我正在使用python multiprocessing库执行Selenium脚本。我的代码如下:

#-- start and join multiple threads ---
thread_list = []
total_threads=10 #-- no of parallel threads
for i in range(total_threads):
    t = Process(target=get_browser_and_start, args=[url,nlp,pixel])
    thread_list.append(t)
    print "starting thread..."
    t.start()

for t in thread_list:
    print "joining existing thread..."
    t.join()


据我了解join()函数,它将等待每个进程完成。但是,我希望进程一旦发布,将被分配另一个任务以执行新功能。

可以这样理解:

假设有8个进程从一开始就启动了。

no_of_tasks_to_perform = 100

for i in range(no_of_tasks_to_perform):
    processes start(8)
    if process no 2 finished executing, start new process
    maintain 8 process at any point of time till
    "i" is <= no_of_tasks_to_perform

最佳答案

与其时不时启动新进程,不如尝试将所有任务放入multiprocessing.Queue(),并启动8个长期运行的进程,在每个进程中继续访问任务队列以获取新任务,然后执行工作,直到出现没有任务了。

在您的情况下,它更像是这样:

from multiprocessing import Queue, Process

def worker(queue):
    while not queue.empty():
        task = queue.get()

        # now start to work on your task
        get_browser_and_start(url,nlp,pixel) # url, nlp, pixel can be unpacked from task

def main():
    queue = Queue()

    # Now put tasks into queue
    no_of_tasks_to_perform = 100

    for i in range(no_of_tasks_to_perform):
        queue.put([url, nlp, pixel, ...])

    # Now start all processes
    process = Process(target=worker, args=(queue, ))
    process.start()
    ...
    process.join()

10-05 22:17