text = "test"

def unpack(args):
    return args[0](*args[1:])

def harvester(text, case):
    X = case[0]
    text+ str(X)

if __name__ == '__main__':
    pool = multiprocessing.Pool(processes=6)
    case = RAW_DATASET
    # args is a list of tuples 
    # with the function to execute as the first item in each tuple
    args = [(harvester, text, c) for c in case]
    # doing it this way, we can pass any function
    # and we don't need to define a wrapper for each different function
    # if we need to use more than one
    pool.map(unpack, args)
    pool.close()
    pool.join()

有一个叫做pathos的多处理分支（注意：使用GitHub上的版本），它不需要starmap——map函数镜像Python map的API，因此map可以接受多个参数。

使用pathos，您通常也可以在解释器中执行多处理，而不是陷入__main__块。Pathos将在经过一些轻微的更新后发布——主要是转换为Python3.x。

  Python 2.7.5 (default, Sep 30 2013, 20:15:49)
  [GCC 4.2.1 (Apple Inc. build 5566)] on darwin
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def func(a,b):
  ...     print a,b
  ...
  >>>
  >>> from pathos.multiprocessing import ProcessingPool
  >>> pool = ProcessingPool(nodes=4)
  >>> pool.map(func, [1,2,3], [1,1,1])
  1 1
  2 1
  3 1
  [None, None, None]
  >>>
  >>> # also can pickle stuff like lambdas
  >>> result = pool.map(lambda x: x**2, range(10))
  >>> result
  [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
  >>>
  >>> # also does asynchronous map
  >>> result = pool.amap(pow, [1,2,3], [4,5,6])
  >>> result.get()
  [1, 32, 729]
  >>>
  >>> # or can return a map iterator
  >>> result = pool.imap(pow, [1,2,3], [4,5,6])
  >>> result
  <processing.pool.IMapIterator object at 0x110c2ffd0>
  >>> list(result)
  [1, 32, 729]

pathos有几种方法可以让你得到星图的精确行为。

>>> def add(*x):
...   return sum(x)
...
>>> x = [[1,2,3],[4,5,6]]
>>> import pathos
>>> import numpy as np
>>> # use ProcessPool's map and transposing the inputs
>>> pp = pathos.pools.ProcessPool()
>>> pp.map(add, *np.array(x).T)
[6, 15]
>>> # use ProcessPool's map and a lambda to apply the star
>>> pp.map(lambda x: add(*x), x)
[6, 15]
>>> # use a _ProcessPool, which has starmap
>>> _pp = pathos.pools._ProcessPool()
>>> _pp.starmap(add, x)
[6, 15]
>>>

将Python 3.3+与pool.starmap（）一起使用：

from multiprocessing.dummy import Pool as ThreadPool 

def write(i, x):
    print(i, "---", x)

a = ["1","2","3"]
b = ["4","5","6"] 

pool = ThreadPool(2)
pool.starmap(write, zip(a,b)) 
pool.close() 
pool.join()

结果：

1 --- 4
2 --- 5
3 --- 6

如果您喜欢，还可以zip（）更多参数：zip（a，b，c，d，e）

如果希望将常量值作为参数传递：

import itertools

zip(itertools.repeat(constant), a)

如果您的函数应该返回以下内容：

results = pool.starmap(write, zip(a,b))

这将提供一个包含返回值的列表。

这里有另一种方法，IMHO比提供的任何其他答案都更简单和优雅。

该程序有一个函数，它获取两个参数，打印它们并打印总和：

import multiprocessing

def main():

    with multiprocessing.Pool(10) as pool:
        params = [ (2, 2), (3, 3), (4, 4) ]
        pool.starmap(printSum, params)
    # end with

# end function

def printSum(num1, num2):
    mySum = num1 + num2
    print('num1 = ' + str(num1) + ', num2 = ' + str(num2) + ', sum = ' + str(mySum))
# end function

if __name__ == '__main__':
    main()

输出为：

num1 = 2, num2 = 2, sum = 4
num1 = 3, num2 = 3, sum = 6
num1 = 4, num2 = 4, sum = 8

有关更多信息，请参阅python文档：

https://docs.python.org/3/library/multiprocessing.html#module-多处理工具

特别是要检查星图功能。

我使用的是Python 3.6，我不确定这是否适用于较旧的Python版本

为什么在文档中没有这样一个非常直接的例子，我不确定。

答案取决于版本和情况。最近版本的Python（从3.3开始）的最一般的答案首先由J.F.Sebastian在下面描述。1它使用Pool.starmap方法，接受一系列参数元组。然后，它会自动将每个元组中的参数解包，并将它们传递给给定的函数：

import multiprocessing
from itertools import product

def merge_names(a, b):
    return '{} & {}'.format(a, b)

if __name__ == '__main__':
    names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']
    with multiprocessing.Pool(processes=3) as pool:
        results = pool.starmap(merge_names, product(names, repeat=2))
    print(results)

# Output: ['Brown & Brown', 'Brown & Wilson', 'Brown & Bartlett', ...

对于早期版本的Python，您需要编写一个助手函数来显式地解包参数。如果要与一起使用，还需要编写一个包装器，将Pool转换为上下文管理器。（感谢穆恩指出了这一点。）

import multiprocessing
from itertools import product
from contextlib import contextmanager

def merge_names(a, b):
    return '{} & {}'.format(a, b)

def merge_names_unpack(args):
    return merge_names(*args)

@contextmanager
def poolcontext(*args, **kwargs):
    pool = multiprocessing.Pool(*args, **kwargs)
    yield pool
    pool.terminate()

if __name__ == '__main__':
    names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']
    with poolcontext(processes=3) as pool:
        results = pool.map(merge_names_unpack, product(names, repeat=2))
    print(results)

# Output: ['Brown & Brown', 'Brown & Wilson', 'Brown & Bartlett', ...

在更简单的情况下，使用固定的第二个参数，也可以使用partial，但仅在Python 2.7+中使用。

import multiprocessing
from functools import partial
from contextlib import contextmanager

@contextmanager
def poolcontext(*args, **kwargs):
    pool = multiprocessing.Pool(*args, **kwargs)
    yield pool
    pool.terminate()

def merge_names(a, b):
    return '{} & {}'.format(a, b)

if __name__ == '__main__':
    names = ['Brown', 'Wilson', 'Bartlett', 'Rivera', 'Molloy', 'Opie']
    with poolcontext(processes=3) as pool:
        results = pool.map(partial(merge_names, b='Sons'), names)
    print(results)

# Output: ['Brown & Sons', 'Wilson & Sons', 'Bartlett & Sons', ...

1.这大部分都是由他的答案激发的，而他的答案很可能应该被接受。但由于这本书一直停留在顶端，似乎最好为未来读者改进它。

另一个简单的选择是将函数参数包装在元组中，然后包装应该在元组中传递的参数。在处理大量数据时，这可能并不理想。我相信它会为每个元组创建副本。

from multiprocessing import Pool

def f((a,b,c,d)):
    print a,b,c,d
    return a + b + c +d

if __name__ == '__main__':
    p = Pool(10)
    data = [(i+0,i+1,i+2,i+3) for i in xrange(10)]
    print(p.map(f, data))
    p.close()
    p.join()

以某种随机顺序给出输出：

0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
7 8 9 10
6 7 8 9
8 9 10 11
9 10 11 12
[6, 10, 14, 18, 22, 26, 30, 34, 38, 42]