Develop Reference

How can I use async Python to allow Jupyter button widget clicks to process while running an async recursive function?

I set up a Jupyter Notebook where I'm running a recursive function that clears the output like so, creating an animated output:
from IPython.display import clear_output
active = True
def animate:
myOutput = ""
# do stuff
clear_output(wait=True)
print(myOutput)
sleep(0.2)
if active:
animate()
and that's working perfectly.
But now I want to add in one more step: A speed toggle. What I'm animating is a debugging visualization of a cursor moving through interpreted code as an interpreter I'm writing parses that code. I tried conditional slow-downs to have more time to read what's going on as the parsing continues, but what I really need is to be able to click a button to toggle the speed between fast and slow. Maybe I'll use a slider, but for now I just want a button for proof of concept.
This sounds simple enough. Note that I'm writing this statefully as a class because I need to read / write the state from within another imported class.
Jupyter block 1:
import ipywidgets as widgets
from IPython.display import display
out = widgets.Output()
class ToggleState():
def __init__(self):
self.button = widgets.Button(description="Toggle")
self.button.on_click(self.toggle)
display(self.button)
self.toggleState = False
print("Toggle State:", self.toggleState)
def toggle(self, arg): # arg has to be accepted here to meet on_click requirements
self.toggleState = not self.toggleState
print("Toggle State:", self.toggleState)
def read(self):
return self.toggleState
toggleState = ToggleState()
Then, in Jupyter block 2, note I decided to to do this in a separate block because the clear_output I'm doing with the animate func clears the button if it's in the same block, and therein lies the problem:
active = True
def animate:
myOutput = ""
# do stuff
clear_output(wait=True)
print(myOutput)
if toggleState.read():
sleep(5)
else:
sleep(0.2)
if active:
animate()
But the problem with this approach was that two blocks don't actually run at the same time (without using parallel kernels which is way more complexity than I care for) so that button can't keep receiving input in the previous block. Seems obvious now, but I didn't think about it.
How can I clear input in a way that doesn't delete my button too (so I can put the button in the animating block)?
Edit:
I thought I figured the solution, but only part of it:
Using the Output widget:
out = widgets.Output()
and
with out:
clear_output(wait=True) # clears only the logged output
# logging code
We can render to two separate stdouts within the same block. This works to an extent, as in the animation renders and the button isn't cleared. But still while the animation loop is running the button seems to be incapable of processing input. So it does seem like a synchronous code / event loop blocking problem. What's the issue here?
Do I need an alternative to sleep that frees up the event loop?
Edit 2:
After searching async code in Python, I learned about asyncio but I'm still struggling. Jupyter already runs the code via asyncio.run(), but the components obviously have to be defined as async for that to matter. I defined animate as async and tried using async sleeps, but the event loops still seems to be locked for the button.

Unable to exit tkinter app when using "wait_variable()"

I have a python code that includes tkinter window and other running tasks.
I've been trying to bind "WM_DELETE_WINDOW" event to a function that exits my python code when I close the window but can't achieve that.
This is what I try:
def on_exit():
root.destroy()
sys.exit()
root.protocol('WM_DELETE_WINDOW', on_exit)
The window is destroyed successfully but the python code doesn't exit. Any possible reason for sys.exit() not to work?
What am I doing wrong? any alternative approach should I try?
Doing some testing I figured out what can be the problem.
Here's a small code that summarizes my code which is much bigger.
import tkinter as tk
import sys
root = tk.Tk()
submitted = tk.IntVar()
def on_exit():
root.destroy()
sys.exit()
root.protocol('WM_DELETE_WINDOW', on_exit)
def submit():
submitted.set(1)
print("submitted")
button= tk.Button(root, text="Submit",command=submit)
button.pack()
button.wait_variable(submitted)
root.mainloop()
I believe now that wait_variable is the source of the problem.
And the code actually exits when I added submitted.set(1) to on_exit() ( or if I clicked the button first before closing the window ) but if I tried closing the window without pressing the button, the code won't exit.
So does this mean that wait_variable not only makes tkinter app wait, but also prevents python code exiting?!
I tried os._exit(1) and it worked, but I think it's not clean.

As your updated question points out the problem is wait_variable(). Going off the documentation for this method wait_variable() enters a local event loop that wont interrupt the mainloop however it appears that until that local event loop is terminated (the variable is updated in some way) it will prevent the python instance from terminating as there is still an active loop. So in order to prevent this you have also correctly pointed out you need to update this variable right before you terminate the tk instance.
This might seam odd but it is the behavior I would expect. It is my understanding that an active loop needs to be terminated before a python instance can exit.
As Bryan has pointed out in the comments the wait_variable() method is "a function which calls the vwait command inside the embedded tcl interpreter. This tcl interpreter knows nothing about python exceptions which is likely why it doesn't recognize the python exception raised by sys.exit()"
Link to relevant documentation:
wait_variable()
Relevant text from link:
wait_variable(name)
Waits for the given Tkinter variable to
change. This method enters a local event loop, so other parts of the
application will still be responsive. The local event loop is
terminated when the variable is updated (setting it to it’s current
value also counts).
You can also set the variable to whatever it is currently set as to terminate this event loop.
This line should work for you:
submitted.set(submitted.get())
That said you do not actually need sys.exit(). You can simply use root.destroy().
You new function should look like this:
def on_exit():
submitted.set(submitted.get())
root.destroy()
The python instance will automatically close if there is no more code after the mainloop.

Simple animation with Tkinter Python

I've searched for a simple animation code with Tkinter but I've found very different examples and I can't understand the correct way to write an animation.
Here my working code to display a simple moving circle:
import tkinter as tk
import time
root=tk.Tk()
canvas=tk.Canvas(root,width=400,height=400)
canvas.pack()
circle=canvas.create_oval(50,50,80,80,outline="white",fill="blue")
def redraw():
canvas.after(100,redraw)
canvas.move(circle,5,5)
canvas.update()
canvas.after(100,redraw)
root.mainloop()
In this code I can't correctly understand: how the after method works, where correctly put the update and the move method (before after method ?), is there another way to write an animation code? may you post me another example and comment the code please?
Thanks :)

Calling update
You should not call canvas.update(). As a general rule of thumb you should never call update. For a short essay on why, see this essay written by one of the original developers of the underlying tcl interpreter.
If you take out the call to canvas.update(), you have the proper way to do animation in a tkinter program.
Calling after to start the animation
You don't need to call after immediately before calling root.mainloop(). This works just as well:
...
redraw()
root.mainloop()
The choice to use or not use after in this specific case is dependent on if you want the animation to start immediately (possibly even before the widget is visible) or if you want it to happen after a short delay (possibly after the widget is made visible)
How after works
mainloop is nothing more than an infinite loop that checks the event queue for events. When it finds an event, it pops it off of the list and processes it. after is nothing more than making a request that says "in 100 ms, please add a new event to the queue". When the time limit expires, an event is added to the queue that says, in effect, "run this command". The next time the loop checks for events, it sees this event, pulls it off of the queue, and runs the command.
When you call after from within a method that itself was called by after, you're saying in effect "wait 100ms and do it again", creating an infinite loop. If you put the call to after before moving the object, you're saying "every 100ms run this function". If you put it after you're saying "run this function 100 ms after the last time it was run". The difference is very subtle and usually not perceptible unless your function takes a long time to run.

my code is:
from tkinter import *
import time
tk = Tk()
płótno = Canvas(tk, width=500, height=500)
płótno.pack()
płótno.create_polygon(10,10,10,70,70,10,fill="blue",outline="black")
for x in range(0,51):
płótno.move(1,5,0)
płótno.update()
rest(0.05)
płótno means canvas

How to pause and restart an animation in VTK

I am trying to do an animation program in VTK, in which I could make the VTK objects animate
but I failed to do pausing animation and restart animation, I came to know recently to stop the VTK timer but after if I click the animate button again to start then the program got crashed with following error, I have only one clue that the following line is crashing but this line will work at the first time of animation button click but fails on the second button click!!. The second function "animation_Stop" is my attempt of stopping the function by destroying the whole function, so I hoped I could avoid the program crash but it was also a failure!!
Error:
python.exe has stopped working
Error line:
self.renderWindowInteractor.SetRenderWindow(obj_renwin.renwin)
Please note my detailed code lines for animation and someone please help me to restart
and pause the animation in vtk python
def animation(self,obj_renwin,X):
if X==1:
print "start or restart animation"
self.renderWindowInteractor = vtk.vtkRenderWindowInteractor()
objRen=self.renderWindowInteractor.GetRenderWindow()
self.renderWindowInteractor.SetRenderWindow(obj_renwin.renwin)
obj_renwin.renwin.Render()
self.renderWindowInteractor.Initialize()
cb = vtkTimerCallback()
cb.actor = obj_renwin.actor
self.renderWindowInteractor.AddObserver('TimerEvent', cb.execute)
self.timerId = self.renderWindowInteractor.CreateRepeatingTimer(5);
if X==2:
print "stop animation"
self.renderWindowInteractor.DestroyTimer(self.timerId)
def animation_Stop(self,obj_renwin):
print "stop animation"
#self.animation(obj_renwin,1).destroy()
del (ConeRender.Cone.animation)

If you start the vtkTimer like that:
vtkSmartPointer<vtkTimerCallback> cb =
vtkSmartPointer<vtkTimerCallback>::New();
interactor->AddObserver(vtkCommand::TimerEvent, cb);
you may consider to stop/pause the timer with
vtkCommand::EndInteraction
i.e.
interactor->InvokeEvent(vtkCommand::TimerEvent, cb);
[It is just a moment thought, you can try it] ... :)

Is there a way to detach matplotlib plots so that the computation can continue?

After these instructions in the Python interpreter one gets a window with a plot:
from matplotlib.pyplot import *
plot([1,2,3])
show()
# other code
Unfortunately, I don't know how to continue to interactively explore the figure created by show() while the program does further calculations.
Is it possible at all? Sometimes calculations are long and it would help if they would proceed during examination of intermediate results.

Use matplotlib's calls that won't block:
Using draw():
from matplotlib.pyplot import plot, draw, show
plot([1,2,3])
draw()
print('continue computation')
# at the end call show to ensure window won't close.
show()
Using interactive mode:
from matplotlib.pyplot import plot, ion, show
ion() # enables interactive mode
plot([1,2,3]) # result shows immediatelly (implicit draw())
print('continue computation')
# at the end call show to ensure window won't close.
show()

Use the keyword 'block' to override the blocking behavior, e.g.
from matplotlib.pyplot import show, plot
plot(1)
show(block=False)
# your code
to continue your code.

It is better to always check with the library you are using if it supports usage in a non-blocking way.
But if you want a more generic solution, or if there is no other way, you can run anything that blocks in a separated process by using the multprocessing module included in python. Computation will continue:
from multiprocessing import Process
from matplotlib.pyplot import plot, show
def plot_graph(*args):
for data in args:
plot(data)
show()
p = Process(target=plot_graph, args=([1, 2, 3],))
p.start()
print 'yay'
print 'computation continues...'
print 'that rocks.'
print 'Now lets wait for the graph be closed to continue...:'
p.join()
That has the overhead of launching a new process, and is sometimes harder to debug on complex scenarios, so I'd prefer the other solution (using matplotlib's nonblocking API calls)

Try
import matplotlib.pyplot as plt
plt.plot([1,2,3])
plt.show(block=False)
# other code
# [...]
# Put
plt.show()
# at the very end of your script to make sure Python doesn't bail out
# before you finished examining.
The show() documentation says:
In non-interactive mode, display all figures and block until the figures have been closed; in interactive mode it has no effect unless figures were created prior to a change from non-interactive to interactive mode (not recommended). In that case it displays the figures but does not block.
A single experimental keyword argument, block, may be set to True or False to override the blocking behavior described above.

IMPORTANT: Just to make something clear. I assume that the commands are inside a .py script and the script is called using e.g. python script.py from the console.
A simple way that works for me is:
Use the block = False inside show : plt.show(block = False)
Use another show() at the end of the .py script.
Example of script.py file:
plt.imshow(*something*)
plt.colorbar()
plt.xlabel("true ")
plt.ylabel("predicted ")
plt.title(" the matrix")
# Add block = False
plt.show(block = False)
################################
# OTHER CALCULATIONS AND CODE HERE ! ! !
################################
# the next command is the last line of my script
plt.show()

You may want to read this document in matplotlib's documentation, titled:
Using matplotlib in a python shell

In my case, I wanted to have several windows pop up as they are being computed. For reference, this is the way:
from matplotlib.pyplot import draw, figure, show
f1, f2 = figure(), figure()
af1 = f1.add_subplot(111)
af2 = f2.add_subplot(111)
af1.plot([1,2,3])
af2.plot([6,5,4])
draw()
print 'continuing computation'
show()
PS. A quite useful guide to matplotlib's OO interface.

Well, I had great trouble figuring out the non-blocking commands... But finally, I managed to rework the "Cookbook/Matplotlib/Animations - Animating selected plot elements" example, so it works with threads (and passes data between threads either via global variables, or through a multiprocess Pipe) on Python 2.6.5 on Ubuntu 10.04.
The script can be found here: Animating_selected_plot_elements-thread.py - otherwise pasted below (with fewer comments) for reference:
import sys
import gtk, gobject
import matplotlib
matplotlib.use('GTKAgg')
import pylab as p
import numpy as nx
import time
import threading
ax = p.subplot(111)
canvas = ax.figure.canvas
# for profiling
tstart = time.time()
# create the initial line
x = nx.arange(0,2*nx.pi,0.01)
line, = ax.plot(x, nx.sin(x), animated=True)
# save the clean slate background -- everything but the animated line
# is drawn and saved in the pixel buffer background
background = canvas.copy_from_bbox(ax.bbox)
# just a plain global var to pass data (from main, to plot update thread)
global mypass
# http://docs.python.org/library/multiprocessing.html#pipes-and-queues
from multiprocessing import Pipe
global pipe1main, pipe1upd
pipe1main, pipe1upd = Pipe()
# the kind of processing we might want to do in a main() function,
# will now be done in a "main thread" - so it can run in
# parallel with gobject.idle_add(update_line)
def threadMainTest():
global mypass
global runthread
global pipe1main
print "tt"
interncount = 1
while runthread:
mypass += 1
if mypass > 100: # start "speeding up" animation, only after 100 counts have passed
interncount *= 1.03
pipe1main.send(interncount)
time.sleep(0.01)
return
# main plot / GUI update
def update_line(*args):
global mypass
global t0
global runthread
global pipe1upd
if not runthread:
return False
if pipe1upd.poll(): # check first if there is anything to receive
myinterncount = pipe1upd.recv()
update_line.cnt = mypass
# restore the clean slate background
canvas.restore_region(background)
# update the data
line.set_ydata(nx.sin(x+(update_line.cnt+myinterncount)/10.0))
# just draw the animated artist
ax.draw_artist(line)
# just redraw the axes rectangle
canvas.blit(ax.bbox)
if update_line.cnt>=500:
# print the timing info and quit
print 'FPS:' , update_line.cnt/(time.time()-tstart)
runthread=0
t0.join(1)
print "exiting"
sys.exit(0)
return True
global runthread
update_line.cnt = 0
mypass = 0
runthread=1
gobject.idle_add(update_line)
global t0
t0 = threading.Thread(target=threadMainTest)
t0.start()
# start the graphics update thread
p.show()
print "out" # will never print - show() blocks indefinitely!
Hope this helps someone,
Cheers!

In many cases it is more convenient til save the image as a .png file on the hard drive. Here is why:
Advantages:
You can open it, have a look at it and close it down any time in the process. This is particularly convenient when your application is running for a long
time.
Nothing pops up and you are not forced to have the windows open. This is particularly convenient when you are dealing with many figures.
Your image is accessible for later reference and is not lost when closing the figure window.
Drawback:
The only thing I can think of is that you will have to go and finder the folder and open the image yourself.

If you are working in console, i.e. IPython you could use plt.show(block=False) as pointed out in the other answers. But if you're lazy you could just type:
plt.show(0)
Which will be the same.

I had to also add plt.pause(0.001) to my code to really make it working inside a for loop (otherwise it would only show the first and last plot):
import matplotlib.pyplot as plt
plt.scatter([0], [1])
plt.draw()
plt.show(block=False)
for i in range(10):
plt.scatter([i], [i+1])
plt.draw()
plt.pause(0.001)

On my system show() does not block, although I wanted the script to wait for the user to interact with the graph (and collect data using 'pick_event' callbacks) before continuing.
In order to block execution until the plot window is closed, I used the following:
fig = plt.figure()
ax = fig.add_subplot(1,1,1)
ax.plot(x,y)
# set processing to continue when window closed
def onclose(event):
fig.canvas.stop_event_loop()
fig.canvas.mpl_connect('close_event', onclose)
fig.show() # this call does not block on my system
fig.canvas.start_event_loop_default() # block here until window closed
# continue with further processing, perhaps using result from callbacks
Note, however, that canvas.start_event_loop_default() produced the following warning:
C:\Python26\lib\site-packages\matplotlib\backend_bases.py:2051: DeprecationWarning: Using default event loop until function specific to this GUI is implemented
warnings.warn(str,DeprecationWarning)
although the script still ran.

I also wanted my plots to display run the rest of the code (and then keep on displaying) even if there is an error (I sometimes use plots for debugging). I coded up this little hack so that any plots inside this with statement behave as such.
This is probably a bit too non-standard and not advisable for production code. There is probably a lot of hidden "gotchas" in this code.
from contextlib import contextmanager
#contextmanager
def keep_plots_open(keep_show_open_on_exit=True, even_when_error=True):
'''
To continue excecuting code when plt.show() is called
and keep the plot on displaying before this contex manager exits
(even if an error caused the exit).
'''
import matplotlib.pyplot
show_original = matplotlib.pyplot.show
def show_replacement(*args, **kwargs):
kwargs['block'] = False
show_original(*args, **kwargs)
matplotlib.pyplot.show = show_replacement
pylab_exists = True
try:
import pylab
except ImportError:
pylab_exists = False
if pylab_exists:
pylab.show = show_replacement
try:
yield
except Exception, err:
if keep_show_open_on_exit and even_when_error:
print "*********************************************"
print "Error early edition while waiting for show():"
print "*********************************************"
import traceback
print traceback.format_exc()
show_original()
print "*********************************************"
raise
finally:
matplotlib.pyplot.show = show_original
if pylab_exists:
pylab.show = show_original
if keep_show_open_on_exit:
show_original()
# ***********************
# Running example
# ***********************
import pylab as pl
import time
if __name__ == '__main__':
with keep_plots_open():
pl.figure('a')
pl.plot([1,2,3], [4,5,6])
pl.plot([3,2,1], [4,5,6])
pl.show()
pl.figure('b')
pl.plot([1,2,3], [4,5,6])
pl.show()
time.sleep(1)
print '...'
time.sleep(1)
print '...'
time.sleep(1)
print '...'
this_will_surely_cause_an_error
If/when I implement a proper "keep the plots open (even if an error occurs) and allow new plots to be shown", I would want the script to properly exit if no user interference tells it otherwise (for batch execution purposes).
I may use something like a time-out-question "End of script! \nPress p if you want the plotting output to be paused (you have 5 seconds): " from https://stackoverflow.com/questions/26704840/corner-cases-for-my-wait-for-user-input-interruption-implementation.

plt.figure(1)
plt.imshow(your_first_image)
plt.figure(2)
plt.imshow(your_second_image)
plt.show(block=False) # That's important
raw_input("Press ENTER to exist") # Useful when you run your Python script from the terminal and you want to hold the running to see your figures until you press Enter

The OP asks about detatching matplotlib plots. Most answers assume command execution from within a python interpreter. The use-case presented here is my preference for testing code in a terminal (e.g. bash) where a file.py is run and you want the plot(s) to come up but the python script to complete and return to a command prompt.
This stand-alone file uses multiprocessing to launch a separate process for plotting data with matplotlib. The main thread exits using the os._exit(1) mentioned in this post. The os._exit() forces main to exit but leaves the matplotlib child process alive and responsive until the plot window is closed. It's a separate process entirely.
This approach is a bit like a Matlab development session with figure windows that come up with a responsive command prompt. With this approach, you have lost all contact with the figure window process, but, that's ok for development and debugging. Just close the window and keep testing.
multiprocessing is designed for python-only code execution which makes it perhaps better suited than subprocess. multiprocessing is cross-platform so this should work well in Windows or Mac with little or no adjustment. There is no need to check the underlying operating system. This was tested on linux, Ubuntu 18.04LTS.
#!/usr/bin/python3
import time
import multiprocessing
import os
def plot_graph(data):
from matplotlib.pyplot import plot, draw, show
print("entered plot_graph()")
plot(data)
show() # this will block and remain a viable process as long as the figure window is open
print("exiting plot_graph() process")
if __name__ == "__main__":
print("starting __main__")
multiprocessing.Process(target=plot_graph, args=([1, 2, 3],)).start()
time.sleep(5)
print("exiting main")
os._exit(0) # this exits immediately with no cleanup or buffer flushing
Running file.py brings up a figure window, then __main__ exits but the multiprocessing + matplotlib figure window remains responsive with zoom, pan, and other buttons because it is an independent process.
Check the processes at the bash command prompt with:
ps ax|grep -v grep |grep file.py

In my opinion, the answers in this thread provide methods which don't work for every systems and in more complex situations like animations. I suggest to have a look at the answer of MiKTeX in the following thread, where a robust method has been found:
How to wait until matplotlib animation ends?

Here is the simplest solution I found (thread blocking code)
plt.show(block=False) # this avoids blocking your thread
plt.pause(1) # comment this if you do not want a time delay
# do more stuff
plt.show(block=True) # this prevents the window from closing on you

If you want to open multiple figures, while keeping them all opened, this code worked for me:
show(block=False)
draw()

While not directly answering OPs request, Im posting this workaround since it may help somebody in this situation:
Im creating an .exe with pyinstaller since I cannot install python where I need to generate the plots, so I need the python script to generate the plot, save it as .png, close it and continue with the next, implemented as several plots in a loop or using a function.
for this Im using:
import matplotlib.pyplot as plt
#code generating the plot in a loop or function
#saving the plot
plt.savefig(var+'_plot.png',bbox_inches='tight', dpi=250)
#you can allways reopen the plot using
os.system(var+'_plot.png') # unfortunately .png allows no interaction.
#the following avoids plot blocking the execution while in non-interactive mode
plt.show(block=False)
#and the following closes the plot while next iteration will generate new instance.
plt.close()
Where "var" identifies the plot in the loop so it wont be overwritten.

What I have found as the best solution so the program does not wait for you to close the figure and have all your plots together so you can examine them side by side is to show all the plots at the end.
But this way you cannot examine the plots while program is running.
# stuff
numFig = 1
plt.figure(numFig)
numFig += 1
plt.plot(x1, y1)
# other stuff
plt.figure(numFig)
numFig += 1
plt.plot(x2, y2)
# more stuff
plt.show()

Use plt.show(block=False), and at the end of your script call plt.show().
This will ensure that the window won't be closed when the script is finished.