Python:exec与ProcessPoolExecutor
mainWork.pyPython:exec与ProcessPoolExecutor,python,python-3.x,parallel-processing,concurrent.futures,Python,Python 3.x,Parallel Processing,Concurrent.futures,mainWork.py from concurrent.futures import ProcessPoolExecutor import os import time def parInnerLoop(item): a = 2+item print(f'A. {a} Processing {os.getpid()} done on {item}\n') exec(open('mainWork.py').read()) print(f'D. {a} Proces
from concurrent.futures import ProcessPoolExecutor
import os
import time
def parInnerLoop(item):
a = 2+item
print(f'A. {a} Processing {os.getpid()} done on {item}\n')
exec(open('mainWork.py').read())
print(f'D. {a} Processing {os.getpid()} done on {item}\n')
def main():
with ProcessPoolExecutor(max_workers=4) as executor:
for itemNo in range(10):
executor.submit(parInnerLoop, itemNo)
print('done')
if __name__ == '__main__':
main()
错误:
print(f'B. {a} Processing {os.getpid()} done on {item}\n')
a = 12
print(f'C. {a} Processing {os.getpid()} done on {item}\n')
问题:
mainWork.py
的所有代码作为parInnerLoop
的一部分执行,这样a
的值的变化应该反映在parInnerLoop
以及mainWork.py
中,反之亦然。我得到如下输出。a
的值在打印以D开始的位置不会改变…
,我希望它是120
Error in atexit._run_exitfuncs:
Traceback (most recent call last):
File "D:\Program Files\Python\Python37\Lib\concurrent\futures\process.py", line 101, in _python_exit
thread_wakeup.wakeup()
File "D:\Program Files\Python\Python37\Lib\concurrent\futures\process.py", line 89, in wakeup
self._writer.send_bytes(b"")
File "D:\Program Files\Python\Python37\Lib\multiprocessing\connection.py", line 183, in send_bytes
self._check_closed()
File "D:\Program Files\Python\Python37\Lib\multiprocessing\connection.py", line 136, in _check_closed
raise OSError("handle is closed")
OSError: handle is closed
发生了什么事?请帮忙
我最终想要实现的是:以不同的设置并行运行相同的代码,而不相互干扰
额外:
A. 2 Processing 19784 done on 0
B. 2 Processing 19784 done on 0
C. 12 Processing 19784 done on 0
D. 2 Processing 19784 done on 0
run.py
def myFun():
a = 1
b = 2
print(a)
exec(open('run.py').read())
#execfile("run.py")
print(a)
print(b)
print(c)
def main():
myFun()
if __name__ == '__main__':
main()
输出
a = a+5
b = 10
c = 15
不是一个完整的答案,但它似乎与 谢谢你的报道!经过快速调查,问题似乎是因为concurrent.futures模块最近发生了更改。简而言之,它们维护一个对某些服务线程(名为“QueueManagerThread”)的弱引用字典,并在退出Python进程时尝试关闭字典中的任何线程。字典通常为空,但如果存在对线程的引用,则可能会导致尝试关闭不活动的线程。在这种情况下,将引发原始异常
我在Linux上遇到了与3.7.6相同的问题。这似乎是一个开放的市场bug@nodakai,谢谢您的确认。@nokada,我在Windows上使用Python 3.7.8时遇到了同样的问题。看起来像,但尚未修复。请更新到python 3.9+以解决此问题
1
6
10
15