ano*_*its 6 python multithreading opencv
我一直在 python 中使用 cv2 和多处理,我终于有了一个工作脚本,一旦各个帧已经在输入队列中,它就会对它们进行处理。但是,我想首先通过使用多个核心来加快将帧放入队列的速度,因此我尝试使用相同的多处理方法将每个图像读入队列。但我似乎无法让它发挥作用,我也不知道为什么。我想也许是因为我试图写入一个队列,所以我将它们分开,但现在我想知道这是否是因为我试图同时读取同一个视频文件。
这是我希望用伪代码完成的事情:
for process in range(processCount):
start a process that does this:
for frame in range(startFrame,endFrame):
set next frame to startFrame
read frame
add frame to queue
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这是我当前的代码。我尝试过使用池和单独的进程,但现在我坚持使用单独的进程,因为我不确定问题是否出在队列管理上。如果我手动调用 getFrame,我会将正确的内容放入队列中,所以我认为该函数本身可以正常工作。
我确信我正在做一些非常愚蠢(或者非常奇怪)的事情。有人可以提出解决方案吗?如果也只有一个队列就太好了……我只有两个队列来尝试解决问题。
提前致谢。
import numpy as np
import cv2
import multiprocessing as mp
import time
def getFrame(queue, startFrame, endFrame):
for frame in range(startFrame, endFrame):
cap.set(1,frame)
frameNo = int(cap.get(0))
ret, frame = cap.read()
queue.put((frameNo,frame))
file = 'video.mov'
cap = cv2.VideoCapture(file)
fileLen = int(cap.get(7))
# get cpuCount for processCount
processCount = mp.cpu_count()/3
inQ1 = mp.JoinableQueue() # not sure if this is right queue type, but I also tried mp.Queue()
inQ2 = mp.JoinableQueue()
qList = [inQ1,inQ2]
# set up bunches
bunches = []
for startFrame in range(0,fileLen,fileLen/processCount):
endFrame = startFrame + fileLen/processCount
bunches.append((startFrame,endFrame))
getFrames = []
for i in range(processCount):
getFrames.append(mp.Process(target=getFrame, args=(qList[i], bunches[i][0],bunches[i][1],)))
for process in getFrames:
process.start()
results1 = [inQ1.get() for p in range(bunches[0][0],bunches[0][1])]
results2 = [inQ2.get() for p in range(bunches[1][0],bunches[1][1])]
inQ1.close()
inQ2.close()
cap.release()
for process in getFrames:
process.terminate()
process.join()
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代码中确实有一个错误:\xc2\xa0VideoCapture跨进程使用同一个对象。显然文件中当前读取的位置存在冲突。
话虽这么说,当尝试为每个进程实例化一个 VideoCapture 时,我的解释器崩溃(使用python3.4.2+opencv3.0.0-beta和python2.7.6+进行测试opencv2.4.8)。如果您想检查/进一步了解,这是我迄今为止的尝试。
import cv2\nimport multiprocessing as mp\n\ndef getFrame(queue, startFrame, endFrame):\n cap = cv2.VideoCapture(file) # crashes here\n print("opened capture {}".format(mp.current_process()))\n for frame in range(startFrame, endFrame):\n # cap.set(cv2.CAP_PROP_POS_FRAMES, frame) # opencv3 \n cap.set(cv2.cv.CV_CAP_PROP_POS_FRAMES, frame)\n # frameNo = int(cap.get(cv2.CAP_PROP_POS_FRAMES)) # opencv3\n frameNo = int(cap.get(cv2.cv.CV_CAP_PROP_POS_FRAMES))\n ret, f = cap.read()\n if ret:\n print("{} - put ({})".format(mp.current_process(), frameNo))\n queue.put((frameNo, f))\n cap.release()\n\nfile = "video.mov"\ncapture_temp = cv2.VideoCapture(file)\n# fileLen = int((capture_temp).get(cv2.CAP_PROP_FRAME_COUNT)) # opencv3\nfileLen = int((capture_temp).get(cv2.cv.CV_CAP_PROP_FRAME_COUNT))\ncapture_temp.release()\n\n# get cpuCount for processCount\n# processCount = mp.cpu_count() / 3\nprocessCount = 2\n\ninQ1 = mp.JoinableQueue() # not sure if this is right queue type, but I also tried mp.Queue()\ninQ2 = mp.JoinableQueue()\nqList = [inQ1, inQ2]\n\n# set up bunches\nbunches = []\nfor startFrame in range(0, fileLen, int(fileLen / processCount)):\n endFrame = startFrame + int(fileLen / processCount)\n bunches.append((startFrame, endFrame))\n\ngetFrames = []\nfor i in range(processCount):\n getFrames.append(mp.Process(target=getFrame, args=(qList[i], bunches[i][0], bunches[i][1])))\n\nfor process in getFrames:\n process.start()\n\nresults1 = [inQ1.get() for p in range(bunches[0][0], bunches[0][1])]\nresults2 = [inQ2.get() for p in range(bunches[1][0], bunches[1][1])]\n\ninQ1.close()\ninQ2.close()\n\nfor process in getFrames:\n process.terminate()\n process.join()\nRun Code Online (Sandbox Code Playgroud)\n
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