Sterling, I have version 1.2.0. I read the legend guide. Which thing should be possible? If you mean giving the points as a tuple, here's some simple test code:
import matplotlib.pyplot as plt import numpy as np fig = plt.figure() ax = fig.add_subplot(1,1,1) p1 = ax.scatter([0.1],[0.5],c='r',marker='s') p2 = ax.scatter([0.3],[0.2],c='b',marker='o') l = ax.legend((p1,p2),['points'],scatterpoints=2) plt.show() I get a legend with only red squares. What I want is a legend line with a red square and a blue circle. I can get points with different colors if I do: p1 = ax.scatter([0.1,0.3],[0.5,0.2],c=('r','b'),marker='s') l = ax.legend(['points'],scatterpoints=2) (though as far as I can tell, you can't have a sequence of marker types). Jon On Wed, 2013-01-23 at 09:57 -0800, Sterling Smith wrote: > Jon, > > What version of matplotlib do you have? According to > http://matplotlib.org/users/legend_guide.html > what you say should be possible in 1.2.0. > > -Sterling > > On Jan 23, 2013, at 9:35AM, Jonathan Slavin wrote: > > > Hi, > > > > I'm wondering if there is some straightforward way to combine two > > PathCollection objects to create a new PathCollection object. My goal > > is to include two points that use different axes (one twin'ed to the > > other) into a single legend item (different point types, same label). > > Each call to scatter creates a new PathCollection object. If I could > > just combine two (or more) PathCollection objects -- either extending > > them or concatenating or whatever -- then I could give that to legend > > and it would work correctly. But it's not clear to me how I can do > > that. I tried providing the two objects as a tuple to legend, but it > > doesn't work (only point characteristics of one of them is used). > > > > Really, I think this should be easier -- both the combining of such > > objects (which after all, are collections) and providing legend with > > simple arguments to produce what you want. As far as I can tell I can't > > just tell leged directly the symbol type(s) and colors of point(s) per > > each label. Though in general the automatic method whereby you provide > > the object to the legend is nice and easy, a more crude and direct way > > would be a good option for special cases. > > > > Jon > > -- > > ______________________________________________________________ > > Jonathan D. Slavin Harvard-Smithsonian CfA > > jsla...@cfa.harvard.edu 60 Garden Street, MS 83 > > phone: (617) 496-7981 Cambridge, MA 02138-1516 > > cell: (781) 363-0035 USA > > ______________________________________________________________ > > > > > > ------------------------------------------------------------------------------ > > Master Visual Studio, SharePoint, SQL, ASP.NET, C# 2012, HTML5, CSS, > > MVC, Windows 8 Apps, JavaScript and much more. Keep your skills current > > with LearnDevNow - 3,200 step-by-step video tutorials by Microsoft > > MVPs and experts. ON SALE this month only -- learn more at: > > http://p.sf.net/sfu/learnnow-d2d > > _______________________________________________ > > Matplotlib-users mailing list > > Matplotlib-users@lists.sourceforge.net > > https://lists.sourceforge.net/lists/listinfo/matplotlib-users > -- ______________________________________________________________ Jonathan D. Slavin Harvard-Smithsonian CfA jsla...@cfa.harvard.edu 60 Garden Street, MS 83 phone: (617) 496-7981 Cambridge, MA 02138-1516 cell: (781) 363-0035 USA ______________________________________________________________ ------------------------------------------------------------------------------ Master Visual Studio, SharePoint, SQL, ASP.NET, C# 2012, HTML5, CSS, MVC, Windows 8 Apps, JavaScript and much more. Keep your skills current with LearnDevNow - 3,200 step-by-step video tutorials by Microsoft MVPs and experts. ON SALE this month only -- learn more at: http://p.sf.net/sfu/learnnow-d2d _______________________________________________ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users