ipython : get access to current figure()

plt.gcf() to get current figure

plt.gca() to get current axis

With the plt? example (assuming ipython --pylab)

In [44]: x=np.arange(0,5,.1)In [45]: y=np.sin(x)In [46]: plt.plot(x,y)Out[46]: [<matplotlib.lines.Line2D at 0xb09418cc>]

displays figure 1; get its handle with:

In [47]: f=plt.figure(1)In [48]: fOut[48]: <matplotlib.figure.Figure at 0xb17acb2c>

and a list of its axes:

In [49]: f.axesOut[49]: [<matplotlib.axes._subplots.AxesSubplot at 0xb091198c>]

turn the grid on for the current (and only) axis:

In [51]: a=f.axes[0]In [52]: a.grid(True)

I haven't used the plt in a while, so found this stuff by just making the plot and searching the tab completion and ? for likely stuff. I'm pretty sure this is also available in the plt documentation.

Or you can create the figure first, and hang on to its handle

In [53]: fig=plt.figure()In [55]: ax1=fig.add_subplot(2,1,1)In [56]: ax2=fig.add_subplot(2,1,2)In [57]: plt.plot(x,y)Out[57]: [<matplotlib.lines.Line2D at 0xb12ed5ec>]In [58]: fig.axesOut[58]: [<matplotlib.axes._subplots.AxesSubplot at 0xb0917e2c>, <matplotlib.axes._subplots.AxesSubplot at 0xb17a35cc>]

And there's gcf and gca (get current figure/axis). Same as in MATLAB if my memory is correct.

In [68]: plt.gca()Out[68]: <matplotlib.axes._subplots.AxesSubplot at 0xb17a35cc>In [66]: plt.gcf()Out[66]: <matplotlib.figure.Figure at 0xb091eeec>

(these are used in the sidebar link: Matplotlib.pyplot - Deactivate axes in figure. /Axis of figure overlap with axes of subplot)