[英]Python Concurrency ThreadPoolExecutor - stop execution if condition is met
我坚持尝试使用 ThreadPoolExecutor 进行多线程,如果其中一个进程满足标准,则执行将安全停止,以便不必完成剩余的线程。
我有以下概念,但它不起作用,它一直在运行:
import time, sys
from concurrent.futures import ThreadPoolExecutor
data = {"future1": 'blank', "future2": 'blank', "future3": 'blank'}
def function1(n):
global data #not sure if this is necessary, as it seems to be able to access this anyway?
print(n)
time.sleep(1)
data['future1'] = n
def function2(n):
global data
print(n)
time.sleep(2)
data['future2'] = n
def function3(n):
global data
print(n)
time.sleep(3)
data['future3'] = n
with ThreadPoolExecutor(max_workers=4) as executor:
while True:
future1=executor.submit(function1, 'test1')
future2=executor.submit(function2, 'test2')
future3=executor.submit(function3, 'test3')
if data['future2']!='blank':
executor.shutdown(wait=False)
sys.exit()
不确定我在这里做错了什么,任何帮助将不胜感激。
这是完整的答案,事实证明executor.shutdown(wait=False)
不是 Sachin 提到的方法。 完全归功于https://gist.github.com/clchiou/f2608cbe54403edb0b13
import time, sys
from concurrent.futures import ThreadPoolExecutor
import concurrent.futures.thread
data = {"future1": None, "future2": None, "future3": None}
def function1(n):
time.sleep(1)
data['future1'] = n
print(n)
def function2(n):
time.sleep(2)
data['future2'] = n
print(n)
def function3(n):
time.sleep(3)
data['future3'] = n
print(n)
with ThreadPoolExecutor(max_workers=4) as executor:
executor.submit(function1, 'test1')
executor.submit(function2, 'test2')
executor.submit(function3, 'test3')
while True:
if any(v is not None for v in data.values()):
executor._threads.clear()
concurrent.futures.thread._threads_queues.clear()
break
print(data)
您正在while True:
循环中运行线程,它一次又一次地启动调用。
with ThreadPoolExecutor(max_workers=4) as executor:
future1=executor.submit(function1, 'test1')
future2=executor.submit(function2, 'test2')
future3=executor.submit(function3, 'test3')
while True:
if data['future2']!='blank':
executor.shutdown(wait=False)
sys.exit()
仍然executor.shutdown(wait=False)
会等待它的子进程完成
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.