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]
>>>

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)

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]
>>>

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。

这可能是另一种选择。技巧在于包装器函数，它返回传递给pool.map的另一个函数。下面的代码读取一个输入数组，对于其中的每个（唯一）元素，返回该元素在数组中出现的次数（即计数）。例如，如果输入是

np.eye(3) = [ [1. 0. 0.]
              [0. 1. 0.]
              [0. 0. 1.]]

然后零出现6次，一出现3次

import numpy as np
from multiprocessing.dummy import Pool as ThreadPool
from multiprocessing import cpu_count


def extract_counts(label_array):
    labels = np.unique(label_array)
    out = extract_counts_helper([label_array], labels)
    return out

def extract_counts_helper(args, labels):
    n = max(1, cpu_count() - 1)
    pool = ThreadPool(n)
    results = {}
    pool.map(wrapper(args, results), labels)
    pool.close()
    pool.join()
    return results

def wrapper(argsin, results):
    def inner_fun(label):
        label_array = argsin[0]
        counts = get_label_counts(label_array, label)
        results[label] = counts
    return inner_fun

def get_label_counts(label_array, label):
    return sum(label_array.flatten() == label)

if __name__ == "__main__":
    img = np.ones([2,2])
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")
           
    img = np.eye(3)
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")
    
    img = np.random.randint(5, size=(3, 3))
    out = extract_counts(img)
    print('input array: \n', img)
    print('label counts: ', out)
    print("========")

你应该得到：

input array: 
 [[1. 1.]
 [1. 1.]]
label counts:  {1.0: 4}
========
input array: 
 [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
label counts:  {0.0: 6, 1.0: 3}
========
input array: 
 [[4 4 0]
 [2 4 3]
 [2 3 1]]
label counts:  {0: 1, 1: 1, 2: 2, 3: 2, 4: 3}
========

pool.map是否有支持多个参数的变体？

Python 3.3包含pool.starmap（）方法：

#!/usr/bin/env python3
from functools import partial
from itertools import repeat
from multiprocessing import Pool, freeze_support

def func(a, b):
    return a + b

def main():
    a_args = [1,2,3]
    second_arg = 1
    with Pool() as pool:
        L = pool.starmap(func, [(1, 1), (2, 1), (3, 1)])
        M = pool.starmap(func, zip(a_args, repeat(second_arg)))
        N = pool.map(partial(func, b=second_arg), a_args)
        assert L == M == N

if __name__=="__main__":
    freeze_support()
    main()

对于旧版本：

#!/usr/bin/env python2
import itertools
from multiprocessing import Pool, freeze_support

def func(a, b):
    print a, b

def func_star(a_b):
    """Convert `f([1,2])` to `f(1,2)` call."""
    return func(*a_b)

def main():
    pool = Pool()
    a_args = [1,2,3]
    second_arg = 1
    pool.map(func_star, itertools.izip(a_args, itertools.repeat(second_arg)))

if __name__=="__main__":
    freeze_support()
    main()

输出

1 1
2 1
3 1

注意这里是如何使用itertools.izip（）和itertools.crepeat（）的。

由于@unsubu提到的错误，您不能在Python 2.6上使用functools.partial（）或类似功能，因此应该显式定义简单包装函数func_tar（）。另请参阅uptimebox建议的解决方法。