by default) just set fixed theta ticks at multiples of pi/4. > I would rather suggest to insert a new Locator class being aware of > radians. It would suffice to return tick positions dividing 2 pi into > an integer number of bins. It's not necessary to cover all the > peculiarities of the old historic division system into 360 parts. > Perhaps using FixedLocator, rather than explicitly setting the ticks using set_xticks (as polar plots currently do) would be better. However, the locator could still be changed, not really addressing your problem. For convenience, however, we could add a locator that given n, divides 2pi into n parts. > Accompanying would be formatters in radians and degrees with > adjustable precision (no autodetect necessary). > Sure. Adding a radian formatter makes sense. Mike -- Michael Droettboom Science Software Branch Space Telescope Science Institute Baltimore, Maryland, USA -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
2010/8/19 Michael Droettboom : > On 08/18/2010 06:03 PM, Friedrich Romstedt wrote: >> Is the attached issue with a plain polar axes already fixed? I never >> encountered this before. 344 degrees happens to be 6.0 rad. I'm on >> svn 8626. > > How are you creating that graph? By default, polar plots don't do that. Yeah, it's my issue, but I'm not happy with fixing it. Currently, matplotlib forces the xticks (i.e., the theta ticks) to be at sensible values via .set_xticks() and .set_xlabels() (projections/polar.py). I'm coding a matplotlib extension package which has to clear the axes often, but restoring the major locators, the title and stuff after clearing. It was agnostic to the specialities of polar axes so far. I could say, "if nothing is requested specially, treat it as a running system", but I see this as clumsy and error-prone at all. I would rather suggest to insert a new Locator class being aware of radians. It would suffice to return tick positions dividing 2 pi into an integer number of bins. It's not necessary to cover all the peculiarities of the old historic division system into 360 parts. Accompanying would be formatters in radians and degrees with adjustable precision (no autodetect necessary). Friedrich -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On 08/18/2010 06:03 PM, Friedrich Romstedt wrote: > 2010/8/18 Michael Droettboom: > >> This bug (that the r-axis labels are in the wrong place) should now be fixed >> in r8651. This doesn't, unfortunately, address the original question about >> annular plots. >> > Is the attached issue with a plain polar axes already fixed? I never > encountered this before. 344 degrees happens to be 6.0 rad. I'm on > svn 8626. > How are you creating that graph? By default, polar plots don't do that. Mike -- Michael Droettboom Science Software Branch Space Telescope Science Institute Baltimore, Maryland, USA -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On Wed, Aug 18, 2010 at 10:03 PM, Nils Wagner wrote: > I would like to plot an annulus. > With mpl_toolkits.axisartist, it is possible to make an axes of annulus. But, the resulting axes is not fully compatible with the original matplotlib axes. Most of the tick-related commands won't work and you need to use new commands provided by the axisartist module. Attached is a sample script and a screeshot. Also see these examples, http://matplotlib.sourceforge.net/examples/axes_grid/demo_floating_axes.html -JJ <> demo_floating_axes.py Description: Binary data -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
2010/8/18 Michael Droettboom : > This bug (that the r-axis labels are in the wrong place) should now be fixed > in r8651. This doesn't, unfortunately, address the original question about > annular plots. Is the attached issue with a plain polar axes already fixed? I never encountered this before. 344 degrees happens to be 6.0 rad. I'm on svn 8626. polar.pdf Description: Adobe PDF document -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On 08/18/2010 09:51 AM, Benjamin Root wrote: On Wed, Aug 18, 2010 at 8:03 AM, Nils Wagner mailto:nwag...@iam.uni-stuttgart.de>> wrote: Hi all, Is it possible to create a polar plot, where the lower bound of the radius is larger than zero ? I would like to plot an annulus. Any pointer would be appreciated. Thanks in advance Nils Nils, It appears that there is a .set_rmin() function, however, I don't think it does what we expect it to. I can't get an annulus, but it only plots the parts that are r >= r_min (but r_min is at the origin, and the axis labels are in the wrong places...) Ben Root This bug (that the r-axis labels are in the wrong place) should now be fixed in r8651. This doesn't, unfortunately, address the original question about annular plots. Mike -- Michael Droettboom Science Software Branch Space Telescope Science Institute Baltimore, Maryland, USA -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On Wed, 18 Aug 2010 08:51:31 -0500 Benjamin Root wrote: > On Wed, Aug 18, 2010 at 8:03 AM, Nils Wagner > wrote: > >> Hi all, >> >> Is it possible to create a polar plot, where the lower >> bound of the radius is larger than zero ? >> I would like to plot an annulus. >> >> Any pointer would be appreciated. >> >> Thanks in advance >> >> Nils >> >> > Nils, > > It appears that there is a .set_rmin() function, >however, I don't think it > does what we expect it to. I can't get an annulus, but >it only plots the > parts that are r >= r_min (but r_min is at the origin, >and the axis labels > are in the wrong places...) > > Ben Root Ben, Thank you for your reply. Please can you send me your example. Nils -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot
On Wed, Aug 18, 2010 at 8:03 AM, Nils Wagner wrote: > Hi all, > > Is it possible to create a polar plot, where the lower > bound of the radius is larger than zero ? > I would like to plot an annulus. > > Any pointer would be appreciated. > > Thanks in advance > > Nils > > Nils, It appears that there is a .set_rmin() function, however, I don't think it does what we expect it to. I can't get an annulus, but it only plots the parts that are r >= r_min (but r_min is at the origin, and the axis labels are in the wrong places...) Ben Root -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] polar plot
Hi all, Is it possible to create a polar plot, where the lower bound of the radius is larger than zero ? I would like to plot an annulus. Any pointer would be appreciated. Thanks in advance Nils -- This SF.net email is sponsored by Make an app they can't live without Enter the BlackBerry Developer Challenge http://p.sf.net/sfu/RIM-dev2dev ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot fills entire plot while drawing line, bug or feature?
azerith wrote: > Hi, I'm new to matplotlib and in need to draw a single impulse on a polar > plot, but don't know how to do it, so i just draw a line using > pylab.polar([0,0],[0,100],'g-') > but when i draw shorter lines or > pylab.polar([0,0],[0,0],'g-') > i get the whole plot filled with green color, instead of a single dot or a > very short line > does anyone else has similar issue? is this a bug or a feature that i can > turned off somewhere? > i'm using ubuntu. > > Thank you for any help It's a bug. You probably have quite an old version of mpl. A shiny new version was just released, and if you can install it in place of the ubuntu package, that particular bug will go away. Eric -- Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day trial. Simplify your report design, integration and deployment - and focus on what you do best, core application coding. Discover what's new with Crystal Reports now. http://p.sf.net/sfu/bobj-july ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] polar plot fills entire plot while drawing line, bug or feature?
Hi, I'm new to matplotlib and in need to draw a single impulse on a polar plot, but don't know how to do it, so i just draw a line using pylab.polar([0,0],[0,100],'g-') but when i draw shorter lines or pylab.polar([0,0],[0,0],'g-') i get the whole plot filled with green color, instead of a single dot or a very short line does anyone else has similar issue? is this a bug or a feature that i can turned off somewhere? i'm using ubuntu. Thank you for any help -- View this message in context: http://www.nabble.com/polar-plot-fills-entire-plot-while-drawing-line%2C-bug-or-feature--tp24840620p24840620.html Sent from the matplotlib - users mailing list archive at Nabble.com. -- Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day trial. Simplify your report design, integration and deployment - and focus on what you do best, core application coding. Discover what's new with Crystal Reports now. http://p.sf.net/sfu/bobj-july ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot: problem with negative angles
The grid line will reappear if you set high enough resolution. plt.subplot(111, polar=True, resolution=100) This should be filed as a bug, though. I guess the current default for resolution is 1. I think this was to enable to draw a straight line in polar projection. However, my guess is that it has a side-effect that a angular gridline became a 0-length line connecting two identical points. -JJ On Sun, May 17, 2009 at 2:51 PM, Magnus Benjes wrote: >> Magnus Benjes wrote: >>> Hello, >>> in version 0.98.5.2 the polar plot still has a problem with negativ >>> angles. >>> The polarplot is drawing a circle when the angle changes from negativ to >>> positiv (e.g. from -0.01 to +0.01). >> >> Your example works fine with svn. I don't recall whether the problem was >> fixed before the last release. I think it was. >> > Thank you for the hint, in version 0.98.6 the polar plot has no problems > with negativ angles any more. > But now there are only gridlines in radial direction and the gridlines in > angular direction are missing. > > Magnus > > > -- > Crystal Reports - New Free Runtime and 30 Day Trial > Check out the new simplified licensing option that enables > unlimited royalty-free distribution of the report engine > for externally facing server and web deployment. > http://p.sf.net/sfu/businessobjects > ___ > Matplotlib-users mailing list > Matplotlib-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/matplotlib-users > -- Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot: problem with negative angles
> Magnus Benjes wrote: >> Hello, >> in version 0.98.5.2 the polar plot still has a problem with negativ >> angles. >> The polarplot is drawing a circle when the angle changes from negativ to >> positiv (e.g. from -0.01 to +0.01). > > Your example works fine with svn. I don't recall whether the problem was > fixed before the last release. I think it was. > Thank you for the hint, in version 0.98.6 the polar plot has no problems with negativ angles any more. But now there are only gridlines in radial direction and the gridlines in angular direction are missing. Magnus -- Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] polar plot: problem with negative angles
Magnus Benjes wrote: > Hello, > in version 0.98.5.2 the polar plot still has a problem with negativ angles. > The polarplot is drawing a circle when the angle changes from negativ to > positiv (e.g. from -0.01 to +0.01). Your example works fine with svn. I don't recall whether the problem was fixed before the last release. I think it was. > > But in "What new in 0.98.4" > (http://matplotlib.sourceforge.net/users/whats_new.html) I can read: > "Fix polar interpolation to handle negative values of theta - MGD" I think that commit actually introduced the problem. > > Is there a workaround for this problem? A release appears to be imminent--it was tagged hours ago. Can you update when it appears? Eric > > Regards > Magnus > > > import numpy as np > import matplotlib > import matplotlib.pyplot as plt > > x = np.linspace(-np.pi/3, np.pi/3, 100) > y = np.sin((10*x)**2)+2 > > plt.subplot(111, polar=True) > plt.plot(x,y) > plt.title(matplotlib.__version__) > plt.show() > > > > > > > -- > Crystal Reports - New Free Runtime and 30 Day Trial > Check out the new simplified licensing option that enables > unlimited royalty-free distribution of the report engine > for externally facing server and web deployment. > http://p.sf.net/sfu/businessobjects > > > > > ___ > Matplotlib-users mailing list > Matplotlib-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/matplotlib-users -- Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] polar plot: problem with negative angles
Hello, in version 0.98.5.2 the polar plot still has a problem with negativ angles. The polarplot is drawing a circle when the angle changes from negativ to positiv (e.g. from -0.01 to +0.01). But in "What new in 0.98.4" (http://matplotlib.sourceforge.net/users/whats_new.html) I can read: "Fix polar interpolation to handle negative values of theta - MGD" Is there a workaround for this problem? Regards Magnus import numpy as np import matplotlib import matplotlib.pyplot as plt x = np.linspace(-np.pi/3, np.pi/3, 100) y = np.sin((10*x)**2)+2 plt.subplot(111, polar=True) plt.plot(x,y) plt.title(matplotlib.__version__) plt.show() -- Crystal Reports - New Free Runtime and 30 Day Trial Check out the new simplified licensing option that enables unlimited royalty-free distribution of the report engine for externally facing server and web deployment. http://p.sf.net/sfu/businessobjects___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Polar Plot: Extraneous Circles
mcdevitts wrote: > Hello, > > I've been trying to implement a smith chart like a previous gentlemen on the > list and have run into a similar problem. When my angle value goes from -pi > to pi I get an extraneous circle from the interpolation. From the previous > posts, I am under the impression that this is a known issue and has been > corrected. Am I correct? If so, I am still having this issue using version > 0.98.5.2 of matplotlib. > > Regards, > Sean McDevitt Yes, this is known and was fixed on January 2. Can you update from svn? Eric -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Polar Plot: Extraneous Circles
Hello, I've been trying to implement a smith chart like a previous gentlemen on the list and have run into a similar problem. When my angle value goes from -pi to pi I get an extraneous circle from the interpolation. From the previous posts, I am under the impression that this is a known issue and has been corrected. Am I correct? If so, I am still having this issue using version 0.98.5.2 of matplotlib. Regards, Sean McDevitt -- View this message in context: http://www.nabble.com/Polar-Plot%3A--Extraneous-Circles-tp22156864p22156864.html Sent from the matplotlib - users mailing list archive at Nabble.com. -- Open Source Business Conference (OSBC), March 24-25, 2009, San Francisco, CA -OSBC tackles the biggest issue in open source: Open Sourcing the Enterprise -Strategies to boost innovation and cut costs with open source participation -Receive a $600 discount off the registration fee with the source code: SFAD http://p.sf.net/sfu/XcvMzF8H ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] Polar plot with y-lim set to non-zero
Hi All - is there any way to make a polar plot with the center of the plot *not* set to 0? I tried resetting ylim, but that just changes the grid laid over the plot, not the location of the markers. Thanks -- Ariel -- Create and Deploy Rich Internet Apps outside the browser with Adobe(R)AIR(TM) software. With Adobe AIR, Ajax developers can use existing skills and code to build responsive, highly engaging applications that combine the power of local resources and data with the reach of the web. Download the Adobe AIR SDK and Ajax docs to start building applications today-http://p.sf.net/sfu/adobe-com ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Polar plot
It seems you're getting bitten by the brain-dead interpolation of points
(which is meant to draw line segments as curves that follow the radii).
I have fixed the polar drawing code in SVN r6106 to normalize theta to
(0.0 <= theta <= 2pi) before doing the interpolation, which seems to fix
your example. The user doesn't/shouldn't have to care about whether
theta is negative.
Cheers,
Mike
jan gillis wrote:
> Hi Tony,
>
> Thank you for the reply, the solutions you propose are fine in this
> case. But I'm trying to use the polar plot
> as a smith chart for an instrument and there i will receive data that is
> unknown but can be something like this:
>
> r = np.transpose(.1+np.arange ( 0 , 0.7 , 0.001))
> theta = -4.5 * np.pi *r
> freq = r*10e9
> data = np.multiply(r,np.exp(1j*theta))
> ax.plot(angle(data),abs(data))
>
> Any idea why Polar plot can't handle theta going from negative to
> positive radians?
>
> Jan
>
> Tony S Yu wrote:
>
>> On Sep 17, 2008, at 1:59 AM, jan gillis wrote:
>>
>>
>>> Hello,
>>>
>>> I have a problem with polar plot, if i run the following code in
>>> matplotlib 0.98.3, polar plot is drawing a extra circle to go from
>>> angle -3.14159265 to angle 3.03753126. Is there a solution for this
>>> problem?
>>>
>>>
>>> import numpy as np
>>> from matplotlib.pyplot import figure, show, rc, grid
>>>
>>> # radar green, solid grid lines
>>> rc('grid', color='#316931', linewidth=1, linestyle='-')
>>> rc('xtick', labelsize=15)
>>> rc('ytick', labelsize=15)
>>>
>>> # force square figure and square axes looks better for polar, IMO
>>> fig = figure(figsize=(8,8))
>>> ax = fig.add_axes([0.1, 0.1, 0.8, 0.8], polar=True, axisbg='#d5de9c')
>>>
>>> z = np.zeros((1,2000),complex)
>>> z.real = 0.2
>>> z.imag = np.arange(-50,50,0.05)
>>> gamma_r = np.transpose((z-1)/(z+1))
>>>
>>> ax.plot(np.angle(gamma_r), np.abs(gamma_r), '.-', zorder=0)
>>>
>> Hi Jan,
>>
>> It looks like you get the circle because the angles you're plotting go
>> from negative to positive radians in a weird way. The circle being
>> drawn starts around 0 radians and goes clockwise by negative values.
>> Then when it gets to - pi, it switches to positive indices, i.e. pi.
>> Of course, these are the same points on a polar plot, but different
>> angles, if you want to be consistent.
>>
>> Here are a couple of quick solutions, but there but there maybe better
>> ways of handling this.
>>
>> # ~~
>> # get rid of the plot line above, and add the following
>> theta = np.angle(gamma_r)
>> mag = np.abs(gamma_r)
>>
>> # option 1
>> ordered = np.argsort(theta, axis=0).squeeze()
>> ax.plot(theta[ordered], mag[ordered], '.-', zorder=0)
>>
>> # option 2
>> neg_theta = np.where(theta < 0)
>> theta[neg_theta] += 2 * np.pi
>> ax.plot(theta, mag, '.-', zorder=0)
>> # ~~
>>
>> I hope that's helpful,
>> -Tony
>>
>>
>>> ax.set_rmax(2.0)
>>> grid(True)
>>>
>>> show()
>>>
>>>
>>> Kind regards,
>>> Jean
>>>
>>>
>>> -
>>>
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>>
>
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Re: [Matplotlib-users] Polar plot
Hi Tony,
Thank you for the reply, the solutions you propose are fine in this
case. But I'm trying to use the polar plot
as a smith chart for an instrument and there i will receive data that is
unknown but can be something like this:
r = np.transpose(.1+np.arange ( 0 , 0.7 , 0.001))
theta = -4.5 * np.pi *r
freq = r*10e9
data = np.multiply(r,np.exp(1j*theta))
ax.plot(angle(data),abs(data))
Any idea why Polar plot can't handle theta going from negative to
positive radians?
Jan
Tony S Yu wrote:
>
> On Sep 17, 2008, at 1:59 AM, jan gillis wrote:
>
>> Hello,
>>
>> I have a problem with polar plot, if i run the following code in
>> matplotlib 0.98.3, polar plot is drawing a extra circle to go from
>> angle -3.14159265 to angle 3.03753126. Is there a solution for this
>> problem?
>>
>>
>> import numpy as np
>> from matplotlib.pyplot import figure, show, rc, grid
>>
>> # radar green, solid grid lines
>> rc('grid', color='#316931', linewidth=1, linestyle='-')
>> rc('xtick', labelsize=15)
>> rc('ytick', labelsize=15)
>>
>> # force square figure and square axes looks better for polar, IMO
>> fig = figure(figsize=(8,8))
>> ax = fig.add_axes([0.1, 0.1, 0.8, 0.8], polar=True, axisbg='#d5de9c')
>>
>> z = np.zeros((1,2000),complex)
>> z.real = 0.2
>> z.imag = np.arange(-50,50,0.05)
>> gamma_r = np.transpose((z-1)/(z+1))
>>
>> ax.plot(np.angle(gamma_r), np.abs(gamma_r), '.-', zorder=0)
>
> Hi Jan,
>
> It looks like you get the circle because the angles you're plotting go
> from negative to positive radians in a weird way. The circle being
> drawn starts around 0 radians and goes clockwise by negative values.
> Then when it gets to - pi, it switches to positive indices, i.e. pi.
> Of course, these are the same points on a polar plot, but different
> angles, if you want to be consistent.
>
> Here are a couple of quick solutions, but there but there maybe better
> ways of handling this.
>
> # ~~
> # get rid of the plot line above, and add the following
> theta = np.angle(gamma_r)
> mag = np.abs(gamma_r)
>
> # option 1
> ordered = np.argsort(theta, axis=0).squeeze()
> ax.plot(theta[ordered], mag[ordered], '.-', zorder=0)
>
> # option 2
> neg_theta = np.where(theta < 0)
> theta[neg_theta] += 2 * np.pi
> ax.plot(theta, mag, '.-', zorder=0)
> # ~~
>
> I hope that's helpful,
> -Tony
>
>>
>> ax.set_rmax(2.0)
>> grid(True)
>>
>> show()
>>
>>
>> Kind regards,
>> Jean
>>
>>
>> -
>>
>> 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
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>> ___
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>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>
>
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Re: [Matplotlib-users] Polar plot
On Sep 17, 2008, at 1:59 AM, jan gillis wrote:
> Hello,
>
> I have a problem with polar plot, if i run the following code in
> matplotlib 0.98.3, polar plot is drawing a extra circle to go from
> angle -3.14159265 to angle 3.03753126. Is there a solution for this
> problem?
>
>
> import numpy as np
> from matplotlib.pyplot import figure, show, rc, grid
>
> # radar green, solid grid lines
> rc('grid', color='#316931', linewidth=1, linestyle='-')
> rc('xtick', labelsize=15)
> rc('ytick', labelsize=15)
>
> # force square figure and square axes looks better for polar, IMO
> fig = figure(figsize=(8,8))
> ax = fig.add_axes([0.1, 0.1, 0.8, 0.8], polar=True, axisbg='#d5de9c')
>
> z = np.zeros((1,2000),complex)
> z.real = 0.2
> z.imag = np.arange(-50,50,0.05)
> gamma_r = np.transpose((z-1)/(z+1))
>
> ax.plot(np.angle(gamma_r), np.abs(gamma_r), '.-', zorder=0)
Hi Jan,
It looks like you get the circle because the angles you're plotting go
from negative to positive radians in a weird way. The circle being
drawn starts around 0 radians and goes clockwise by negative values.
Then when it gets to - pi, it switches to positive indices, i.e. pi.
Of course, these are the same points on a polar plot, but different
angles, if you want to be consistent.
Here are a couple of quick solutions, but there but there maybe better
ways of handling this.
# ~~
# get rid of the plot line above, and add the following
theta = np.angle(gamma_r)
mag = np.abs(gamma_r)
# option 1
ordered = np.argsort(theta, axis=0).squeeze()
ax.plot(theta[ordered], mag[ordered], '.-', zorder=0)
# option 2
neg_theta = np.where(theta < 0)
theta[neg_theta] += 2 * np.pi
ax.plot(theta, mag, '.-', zorder=0)
# ~~
I hope that's helpful,
-Tony
>
> ax.set_rmax(2.0)
> grid(True)
>
> show()
>
>
> Kind regards,
> Jean
>
> --
> Jan Gillis
> Ghent University
> IMEC vzw - INTEC
> Sint-Pietersnieuwstraat 41
> B-9000 Gent
> Belgium
> tel. +32 9 264 33 33
> [EMAIL PROTECTED]
>
>
> -
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> the world
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[Matplotlib-users] Polar plot
Hello,
I have a problem with polar plot, if i run the following code in
matplotlib 0.98.3, polar plot is drawing a extra circle to go from
angle -3.14159265 to angle 3.03753126. Is there a solution for this
problem?
import numpy as np
from matplotlib.pyplot import figure, show, rc, grid
# radar green, solid grid lines
rc('grid', color='#316931', linewidth=1, linestyle='-')
rc('xtick', labelsize=15)
rc('ytick', labelsize=15)
# force square figure and square axes looks better for polar, IMO
fig = figure(figsize=(8,8))
ax = fig.add_axes([0.1, 0.1, 0.8, 0.8], polar=True, axisbg='#d5de9c')
z = np.zeros((1,2000),complex)
z.real = 0.2
z.imag = np.arange(-50,50,0.05)
gamma_r = np.transpose((z-1)/(z+1))
ax.plot(np.angle(gamma_r), np.abs(gamma_r), '.-', zorder=0)
ax.set_rmax(2.0)
grid(True)
show()
Kind regards,
Jean
--
Jan Gillis
Ghent University
IMEC vzw - INTEC
Sint-Pietersnieuwstraat 41
B-9000 Gent
Belgium
tel. +32 9 264 33 33
[EMAIL PROTECTED]
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[Matplotlib-users] polar plot
Hello,
I have a problem with polar plot, if i run the following code in
matplotlib 0.98.3, polar plot is drawing a extra circle to go from
angle -3.14159265 to angle 3.03753126. Is there a solution for this problem?
import numpy as np
from matplotlib.pyplot import figure, show, rc, grid
# radar green, solid grid lines
rc('grid', color='#316931', linewidth=1, linestyle='-')
rc('xtick', labelsize=15)
rc('ytick', labelsize=15)
# force square figure and square axes looks better for polar, IMO
fig = figure(figsize=(8,8))
ax = fig.add_axes([0.1, 0.1, 0.8, 0.8], polar=True, axisbg='#d5de9c')
z = np.zeros((1,2000),complex)
z.real = 0.2
z.imag = np.arange(-50,50,0.05)
gamma_r = np.transpose((z-1)/(z+1))
ax.plot(np.angle(gamma_r), np.abs(gamma_r), '.-', zorder=0)
ax.set_rmax(2.0)
grid(True)
show()
Kind regards,
Jean
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Re: [Matplotlib-users] polar plot with glade
On 4/30/07, Gonzalo A. de la Vega <[EMAIL PROTECTED]> wrote: > Hi all, > I'm trying to embed a polar plot into a glade gui. I modified the > mpl_with_glade.py example script to have something to start with. No > problems with normal plot, but when I try to do polar plot, it fails with > the following: Comment out the SpanSelector code -- this is using a matplotlib widget that assumes separable axes, which polar does not have. JDH - This SF.net email is sponsored by DB2 Express Download DB2 Express C - the FREE version of DB2 express and take control of your XML. No limits. Just data. Click to get it now. http://sourceforge.net/powerbar/db2/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
