Python 多进程无法腌制 opencv 视频捕获对象

Question

我正在尝试创建一个独立的进程来处理从相机获取的图像。但多处理似乎很难从opencv中提取视频捕获模块。有人能提出一个变通办法吗？我使用的是python 3.7.1

from multiprocessing import Process
import multiprocessing as mp
import time
import logging
import logging.handlers
import sys
import logging
from enum import Enum
import cv2

class Logger():
    @property
    def logger(self):
        component = "{}.{}".format(type(self).__module__, type(self).__name__)
        #default log handler to dump output to console
        return logging.getLogger(component)

class MyProcess(Logger):
    def __init__(self, ):
        self.process = Process(target = self.run, args=())
        self.exit = mp.Event()
        self.logger.info("initialize class")
        self.capture = cv2.VideoCapture(0)
    def run(self):
        while not self.exit.is_set():
            pass
        print("You exited!")
    def shutdown(self):
        print("Shutdown initiated")
        self.exit.set()

if __name__ == "__main__":
    p = MyProcess()
    p.process.start()
    print( "Waiting for a while")
    time.sleep(3)
    p.shutdown()
    time.sleep(3)
    print("Child process state: {}".format(p.process.is_alive())) 

返回的错误特定于它的视频捕获对象不能被pickle。

文件"C:\Program Files(x86(\Microsoft Visual Studio\Shared\Anaconda3_64\lib\multiprocessing\reduction.py"，第60行，转储ForkingPickler(文件，协议(.dump(obj(

TypeError：无法pickle cv2.VideoCapture对象

Answer 1

我不是专家，但我遇到了同样的问题，我用这种方式解决了。

而不是在初始功能中使用cv2.VideoCapture(0(

 def __init__(self, ):
    self.process = Process(target = self.run, args=())
    self.exit = mp.Event()
    self.logger.info("initialize class")
    self.capture = cv2.VideoCapture(0)

我将初始化移动到运行功能

 def run(self):
    self.capture = cv2.VideoCapture(0)
    while not self.exit.is_set():

它对我有效。也许它对你也有帮助