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]

import time
from multiprocessing import Pool


def f1(args):
    vfirst, vsecond, vthird = args[0] , args[1] , args[2]
    print(f'First Param: {vfirst}, Second value: {vsecond} and finally third value is: {vthird}')
    pass


if __name__ == '__main__':
    p = Pool()
    result = p.map(f1, [['Dog','Cat','Mouse']])
    p.close()
    p.join()
    print(result)

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。

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]
>>>

对我来说，以下是一个简单明了的解决方案：

from multiprocessing.pool import ThreadPool
from functools import partial
from time import sleep
from random import randint

def dosomething(var,s):
    sleep(randint(1,5))
    print(var)
    return var + s

array = ["a", "b", "c", "d", "e"]
with ThreadPool(processes=5) as pool:
    resp_ = pool.map(partial(dosomething,s="2"), array)
    print(resp_)

输出：

a
b
d
e
c
['a2', 'b2', 'c2', 'd2', 'e2']

Python 2的更好解决方案：

from multiprocessing import Pool
def func((i, (a, b))):
    print i, a, b
    return a + b
pool = Pool(3)
pool.map(func, [(0,(1,2)), (1,(2,3)), (2,(3, 4))])

输出

2 3 4

1 2 3

0 1 2

out[]:

[3, 5, 7]