你好，我第一次发帖，我在python上遇到了一个问题。所以我有这段代码，它是一个键绑定，就像当我按下数字1时，我需要execute1（），但在运行之后，我不希望程序退出，而只是在后台等待和运行，等待另一个键被按下。这个程序不断循环execute（）。我该如何解决这个问题？

import threading
 import pyautogui 
 import time
 import pyscreeze
 from pynput import keyboard

 cmb = [{keyboard.Key.alt_l, keyboard.Key.page_up}] #keybind 
 current = set()

def execute1():
   pyautogui.moveTo(2459,122)
   pyautogui.leftClick()
   pyautogui.moveTo(2368,392)
   pyautogui.leftClick()
   pyautogui.moveTo(1542,447)
   pyautogui.leftClick()
   pyautogui.moveTo(1546,518)
   pyautogui.leftClick()
   pyautogui.moveTo(1649,635)
   pyautogui.leftClick()
   pyautogui.moveTo(2018,320)
   time.sleep(1.35)
   pyautogui.leftClick()
   pyautogui.typewrite("111")
   pyautogui.moveTo(2006,379)
   pyautogui.leftClick()

def execute2():
   pyautogui.moveTo(2459,122)
   pyautogui.leftClick()
   pyautogui.moveTo(2271,505)
   pyautogui.leftClick()
   pyautogui.moveTo(1542,447)
   pyautogui.leftClick()
   pyautogui.moveTo(1546,518)
   pyautogui.leftClick()
   pyautogui.moveTo(1649,635)
   pyautogui.leftClick()
   pyautogui.moveTo(2018,320)
   time.sleep(1.35)
   pyautogui.leftClick()
   pyautogui.typewrite("111")
   pyautogui.moveTo(2006,379)
   pyautogui.leftClick()

def execute3():
   pyautogui.moveTo(2459,122)
   pyautogui.leftClick()
   pyautogui.moveTo(2292,603)
   pyautogui.leftClick()
   pyautogui.moveTo(1542,447)
   pyautogui.leftClick()
   pyautogui.moveTo(1546,518)
   pyautogui.leftClick()
   pyautogui.moveTo(1649,635)
   pyautogui.leftClick()
   pyautogui.moveTo(2018,320)
   time.sleep(1.35)
   pyautogui.leftClick()
   pyautogui.typewrite("111")
   pyautogui.moveTo(2006,379)
   pyautogui.leftClick()

def on_press(key): #check when key presses
    if key == keyboard.KeyCode(char='1'): # check if '1' key is pressed
       execute1()
       current.add(key)
    elif key == keyboard.KeyCode(char='2'): # check if '2' key is pressed
        execute2()
        current.add(key)
    elif key == keyboard.KeyCode(char='3'): # check if '3' key is pressed
        execute3()
        current.add(key)

def on_release(key): #check when key releases
    if any([key in z for z in cmb]):
      current.remove(key)

with keyboard.Listener(on_press=on_press, on_release=on_release) as listener:
        listener.join()

我试着这么做：

def on_press(key): #check when key presses
    if key == keyboard.KeyCode(char='1'): # check if '1' key is pressed
      t = threading.Thread(target=execute1)
      t.start()
      t.join()
      current.add(key)

对于所有的执行，但有另一个问题，当它到达pyautogui.typewrite("111")它只是键入了其中一个字符，但我想要快。
这个程序工作时，我只有1个执行，但我想他们现在6，我找不到一个解决方案