您可能需要阅读matplotlib文档中的这个文档，标题为:

在python shell中使用matplotlib

在我的系统上，show()没有阻塞，尽管我希望脚本在继续之前等待用户与图形交互(并使用'pick_event'回调来收集数据)。

为了阻止执行直到绘图窗口关闭，我使用了以下方法:

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

但是请注意，canvas.start_event_loop_default()产生了以下警告:

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)

尽管脚本仍然在运行。

嗯，我在搞清楚非阻塞命令方面遇到了很大的困难……但最后，我成功地重做了“Cookbook/Matplotlib/Animations -动画选定的绘图元素”的例子，所以它可以在Ubuntu 10.04的Python 2.6.5上与线程一起工作(并通过全局变量或多进程管道在线程之间传递数据)。

脚本可以在这里找到:Animating_selected_plot_elements-thread.py -否则粘贴在下面(注释更少)以供参考:

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! 

希望这能帮助到一些人， 干杯!

OP询问分离matplotlib图。大多数回答假设从python解释器中执行命令。这里给出的用例是我对在终端(例如bash)中测试代码的偏好，在终端中运行file.py，并且您希望出现绘图，但python脚本完成并返回命令提示符。

这个独立文件使用多处理启动一个单独的进程，用matplotlib绘制数据。主线程使用本文中提到的os._exit(1)退出。os._exit()强制main退出，但在plot窗口关闭之前，matplotlib子进程仍然活跃并保持响应。这是一个完全独立的过程。

这种方法有点像带有图形窗口的Matlab开发会话，会产生响应式命令提示符。使用这种方法，您将失去与图形窗口进程的所有联系，但是，这对于开发和调试来说是可以的。只需关闭窗口并继续测试。

多处理是专为python代码执行而设计的，这使得它可能比子进程更适合。multiprocessing是跨平台的，所以这应该在Windows或Mac上工作得很好，很少或没有调整。不需要检查底层操作系统。这是在linux Ubuntu 18.04LTS上测试的。

#!/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

运行file.py会弹出一个图形窗口，然后__main__退出，但是multiprocessing + matplotlib图形窗口仍然对缩放、平移和其他按钮有响应，因为它是一个独立的进程。

在bash命令提示符下检查进程:

Ps ax|grep -v grep |grep file.py

在我的例子中，我希望在计算窗口时弹出几个窗口。作为参考，方法如下:

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()

一个非常有用的matplotlib的OO接口指南。

我还希望我的图显示运行其余的代码(然后继续显示)，即使出现错误(我有时使用图进行调试)。我编写了这个小代码，让这个with语句中的任何plot都像这样。

这可能有点太非标准了，不适合用于生产代码。这段代码中可能有很多隐藏的“陷阱”。

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

如果/当我实现了一个适当的“保持图打开(即使发生错误)并允许显示新的图”，我希望脚本在没有用户干扰的情况下正确退出(用于批处理执行)。

我可能会使用超时问题“脚本结束!”\nPress p如果你想要绘图输出暂停(你有5秒):" from https://stackoverflow.com/questions/26704840/corner-cases-for-my-wait-for-user-input-interruption-implementation。