考虑到这段代码需要在乌藨子上运行.
使用***键盘模块***可能不是一个好主意，因为在Linux上，程序的执行可能会突然要求root访问权限，以便与设备驱动程序交互并侦听键盘事件。
要使命令“sudo python my_script.py”正常工作，可能需要进一步调整。
因此，这里是我的简单解决方案，利用线程和std input（）函数，以避免使用键盘库和关闭Python程序。
正如@jared所建议的那样，最好选择一个任意的键来实现所需的结果，而不是任何键。

片段：

import matplotlib.pyplot as plt
import threading

# Empty data for now
x_data = []
y_data = []

############## Plotting 
# Enable the interactive mode for plotting
plt.ion() 
# The comma after "graph" is fundamental to unpack the first element of the returned tuple 
# and assigning it to the variable graph.
# (Without the comma, graph would be assigned to the entire tuple)
graph, = plt.plot(x_data, y_data)

def keyboard_input():
    """ It is not necessary to specify the 'q' key, 
    since input() waits for the user to enter a value and press enter,  
    but if the value entered is "q", the function return and then the subsequent code is executed.
    """
    input()
    plt.close()

# Start the thread that execute the keyboard input method
thread = threading.Thread(target=keyboard_input)
thread.start()

# Display the plot. It is constantly updated in real-time as new data can be, someway, added to x_data and y_data
while True:
    graph.set_ydata(y_data)
    graph.set_xdata(x_data)
    plt.draw()
    plt.pause(0.01)
    # Check if the plot window (with the specified figure number 1, the default value) has been closed, 
    # Otherwise, waits for the thread object to complete the join() method and exit
    if not plt.fignum_exists(1):
        thread.join()
        break

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在任何按键下退出可能会有点烦人。我建议选择一个特定的键来退出。在下面的例子中，我使用"q"键来“退出”。使用keyboard python library（您可能需要安装）完成检测。

import matplotlib.pyplot as plt
import numpy as np
import keyboard

plt.close("all")

plt.ion()

rng = np.random.default_rng(42)
N = 10
x_data = np.linspace(0, 5, N)
y_data = rng.random(N)

line, = plt.plot(x_data, y_data)

while True:
    if keyboard.is_pressed("q"):
        break    
    
    line.set_ydata(rng.random(N))
    plt.draw()
    plt.pause(0.01)

print("End of code")

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