Python 多处理并行运行2个任务-使用Bluepy在Raspberry Pi上进行项目

我正在进行一个项目，其目标是使用蓝牙低能量将几个nano 33 iot Arduinos（作为外围设备）连接到一个Raspberry pi（作为中心设备）

我的Arduinos不断向Raspberry Pi发送数据。我正在使用“侦听通知”功能实时截获所有数据

我的代码如下所示：

from bluepy import btle

class MyDelegate(btle.DefaultDelegate):
    def __init__(self,params):
        btle.DefaultDelegate.__init__(self)

    def handleNotification(self,cHandle,data):
        print(data)


def connectionArduino1() :
    global Arduino1
    print("trying to connect to Arduino1")
    connected = False
    while(connected == False) :
        try :
            Arduino1 = btle.Peripheral("xx:xx:xx:xx:xx:xx")
            Arduino1.setDelegate(MyDelegate(0))

            servicesArduino1=Arduino1.getServices()

            sArduino1 = Arduino1.getServiceByUUID(UUID)
            chArduino1 = sArduino1.getCharacteristics()[0]

            Arduino1.writeCharacteristic(14,bytes('Start','utf-8'))
            Arduino1.writeCharacteristic(chArduino1.valHandle+1, (1).to_bytes(2, byteorder='little'))
            connected = True
        except :
            connected = False


def connectionArduino2() :
    global Arduino2
    print("trying to connect to Arduino2")
    connected = False
    while(connected == False) :
        try :
            Arduino2 = btle.Peripheral("xx:xx:xx:xx:xx:xx")
            Arduino2.setDelegate(MyDelegate(0))

            servicesArduino2=Arduino2.getServices()

            sArduino2 = Arduino2.getServiceByUUID(UUID)
            chArduino2 = sArduino2.getCharacteristics()[0]

            Arduino2.writeCharacteristic(14,bytes('Start','utf-8'))
            Arduino2.writeCharacteristic(chArduino2.valHandle+1, (1).to_bytes(2, byteorder='little'))
            connected = True
        except :
            connected = False


def mainLoop() :
    while True:

        try :
            if Arduino1.waitForNotifications(5) :
                print("Notification received")
            else :
                try :
                    Arduino1.disconnect()
                    connectionArduino1()
                except :
                    connectionArduino1()
        except :
            connectionArduino1()

        try :
            if Arduino2.waitForNotifications(5) :
                print("Notification received")
            else :
                try :
                    Arduino2.disconnect()
                    connectionArduino2()
                except :
                    connectionArduino2()
        except :
            connectionArduino2()

connectionArduino1()
connectionArduino2()
mainLoop()

main循环中的
try
命令：

确保如果其中一个Arduinos断开，程序不会崩溃
如果Arduino断开连接，则自动将Raspberry Pi重新连接到Arduino（通过启动
连接Arduinox
功能）
问题是当两个Arduinos中的一个断开时：当程序尝试重新连接到它时，它不再能够读取从其他Arduinos接收到的通知。因此，
mainLoop
功能暂停

我想创建一个健壮的程序，该程序能够读取连接的Arduinos接收到的所有通知，并且能够并行地重新连接到Arduinos（如果其中一个断开连接），以便
mainLoop
功能永远不会暂停

我尝试按如下方式创建线程：

from bluepy import btle
from threading import Thread

class MyDelegate(btle.DefaultDelegate):
    def __init__(self,params):
        btle.DefaultDelegate.__init__(self)

    def handleNotification(self,cHandle,data):
        print(data)

Arduino1Connected = False
Arduino2Connected = False

def connectionArduino1() :
    global Arduino1
    global Arduino1Connected
    print("trying to connect to Arduino1")
    connected = False
    while(Arduino1Connected == False) :
        try :
            Arduino1 = btle.Peripheral("xx:xx:xx:xx:xx:xx")
            Arduino1.setDelegate(MyDelegate(0))
            
            servicesArduino1=Arduino1.getServices()
            
            sArduino1 = Arduino1.getServiceByUUID(UUID)
            chArduino1 = sArduino1.getCharacteristics()[0]

            Arduino1.writeCharacteristic(14,bytes('Start','utf-8'))
            Arduino1.writeCharacteristic(chArduino1.valHandle+1, (1).to_bytes(2, byteorder='little'))
            Arduino1Connected = True
        except :
            Arduino1Connected = False
            
            
def connectionArduino2() : 
    global Arduino2
    global Arduino2Connected
    print("trying to connect to Arduino2")
    connected = False
    while(Arduino2Connected == False) :
        try :
            Arduino2 = btle.Peripheral("xx:xx:xx:xx:xx:xx")
            Arduino2.setDelegate(MyDelegate(0))
            
            servicesArduino2=Arduino2.getServices()
            
            sArduino2 = Arduino2.getServiceByUUID(UUID)
            chArduino2 = sArduino2.getCharacteristics()[0]

            Arduino2.writeCharacteristic(14,bytes('Start','utf-8'))
            Arduino2.writeCharacteristic(chArduino2.valHandle+1, (1).to_bytes(2, byteorder='little'))
            Arduino2Connected = True
        except :
            Arduino2Connected = False
            
            
def mainLoop() : 
    while True:
    
        try : 
            if Arduino1.waitForNotifications(5) :
                print("Notification received")
            else :
                try :
                    Arduino1.disconnect()
                    Arduino1Connected = False
                except :
                    Arduino1Connected = False
        except :
            Arduino1Connected = False

        try : 
            if Arduino2.waitForNotifications(5) :
                print("Notification received")
            else :
                try :
                    Arduino2.disconnect()
                    Arduino2Connected = False
                except :
                    Arduino2Connected = False
        except :
            Arduino2Connected = False

thread1 = Thread(target = mainLoop)
thread2 = Thread(target = connectionArduino1)
thread3 = Thread(target = connectionArduino2)

thread1.start()
thread2.start()
thread3.start()

thread1.join()
thread2.join()
thread3.join()

但这不起作用，我认为我的代码要慢得多

所以我考虑使用线程或多处理原理。具有读取从连接的Arduinos接收到的所有通知的
mainLoop
进程，以及能够检测其中一个Arduinos可能断开连接并能够自动重新连接的进程

你知道我怎样才能做到这一点吗

或者您是否有其他解决方案允许连续读取Arduinos接收到的所有通知，并并行检测断开连接并直接重新连接？
我想我应该创建一个
队列
，然后为每个通过队列的Arduino客户端启动一个线程。然后，线程处理其客户机，重新连接并将接收到的任何消息放入队列。主程序可以使用
select
在任一队列上等待消息，或者更简单的是，检查每个队列是否有消息，而不是阻止等待消息。经过几次测试后，它终于工作了，谢谢您的回答