Python 两个函数线程

Question

I want to thread two functions, first function streaming a video and passing frames to second function, second function reading frames with Optical Character Recognition and converting frames to text. The question how to pass frames from first threaded function to second threaded function?

What I have done already, with first function saving video frames to local file 'frame.jpg' and at the same time reading with second function from 'frame.jpg'. Is it possible to define video frames as global variable and pass to reading function?

import cv2
import pytesseract
from multiprocessing import Process

def video_straming():       #Video streaming function, First Function
    vc = cv2.VideoCapture(0)
    cv2.namedWindow("preview")

    if vc.isOpened():
        rval, frame = vc.read()
    else:
        rval = False

    while rval:
        rval, frame = vc.read()
        cv2.imwrite('frame.jpg',frame)
        key = cv2.waitKey(20)
        if key == 27: # exit on ESC
            break
    cv2.destroyWindow("preview")

def reading():  #Reading from frame.jpg function, Second Function
    while:
        frame = cv2.imread('frame.jpg')
        read = Image.fromarray(frame)
        read = pytesseract.image_to_string(read)
        if len(read) > 80:
            break

if __name__ == '__main__':
    video_stream = Process(target=video_streaming)
    video_stream.start()
    frame_read = Process(target=reading)
    frame_read.start()
    video_stream.join()
    frame_read.join()

Answer 1

Hope this answer can still be of some use.

I use multiprocessing.Pipe() to pass video frames from one processes to another with cv2.VideoCapture() to capture frames and write each image to the Pipe.

import multiprocessing

multiprocessing.set_start_method('spawn')

video_outfrompipe, video_intopipe = multiprocessing.Pipe()

vs = multiprocessing.Process(target=VideoSource, args=(video_intopipe))

vs.start()

vc = multiprocessing.Process(target=VideoConsumer, args=(video_outfrompipe))

vc.start()

vs.join()

vc.join()