Re: [Matplotlib-users] No legend using scatter function

[Gökhan S. <gok...@gm...>]

Thanks JJ,

I was playing with different ways to create legends, and found myself a way
just like you suggested.

I guess in legend as long as a sequence is provided it doesn't matter
whether it is a tuple or list. Am I correct?

Additionally, speaking of scatters; is there a way to add a colorbar next to
scatters depicting the variation in colors or size of the scatters?

Thanks.

On Thu, Aug 6, 2009 at 11:47 AM, Jae-Joon Lee <lee...@gm...> wrote:

> This turned out to be a bug introduced recently, which is now fixed in
> the 0.99 maintenance branch.
> The fix is not merged into the head yet. I tried svnmerge.py but it
> gave some merge conflict. While the conflict seems rather trivial,
> I'll leave it to others.
>
> Meanwhile, you can explicitly give artists and labels to create the legend.
>
> s1 = scatter(..)
> s2 = scatter(..)
>
> legend([s1, s2], ["aaaa", "cccc"])
>
> -JJ
>
>
> On Tue, Aug 4, 2009 at 4:51 PM, Gökhan Sever<gok...@gm...> wrote:
> > Hello,
> >
> > Could not get a legend shown on scatter function used plot:
> >
> > In [4]: a = rand(100)
> >
> > In [5]: b = rand(100)
> >
> > In [22]: scatter(a,b, c=a, s=b, label="aaaa")
> > Out[22]: <matplotlib.collections.CircleCollection object at 0xa133dcc>
> >
> > In [23]: scatter(a,b, c=a*5, s=b*5, label="cccc")
> > Out[23]: <matplotlib.collections.CircleCollection object at 0xa311e6c>
> >
> > In [24]: legend()
> >
> > at rev 7355.
> >
> > with
> >
> > Python 2.6 (r26:66714, Jun  8 2009)
> > IPython 0.10.bzr.r1163 -- An enhanced Interactive Python.
> >
> > Any suggestions?
> >
> >
> > --
> > Gökhan
> >
> >
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-- 
Gökhan