Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Michael Droettboom
I've committed both of these things. The subplot()/polar() change seems tricky, so it may produce some breakage even though the "regression tests" are passing. Please let me know if you see anything strange after this change. Mike Ryan May wrote: > Michael Droettboom wrote: >> Thanks for upd

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Ryan May
Michael Droettboom wrote: > Thanks for updating the docstring. I actually saw this as a usability > bug and have come up with a patch such that polar() (et al) will > *replace* the current axes with a polar plot if it isn't already polar. > This is (from the user's perspective) similar to how,

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Michael Droettboom
Ryan May wrote: > On Tue, Dec 2, 2008 at 11:27 AM, Nils Wagner > <[EMAIL PROTECTED] > > wrote: > > Thank you very much ! > It would be nice to have that information in the docstring > > > Done. Thanks for updating the docstring. I actually saw this as a usabilit

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Ryan May
> > The next inquiry is related to xticks. > I have added > > xticks(linspace(0,2*pi,24,endpoint=False)) > > The difference between consecutive xticks is varying between 14 and 16 > degrees. The following works around the roundoff for me: xticks(linspace(0, 360, 24, endpoint=False) * pi/180.) R

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Ryan May
On Tue, Dec 2, 2008 at 11:27 AM, Nils Wagner <[EMAIL PROTECTED]>wrote: > Thank you very much ! > It would be nice to have that information in the docstring > Done. > The next inquiry is related to xticks. > I have added > > xticks(linspace(0,2*pi,24,endpoint=False)) > > The difference between c

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Nils Wagner
On Tue, 2 Dec 2008 11:14:48 -0600 "Ryan May" <[EMAIL PROTECTED]> wrote: On Tue, Dec 2, 2008 at 11:07 AM, Nils Wagner <[EMAIL PROTECTED]>wrote: Hi all, If I run the attached example I obtain no polar plots, but a view like plot(t,r_1) - for what reason ? You need to specify polar=True to

Re: [Matplotlib-users] polar in a subplot

2008-12-02 Thread Ryan May
On Tue, Dec 2, 2008 at 11:07 AM, Nils Wagner <[EMAIL PROTECTED]>wrote: > Hi all, > > If I run the attached example I obtain no polar plots, but a view like > plot(t,r_1) - for what reason ? > You need to specify polar=True to the subplot commands. Try this: from pylab import subplot, polar, lin