Python 多进程共享导入的类

Question

>我在 Python 多处理共享导入的类时遇到了问题。麻烦的部分是这样的：

文件 A：

class Meta:
    db_a = None
    db_b = None
    ...
# class will be initialized at the very beginning of the program and might
# be imported by all other models globally for a global variable/instance
# access, for example, a global DB access instance is in Meta

文件 B：

from file_A import Meta
def runner():
    initialize_meta_db()  # Meta's attributes now have values
    ...
    pool = multiprocessing.Pool(4)
    pool.map(worker, arg_list)
    pool.close()
    pool.join()
    ...
def worker(*args):
    ...
    print(Meta.db_a)  # process will print None
    ...
# a runner function which spawns 4 processes, each process will use class Meta
# to do some work.

但是程序运行时出现错误，对于每个进程，Meta类未初始化，每个属性都None。我知道原因，Meta类只在主进程的内存中初始化，每个子进程都会独立拥有自己的原始类 Meta。

但是有什么方法可以与父进程和子进程共享此类吗？谢谢！

Answer 1

您是否考虑过使用多处理的初始值设定项和初始参数。池？

我稍微修改了您的代码以执行此操作并能够运行。它似乎做了你想做的事。

File_A

class Meta:
    db_a = None
    db_b = None
    @classmethod
    def initialize_meta_db(cls, db_a='a', db_b='b'):
        Meta.db_a = db_a
        Meta.db_b = db_b

File_B

import multiprocessing
from file_A import Meta
def runner():
    Meta.initialize_meta_db()  # Meta's attributes now have values
    pool = multiprocessing.Pool(4, init, '')
    pool.map(worker, (1, 2, 3, 4))
    pool.close()
    pool.join()
def init(*initargs):
    from file_A import Meta
    Meta.initialize_meta_db()
def worker(*args):
    print('Worker {} -- Work {}'.format(args, Meta.db_a))
if __name__ == '__main__':
    runner()