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!

使用matplotlib调用不会阻塞:

使用画():

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()

使用交互模式:

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()

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。

虽然没有直接回答OPs的请求，但我发布了这个变通方法，因为它可能会帮助一些人在这种情况下:

我用pyinstaller创建了一个.exe，因为我不能在我需要生成图形的地方安装python，所以我需要python脚本来生成图形，将其保存为.png，关闭它并继续下一个，在循环中实现为几个图形或使用一个函数。

为此，我使用:

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() 

其中“var”标识循环中的情节，因此它不会被覆盖。

重要提示:只是为了澄清一些事情。我假设命令在.py脚本中，脚本是使用python script.py从控制台调用的。

对我来说，一个简单的方法是:

使用block = False在里面显示:plt。show(block = False) 在.py脚本的末尾使用另一个show()。

script.py文件示例:

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()

最好总是检查您正在使用的库是否支持以非阻塞的方式使用。

但是如果你想要一个更通用的解决方案，或者如果没有其他方法，你可以通过使用python中包含的多处理模块来运行任何在分离进程中阻塞的东西。计算将继续:

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()

这有启动新进程的开销，而且在复杂的场景下有时更难调试，因此我更喜欢其他解决方案(使用matplotlib的非阻塞API调用)