在windows上用python将类传递给多处理池

Question

为了在windows上的交互式python（miniconda）中使用multiprocessing，我发现了一个非常有用的代码。但是，代码不能将类中的self参数传递给要合并的函数。以下是我在google colab上工作但在windows iPython上从未完成的代码：

import multiprocessing
from multiprocessing import Pool
from poolable import make_applicable, make_mappable

def worker(d):
    """worker function"""
    for i in range(10000):
      j = i **(1/3) + d.bias
    return j

class dummy():
  def __init__(self):
    self.bias = 1000
  def calc(self):
    pool = Pool(processes=12)
    results = {}
    for i in range(5):
        results[i] = (pool.apply_async(*make_applicable(worker,self)))
    pool.close()
    pool.join()
    print([results[i].get() for i in range(5)])

d=dummy()
d.calc()

如果我传递其他类型的变量，则该代码在windows上运行良好，例如：

results[i] = (pool.apply_async(*make_applicable(worker,self.bias)))

但是当我将self传递给函数时，这个过程永远不会结束。我不知道该怎么办。你知道吗

poolable.py来自here：

from types import FunctionType
import marshal

def _applicable(*args, **kwargs):
  name = kwargs['__pw_name']
  code = marshal.loads(kwargs['__pw_code'])
  gbls = globals() #gbls = marshal.loads(kwargs['__pw_gbls'])
  defs = marshal.loads(kwargs['__pw_defs'])
  clsr = marshal.loads(kwargs['__pw_clsr'])
  fdct = marshal.loads(kwargs['__pw_fdct'])
  func = FunctionType(code, gbls, name, defs, clsr)
  func.fdct = fdct
  del kwargs['__pw_name']
  del kwargs['__pw_code']
  del kwargs['__pw_defs']
  del kwargs['__pw_clsr']
  del kwargs['__pw_fdct']
  return func(*args, **kwargs)

def make_applicable(f, *args, **kwargs):
  if not isinstance(f, FunctionType): raise ValueError('argument must be a function')
  kwargs['__pw_name'] = f.__name__  # edited
  kwargs['__pw_code'] = marshal.dumps(f.__code__)   # edited
  kwargs['__pw_defs'] = marshal.dumps(f.__defaults__)  # edited
  kwargs['__pw_clsr'] = marshal.dumps(f.__closure__)  # edited
  kwargs['__pw_fdct'] = marshal.dumps(f.__dict__)   # edited
  return _applicable, args, kwargs

def _mappable(x):
  x,name,code,defs,clsr,fdct = x
  code = marshal.loads(code)
  gbls = globals() #gbls = marshal.loads(gbls)
  defs = marshal.loads(defs)
  clsr = marshal.loads(clsr)
  fdct = marshal.loads(fdct)
  func = FunctionType(code, gbls, name, defs, clsr)
  func.fdct = fdct
  return func(x)

def make_mappable(f, iterable):
  if not isinstance(f, FunctionType): raise ValueError('argument must be a function')
  name = f.__name__    # edited
  code = marshal.dumps(f.__code__)   # edited
  defs = marshal.dumps(f.__defaults__)  # edited
  clsr = marshal.dumps(f.__closure__)  # edited
  fdct = marshal.dumps(f.__dict__)  # edited
  return _mappable, ((i,name,code,defs,clsr,fdct) for i in iterable)

编辑：

似乎这个问题不仅存在于self，而且也存在于传递给make_applicable函数的任何其他类。以下代码也未完成：

class dummy2():
  def __init__(self):
    self.bias = 1000  

class dummy():
  def __init__(self):
    self.bias = 1000
  def copy(self):
    return copy.deepcopy(self)
  def calc(self):
    pool = Pool(processes=12)
    results = {}
    for i in range(5):
      d = dummy2()
      results[i] = pool.apply_async(*make_applicable(worker,d))
    pool.close()
    pool.join()
    print([results[i].get() for i in range(5)])

Answer 1

使用IPython控制台： 将代码放入一个模块（mp.py），确保类实例化和方法调用在if __name__ == '__main__':条件中执行

import multiprocessing
from multiprocessing import Pool
from poolable import make_applicable, make_mappable

def worker(d):
    """worker function"""
    for i in range(10000):
      j = i **(1/3) + d.bias
    return j

class Dummy():
  def __init__(self):
#    self.bias = 1000
      self.bias = 10
  def calc(self):
    pool = Pool(processes=12)
    results = {}
    for i in range(5):
        results[i] = (pool.apply_async(*make_applicable(worker,self)))
    pool.close()
    pool.join()
    return [results[i].get() for i in range(5)]

if __name__ == '__main__':
    d=Dummy()
    print(d.calc())

然后在控制台中运行模块：

In [1]: runfile('P:/pyProjects3/mp.py', wdir='P:/pyProjects3')
[31.543628731482663, 31.543628731482663, 31.543628731482663, 31.543628731482663, 31.543628731482663]

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我有Jupyter笔记本（水蟒），但不知道如何使用它，如果我找到它，我会更新这个答案。你知道吗