This is a MWE of what I'm after, adapted from this question:
from matplotlib.pyplot import plot, draw, show
def make_plot():
plot([1,2,3])
draw()
print 'continue computation'
print('Do something before plotting.')
# Now display plot in a window
make_plot()
answer = raw_input('Back to main and window visible? ')
if answer == 'y':
print('Excellent')
else:
print('Nope')
show()
What I want is: I call the function to make the plot, the plot window appears, and then I get to go back to the prompt so I can input some value (based on that image that just displayed) and carry on with the code (the window can then close or remain there, I don't care).
What I get instead is that the window with the plot only appears after the code is completed, which is no good.
I've tried the following with the same results, the plot window appears at the end of the code and not before:
from matplotlib.pyplot import plot, ion, draw
ion() # enables interactive mode
plot([1,2,3]) # result shows immediately (implicit draw())
# at the end call show to ensure window won't close.
draw()
answer = raw_input('Back to main and window visible? ')
if answer == 'y':
print('Excellent')
else:
print('Nope')
The same happens if I change draw() for show().
I've tried the following approach:
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 'computation continues...'
print 'Now lets wait for the graph be closed to continue...:'
p.join()
which results in a Python kernel has crashed error in Canopy with the message:
The kernel (user Python environment) has terminated with error code -6. This may be due to a bug in your code or in the kernel itself.
Output captured from the kernel process is shown below.
[IPKernelApp] To connect another client to this kernel, use:
[IPKernelApp] --existing /tmp/tmp9cshhw.json
QGtkStyle could not resolve GTK. Make sure you have installed the proper libraries.
[xcb] Unknown sequence number while processing queue
[xcb] Most likely this is a multi-threaded client and XInitThreads has not been called
[xcb] Aborting, sorry about that.
python: ../../src/xcb_io.c:274: poll_for_event: La declaración `!xcb_xlib_threads_sequence_lost' no se cumple.
I should mention I'm running Canopy in elementary OS which is based in Ubuntu 12.04.
Also tried solution posted in this question:
import numpy
from matplotlib import pyplot as plt
if __name__ == '__main__':
x = [1, 2, 3]
plt.ion() # turn on interactive mode
for loop in range(0,3):
y = numpy.dot(x, loop)
plt.figure()
plt.plot(x,y)
plt.show()
_ = raw_input("Press [enter] to continue.")
This displays empty plot windows as the code advances (ie: the user hits [enter]) and only displays the images after the code is finished.
This solution (also in the same question) doesn't even display the plot windows:
import numpy
from matplotlib import pyplot as plt
if __name__ == '__main__':
x = [1, 2, 3]
plt.ion() # turn on interactive mode, non-blocking `show`
for loop in range(0,3):
y = numpy.dot(x, loop)
plt.figure() # create a new figure
plt.plot(x,y) # plot the figure
plt.show() # show the figure, non-blocking
_ = raw_input("Press [enter] to continue.") # wait for input from the user
plt.close() # close the figure to show the next one.
You may use plt.show(block=False), which gets rid of the blocking directly.
For your example, this could read
from matplotlib.pyplot import plot, show
def make_plot():
plot([1,2,3])
show(block=False)
print('continue computation')
print('Do something before plotting.')
# Now display plot in a window
make_plot()
answer = input('Back to main and window visible? ')
if answer == 'y':
print('Excellent')
else:
print('Nope')