El 11/02/10 18:28, Jae-Joon Lee escribió:
> 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)])
>
Thanks Jae-Joon. You're right. I tried this before, but I must have done
something wrong then, because when
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,
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
