Re: [Matplotlib-users] How to make a grid of (plot) grids?

On Tue, Jan 4, 2011 at 6:17 PM, Paul Ivanov <piv...@gm...> wrote:

> Gf B, on 2011-01-04 12:31,  wrote:
> > On Mon, Jan 3, 2011 at 3:53 PM, Paul Ivanov <piv...@gm...>
> wrote:
> > Gf B, on 2011-01-03 15:23,  wrote:
> > > > Can such a "grid of grids" be done with matplotlib?  If so, could
> someone
> > > > show me how?
> > >
> > >  You'll be able to group the inner grids visually by adjusting the
> > > spacing.  As far as getting the spines to only outline the outer
> > > grid, and not the inner grid - I think you'll have to do it
> > > manually by hiding the appropriate spines for the inner subplots.
> > >
> >
> > This sort of ad-hoc manual tweaking is what I was hoping to avoid.
> >
> > What would it take to implement a "true" grid-of-grids function in
> > matplotlib?  What I mean by this is a function that can arrange in a grid
> > not only plots but also other grids.  (Is this a question for the devel
> > group?)
>
> I think the true grid-of-grids functunality is already
> implemented. Here's a replication of your Mathematica plots:
>
> ------------
> import numpy as np
> import matplotlib.pyplot as plt
> import matplotlib.gridspec as gridspec
> from itertools import product
>
> def squiggle_xy(a, b, c, d, i=np.linspace(0.0, 2*np.pi, 200)):
>    return np.sin(i*a)*np.cos(i*b), np.sin(i*c)*np.cos(i*d)
>
> f = plt.figure(figsize=(8, 8))
>
> # gridspec inside gridspec
> outer_grid = gridspec.GridSpec(4, 4, wspace=0.0, hspace=0.0)
> for i in xrange(16):
>    inner_grid = gridspec.GridSpecFromSubplotSpec(3, 3,
>            subplot_spec=outer_grid[i], wspace=0.0, hspace=0.0)
>    a, b = int(i/4)+1,i%4+1
>     for j, (c, d) in enumerate(product(range(1, 4), repeat=2)):
>         ax = plt.Subplot(f, inner_grid[j])
>        ax.plot(*squiggle_xy(a, b, c, d))
>         ax.set_xticks([])
>        ax.set_yticks([])
>         f.add_subplot(ax)
>
> all_axes = f.get_axes()
>
> #show only the outside spines
> for ax in all_axes:
>    for sp in ax.spines.values():
>        sp.set_visible(False)
>    if ax.is_first_row():
>        ax.spines['top'].set_visible(True)
>    if ax.is_last_row():
>        ax.spines['bottom'].set_visible(True)
>    if ax.is_first_col():
>        ax.spines['left'].set_visible(True)
>    if ax.is_last_col():
>        ax.spines['right'].set_visible(True)
>
> plt.show()
> ------------
>
> It's a matter of taste, but I think you can get away hiding all
> spines, and just setting the hspace and wspace for the outer_grid
> to some small value (this is what I meant by 'adjusting the
> spacing').
>
> I'll send a patch to the devel list shortly adding this example
> with the following documentation
>
> A Complex Nested GridSpec using SubplotSpec
> ===========================================
> Here's a more sophisticated example of nested gridspect where we put
> a box around outer 4x4 grid, by hiding appropriate spines in each of the
> inner 3x3 grids.
>
> it'll be placed on the gridspec page, after this section:
>
> http://matplotlib.sourceforge.net/users/gridspec.html#gridspec-using-subplotspec
>
> best,
> --
> Paul Ivanov
> 314 address only used for lists,  off-list direct email at:
> http://pirsquared.org | GPG/PGP key id: 0x0F3E28F7
>
>
Just to add, because it is related, another tool that gives you advanced
control over your axes is the AxesGrid toolkit:

http://matplotlib.sourceforge.net/mpl_toolkits/axes_grid/index.html#toolkit-axesgrid-index

However, gridspec should be exactly what you need for this particular
problem.

Ben Root