On Sun, Oct 11, 2009 at 9:07 AM, per freem <perfr...@gmail.com> wrote:

> hi all,
>
> i am trying to share both an axis label (but not the entire axis) and
> a figure legend across a set of subplots. that is, i'd like to have a
> figure where there is a major enlarged ylabel that is meant to label
> the entire row of subplots and a main figure legend, rather than
> individual legends inside subplots. what i have right now is:
>
> import matplotlib.pyplot as plt
> from mpl_toolkits.axes_grid.axislines import SubplotZero
> fig = plt.figure()
> ax = SubplotZero(fig, 3, 1, 1)
> ax1 = fig.add_subplot(ax)
> ax.axis["xzero"].set_visible(True)
> plt.plot([1,2,3], label="line a", c='r')
> plt.plot([1.2, 2.4, 3.01], label="line b", c='b')
> ax = SubplotZero(fig, 3, 1, 2)
> ax2 = fig.add_subplot(ax)
> plt.plot([1,2,3], label="line a", c='r')
> plt.plot([1.2, 2.4, 3.01], label="line b", c='b')
> ax = SubplotZero(fig, 3, 1, 3)
> ax3 = fig.add_subplot(ax)
> plt.plot([1,2,3], label="line a", c='r')
> plt.plot([1.2, 2.01, 3.01], label="line b", c='b')
> plt.figlegend([ax1.lines[0], ax1.lines[1]], ["line a", "line b"], 'upper
> right')
>
> two quick questions about this: first, how can i create a single
> ylabel for the figure?  i still want each subplot to have its own
> yticks and ytick labels but i just want a main label, since the
> quantity plotted on each y axis is the same.
>

I had asked a similar question at
http://www.nabble.com/Common-ylabel-for-subplots-td22416664.html#a22416664

which I had managed to have a common y-label following the suggestions given
on that thread. However I haven't gone further to create a common legend.


>
> finally, plt.figlegend does display a main legend for the figure, like
> i intended, but i was hoping someone can explain how ax1.lines gets
> set. i was expecting ax1.lines to have 2 elements in it, one for each
> line plotted, but it has 4. why is that? right now i simply used
> lines[0] and lines[1] to create the labels for only the plotted
> lines... is there a better way of doing this?
>
> thanks.
>
>
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-- 
Gökhan
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