鉴于以下课程:
from abc import ABCMeta, abstractmethod
from time import sleep
import threading
from threading import active_count, Thread
class ScraperPool(metaclass=ABCMeta):
Queue = []
ResultList = []
def __init__(self, Queue, MaxNumWorkers=0, ItemsPerWorker=50):
# Initialize attributes
self.MaxNumWorkers = MaxNumWorkers
self.ItemsPerWorker = ItemsPerWorker
self.Queue = Queue # For testing purposes.
def initWorkerPool(self, PrintIDs=True):
for w in range(self.NumWorkers()):
Thread(target=self.worker, args=(w + 1, PrintIDs,)).start()
sleep(1) # Explicitly wait one second for this worker to start.
def run(self):
self.initWorkerPool()
# Wait until all workers (i.e. threads) are done.
while active_count() > 1:
print("Active threads: " + str(active_count()))
sleep(5)
self.HandleResults()
def worker(self, id, printID):
if printID:
print("Starting worker " + str(id) + ".")
while (len(self.Queue) > 0):
self.scraperMethod()
if printID:
print("Worker " + str(id) + " is quiting.")
# Todo Kill is this Thread.
return
def NumWorkers(self):
return 1 # Simplified for testing purposes.
@abstractmethod
def scraperMethod(self):
pass
class TestScraper(ScraperPool):
def scraperMethod(self):
# print("I am scraping.")
# print("Scraping. Threads#: " + str(active_count()))
temp_item = self.Queue[-1]
self.Queue.pop()
self.ResultList.append(temp_item)
def HandleResults(self):
print(self.ResultList)
ScraperPool.register(TestScraper)
scraper = TestScraper(Queue=["Jaap", "Piet"])
scraper.run()
print(threading.active_count())
# print(scraper.ResultList)
当所有线程都完成后,仍有一个活动线程 - 最后一行的threading.active_count()
为我提供该号码。
活动帖子为<_MainThread(MainThread, started 12960)>
- 打印为threading.enumerate()
。
我可以假设active_count() == 1
时我的所有线程都已完成吗?
或者,例如,导入的模块可以启动其他线程,以便我的线程实际在active_count() > 1
时完成 - 也是我在run方法中使用的循环的条件。
答案 0 :(得分:2)
根据docs active_count()
包含主线索,因此,如果你在1,那么你最有可能完成,但如果你有另一个新线程来源程序然后你可能会在active_count()
命中之前完成。
我建议您在join
上实施明确的ScraperPool
方法并跟踪您的工作人员,并在需要时明确地将他们加入主线程,而不是检查您是否已完成{{1}调用。
还要记住GIL ......
答案 1 :(得分:2)
您可以假设您的线程在active_count()
达到1时完成。问题是,如果任何其他模块创建了一个线程,您将永远不会达到1.您应该明确管理您的线程。 / p>
示例:您可以将线程放在列表中,并一次加入一个。您的代码的相关更改包括:
def __init__(self, Queue, MaxNumWorkers=0, ItemsPerWorker=50):
# Initialize attributes
self.MaxNumWorkers = MaxNumWorkers
self.ItemsPerWorker = ItemsPerWorker
self.Queue = Queue # For testing purposes.
self.WorkerThreads = []
def initWorkerPool(self, PrintIDs=True):
for w in range(self.NumWorkers()):
thread = Thread(target=self.worker, args=(w + 1, PrintIDs,))
self.WorkerThreads.append(thread)
thread.start()
sleep(1) # Explicitly wait one second for this worker to start.
def run(self):
self.initWorkerPool()
# Wait until all workers (i.e. threads) are done. Waiting in order
# so some threads further in the list may finish first, but we
# will get to all of them eventually
while self.WorkerThreads:
self.WorkerThreads[0].join()
self.HandleResults()