我使用python和OpenCV从RTSP流得到的视频。我从流中获得单帧并将其保存到文件系统。
我写了处理框架获取和保存StreamingWorker
。此外,还有具有所有流对象StreamPool
。我认为,作为StreamingWorker
总是运行,应该只有每个核心一个,以便采取尽可能多地。然后StreamPool
将提供VideoCapture
对象可用StreamingWorker
。
问题是,大多数的脚本运行,阻塞时间:
import os
import time
import threading
import cv2 as cv
class StreamingWorker(object):
def __init__(self, stream_pool):
self.stream_pool = stream_pool
self.start_loop()
def start_loop(self):
while True:
try:
# getting a stream from the read_strategy
stream_object = self.stream_pool.next()
# getting an image from the stream
_, frame = stream_object['stream'].read()
# saving image to file system
cv.imwrite(os.path.join('result', stream_object['feed'], '{}.jpg'.format(time.time())))
except ValueError as e:
print('[error] {}'.format(e))
class StreamPool(object):
def __init__(self, streams):
self.streams = [{'feed': stream, 'stream': cv.VideoCapture(stream)} for stream in streams]
self.current_stream = 0
self.lock = threading.RLock()
def next(self):
self.lock.acquire()
if(self.current_stream + 1 >= len(self.streams)):
self.current_stream = 0
else:
self.current_stream += 1
result = self.streams[self.current_stream]
self.lock.release()
return result
def get_cores():
# This function returns the number of available cores
import multiprocessing
return multiprocessing.cpu_count()
def start(stream_pool):
StreamingWorker(stream_pool)
def divide_list(input_list, amount):
# This function divides the whole list into list of lists
result = [[] for _ in range(amount)]
for i in range(len(input_list)):
result[i % len(result)].append(input_list[i])
return result
if __name__ == '__main__':
stream_list = ['rtsp://some/stream1', 'rtsp://some/stream2', 'rtsp://some/stream3']
num_cores = get_cores()
divided_streams = divide_list(stream_list, num_cores)
for streams in divided_streams:
stream_pool = StreamPool(streams)
thread = threading.Thread(target=start, args=(stream_pool))
thread.start()
当我想到这一点,我并没有考虑到大部分的操作会阻塞操作,如:
# Getting a frame blocks
_, frame = stream_object['stream'].read()
# Writing to the file system blocks
cv.imwrite(os.path.join('result', stream_object['feed'], '{}.jpg'.format(time.time())))
与花费太多时间阻塞问题是,大部分的处理能力被浪费了。我想用期货与ThreadPoolExecutor
的,但我似乎无法达到我的使用处理核心可能的最大数量的目标。也许我没有设置enaugh线程。
是否有处理阻塞操作,以使核处理能力的最佳使用一个标准的方式?我没事具有语言无关的答案。
我结束了使用使用ThreadPoolExecutor
功能add_done_callback(fn)
。
class StreamingWorker(object):
def __init__(self, stream_pool):
self.stream_pool = stream_pool
self.thread_pool = ThreadPoolExecutor(10)
self.start_loop()
def start_loop(self):
def done(fn):
print('[info] future done')
def save_image(stream):
# getting an image from the stream
_, frame = stream['stream'].read()
# saving image to file system
cv.imwrite(os.path.join('result', stream['feed'], '{}.jpg'.format(time.time())))
while True:
try:
# getting a stream from the read_strategy
stream_object = self.stream_pool.next()
# Scheduling the process to the thread pool
self.thread_pool.submit(save_image, (stream_object)).add_done_callback(done)
except ValueError as e:
print('[error] {}'.format(e))
我其实没有什么都想做未来结束后,但如果我用result()
那么while True
将停止,这也击败对子级使用线程池的所有目的。
附注:我有打电话threading.Rlock()
时添加一个self.stream_pool.next()
因为很明显的OpenCV不能处理从多个线程调用。