Another vote here for changing resolution=1 to be the default. It seems strange to plot values on a line plot and not have the values connected. At least in my area, every polar plot of real-world data we make needs to set this flag to 1 or the wrong plot is created.
Ted > -----Original Message----- > From: Michael Droettboom [mailto:md...@stsci.edu] > Sent: Monday, February 02, 2009 8:28 AM > To: James Evans; matplotlib development list > Subject: Re: [matplotlib-devel] FW: Polar Plot Design Issues > > You can get this behavior you are asking for by passing "resolution=1" > to polar. (This has been there for ages, but unfortunately wasn't > documented until recently). We can consider making 1 the default. > > There are cases in which the automatic interpolation is useful, but in > the present implementation the onus is on the user to avoid this "going > the long way" problem. > > Mike > > James Evans wrote: > > Michael, > > > > John said you were the one to go to for this one. I was wondering if > you had any comments about the following? > > > > --James Evans > > > > > >> All, > >> > >> While looking over the polar plot code I came across the following > issue: When plotting something > >> like 'polar( [2*pi/180, 358*pi/180], [2.0, 1.0] )' the plotted line > will actually wrap around the > >> origin of the plot before reaching its destination. Initially I > thought that this was correct > >> behavior. The line numerically passed through all angles between 2 > and 358 degrees in a linear > >> fashion. However after consulting several colleagues and text books > I believe that the behavior is > >> actually wrong. > >> > >> It is my understanding that for polar plots there is no linear > mapping of the axes as it is currently > >> implemented. Rather for a simple two-point line defined in polar > coordinates, the line should > >> essentially take the direct route. This is highlighted by the two- > point equation of a line for polar > >> plots: > >> > >> r = ( r1*r2*sin(t2-t1) ) / ( (r1*sin(t-t1)) - (r2*sin(t-t2)) ) > >> > >> If you were to plug in the two points given above, then increment > theta (t) from 2 degrees to 358 > >> degrees, then convert to Cartesian cords, and plot the results, you > will get the correct line that > >> directly crosses the zero degree line and not one that wraps around > the origin. > >> > >> Is the polar plot function implemented this way on purpose? Which > way should it really be > >> implemented? > >> > > > > > > -- > Michael Droettboom > Science Software Branch > Operations and Engineering Division > Space Telescope Science Institute > Operated by AURA for NASA > > > ----------------------------------------------------------------------- > ------- > This SF.net email is sponsored by: > SourcForge Community > SourceForge wants to tell your story. > http://p.sf.net/sfu/sf-spreadtheword > _______________________________________________ > Matplotlib-devel mailing list > Matplotlib-devel@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/matplotlib-devel > > No virus found in this incoming message. > Checked by AVG - www.avg.com > Version: 8.0.233 / Virus Database: 270.10.17/1931 - Release Date: > 02/02/09 19:21:00 ------------------------------------------------------------------------------ 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-devel mailing list Matplotlib-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-devel
