在Python多处理库中,是否有支持多个参数的pool.map变体?

import multiprocessing

text = "test"

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

if __name__ == '__main__':
    pool = multiprocessing.Pool(processes=6)
    case = RAW_DATASET
    pool.map(harvester(text, case), case, 1)
    pool.close()
    pool.join()

当前回答

另一种方法是将列表列表传递给单参数例程:

import os
from multiprocessing import Pool

def task(args):
    print "PID =", os.getpid(), ", arg1 =", args[0], ", arg2 =", args[1]

pool = Pool()

pool.map(task, [
        [1,2],
        [3,4],
        [5,6],
        [7,8]
    ])

然后可以用自己喜欢的方法构造一个参数列表。

其他回答

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]
>>>

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。

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]

您可以使用以下两个函数,以避免为每个新函数编写包装器:

import itertools
from multiprocessing import Pool

def universal_worker(input_pair):
    function, args = input_pair
    return function(*args)

def pool_args(function, *args):
    return zip(itertools.repeat(function), zip(*args))

将函数函数与参数arg_0、arg_1和arg_2的列表一起使用,如下所示:

pool = Pool(n_core)
list_model = pool.map(universal_worker, pool_args(function, arg_0, arg_1, arg_2)
pool.close()
pool.join()

我认为以下内容会更好:

def multi_run_wrapper(args):
   return add(*args)

def add(x,y):
    return x+y

if __name__ == "__main__":
    from multiprocessing import Pool
    pool = Pool(4)
    results = pool.map(multi_run_wrapper,[(1,2),(2,3),(3,4)])
    print results

输出

[3, 5, 7]