Python 创建对象的副本，而不是在新的多处理过程中重新初始化

这段代码显示了我尝试执行的操作的结构

import multiprocessing
from foo import really_expensive_to_compute_object

## Create a really complicated object that is *hard* to initialise.
T = really_expensive_to_compute_object(10) 

def f(x):
  return T.cheap_calculation(x)

P = multiprocessing.Pool(processes=64)
results = P.map(f, range(1000000))

print results

---

问题是，每个进程开始时都要花费大量时间重新计算T，而不是使用原来计算过一次的T。有没有办法防止这种情况？T有一个快速（深度）复制方法，所以我可以让Python使用它而不是重新计算吗？

为什么不让

f

使用

T

参数而不是引用全局，然后自己进行复制

import multiprocessing, copy
from foo import really_expensive_to_compute_object

## Create a really complicated object that is *hard* to initialise.
T = really_expensive_to_compute_object(10) 

def f(t, x):
  return t.cheap_calculation(x)

P = multiprocessing.Pool(processes=64)
results = P.map(f, (copy.deepcopy(T) for _ in range(1000000)), range(1000000))

print results

文件

显式地将资源传递给子进程

因此，您的代码可以重写为以下内容：

import multiprocessing
import time
import functools

class really_expensive_to_compute_object(object):
    def __init__(self, arg):
        print 'expensive creation'
        time.sleep(3)

    def cheap_calculation(self, x):
        return x * 2

def f(T, x):
    return T.cheap_calculation(x)

if __name__ == '__main__':
    ## Create a really complicated object that is *hard* to initialise.
    T = really_expensive_to_compute_object(10)
    ## helper, to pass expensive object to function
    f_helper = functools.partial(f, T)
    # i've reduced count for tests 
    P = multiprocessing.Pool(processes=4)
    results = P.map(f_helper, range(100))

    print results