You're already using "ax.legend", what kind of OO way do you want?

Instead of calling plt.legend, you may do
   ax.legend([s],[str(i)])

Or, if you know what you're doing, you can do

leg = ax.legend([s],[''], loc=0)

and in the for loop,

   leg.texts[0].set_text(str(i))

Regards,

-JJ


On Thu, Feb 11, 2010 at 5:56 AM, Jorge Scandaliaris
<jorgesmbox...@yahoo.es> wrote:
> Hi,
> I am re-using a scatter plot in the same figure for interactively displaying
> results without ending up with 30 windows open. The legend is relevant, and so
> it must be also updated. So far the only way I found was to use the 
> set_label()
> method and then using plt.legend(). Is this the only way to get a legend 
> updated
> from a label? I am curious, since generally there is the pyplot way and a more
> OO way of achieving things, but I could't find it this time.
>
> The snippet below shows what I am using right now:
>
> import numpy as np
> import scipy as sp
> import matplotlib as mpl
> import matplotlib.pyplot as plt
>
> data = np.random.randn(3,10)
> fig = plt.figure()
> ax = fig.add_subplot(111)
> s, = ax.plot([])
> ax.axis([0,10,-1,1])
> ax.legend([s],[''], loc=0)
> for i in range(data.shape[0]):
>    s.set_data([np.arange(data.shape[1]),data[i]])
>    s.set_label(str(i))
>    plt.legend()
>    plt.draw()
>    plt.ginput(timeout=0)
>
>
>
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