pylab.ion() in python 2, matplotlib 1.1.1 and upda

Question

what I am trying to do is having a script compute something, prepare a plot and show the already obtained results as a pylab.figure - in python 2 (specifically python 2.7) with a stable matplotlib (which is 1.1.1).

In python 3 (python 3.2.3 with a matplotlib git build ... version 1.2.x), this works fine. As a simple example (simulating a lengthy computation by time.sleep()) consider

import pylab
import time
import random

dat=[0,1]
pylab.plot(dat)
pylab.ion()
pylab.draw()    
for i in range (18):
    dat.append(random.uniform(0,1))
    pylab.plot(dat)
    pylab.draw()
    time.sleep(1)

In python 2 (version 2.7.3 vith matplotlib 1.1.1), the code runs cleanly without errors but does not show the figure. A little trial and error with the python2 interpreter seemed to suggest to replace pylab.draw() with pylab.show(); doing this once is apparently sufficient (not, as with draw calling it after every change/addition to the plot). Hence:

import pylab
import time
import random

dat=[0,1]
pylab.plot(dat)
pylab.ion()
pylab.show()    
for i in range (18):
    dat.append(random.uniform(0,1))
    pylab.plot(dat)
    #pylab.draw()
    time.sleep(1)

However, this doesn't work either. Again, it runs cleanly but does not show the figure. It seems to do so only when waiting for user input. It is not clear to me why this is, but the plot is finally shown when a raw_input() is added to the loop

import pylab
import time
import random

dat=[0,1]
pylab.plot(dat)
pylab.ion()
pylab.show()    
for i in range (18):
    dat.append(random.uniform(0,1))
    pylab.plot(dat)
    #pylab.draw()
    time.sleep(1)
    raw_input()

With this, the script will of course wait for user input while showing the plot and will not continue computing the data before the user hits enter. This was, of course, not the intention.

This may be caused by different versions of matplotlib (1.1.1 and 1.2.x) or by different python versions (2.7.3 and 3.2.3).

Is there any way to accomplish with python 2 with a stable (1.1.1) matplotlib what the above script (the first one) does in python 3, matplotlib 1.2.x: - computing data (which takes a while, in the above example simulated by time.sleep()) in a loop or iterated function and - (while still computing) showing what has already been computed in previous iterations - and not bothering the user to continually hit enter for the computation to continue

Thanks; I'd appreciate any help...

Answer 1

You want the pause function to give the gui framework a chance to re-draw the screen:

import pylab
import time
import random
import matplotlib.pyplot as plt

dat=[0,1]
fig = plt.figure()
ax = fig.add_subplot(111)
Ln, = ax.plot(dat)
ax.set_xlim([0,20])
plt.ion()
plt.show()    
for i in range (18):
    dat.append(random.uniform(0,1))
    Ln.set_ydata(dat)
    Ln.set_xdata(range(len(dat)))
    plt.pause(1)

    print 'done with loop'

You don't need to create a new Line2D object every pass through, you can just update the data in the existing one.

Documentation:

pause(interval)
    Pause for *interval* seconds.

    If there is an active figure it will be updated and displayed,
    and the gui event loop will run during the pause.

    If there is no active figure, or if a non-interactive backend
    is in use, this executes time.sleep(interval).

    This can be used for crude animation. For more complex
    animation, see :mod:`matplotlib.animation`.

    This function is experimental; its behavior may be changed
    or extended in a future release.

A really over-kill method to is to use the matplotlib.animate module. On the flip side, it gives you a nice way to save the data if you want (ripped from my answer to Python- 1 second plots continous presentation).

example, api, tutorial

Answer 2

Some backends (in my experience "Qt4Agg") require the pause function, as @tcaswell suggested.

Other backends (in my experience "TkAgg") seem to just update on draw() without requiring a pause. So another solution is to switch your backend, for example with matplotlib.use('TkAgg').