Matplotlib - 强制绘图显示然后返回主代码

Question

这是我所追求的MWE,改编自这个问题:

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()

我想要的是:我调用函数来创建绘图,出现绘图窗口,然后我回到提示符,这样我就可以输入一些值(基于刚才显示的图像)并继续执行代码(窗口可以关闭或保持在那里,我不在乎).

我得到的是,带有绘图的窗口仅在代码完成后出现,这是不好的.

---

加1

我尝试了以下相同的结果,绘图窗口出现在代码的末尾而不是之前:

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')

如果我改变draw(),也会发生同样的情况show().

---

加2

我尝试过以下方法:

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()

这导致消息Python kernel has crashed出错Canopy:

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.

我要提到我跑Canopy在elementary OS总部设在Ubuntu 12.04.

---

加3

也试过这个问题的解决方案:

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.")

当代码前进时(即:用户点击[enter]),这将显示空的绘图窗口,并且仅在代码完成后显示图像.

此解决方案(也在同一问题中)甚至不显示绘图窗口:

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.

Answer 1

您可以使用plt.show(block=False),直接摆脱阻止.

对于你的例子,这可以阅读

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')

Answer 2

所提出的解决方案都不适合我.我用三个不同的IDE PyCharm,Spyder和Pyzo测试它们,使用Python 3.6下的(当前)最新的Matplotlib 2.1.

虽然不是最佳的,但对我有用的是使用plt.pause命令:

import matplotlib.pyplot as plt

def make_plot():
    plt.plot([1, 2, 3])
#    plt.show(block=False)  # The plot does not appear.
#    plt.draw()             # The plot does not appear.
    plt.pause(0.1)          # The plot properly appears.
    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')