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python-3.x - python-3 并行化我的函数问题

问题描述

我正在使用 python-3.x，我想通过使用多处理并行化我的函数来加速我的代码，我应用了多处理但由于某种原因，它可能不起作用，我不确定问题出在哪里？

所以以下是我所做的一个小例子。任何建议表示赞赏

import numpy as np
import math
import multiprocessing as mp

lower_bound = -500
upper_bound =500 
dimension =1000
Base_Value = 10
Popula_size = 3000
MinResolution = 8

population_in = np.random.choice ( np.linspace ( lower_bound , upper_bound , Base_Value ** MinResolution ) , size = ( Popula_size , dimension ) , replace = True )
resolution = np.random.randint(1, 8, size = (1, dimension))

def Discretiz(lower_bound, upper_bound, DiscPopulation, resolution):
        
    pop_size = int(len(DiscPopulation))
    the_new_population = np.zeros ((pop_size, dimension))
    for i in range (pop_size) :
        for ii in range (dimension):          
            decimal = int(np.round((DiscPopulation[i][ii] - lower_bound) / ((upper_bound-lower_bound)/(math.pow(Base_Value,resolution[:,ii])-1))))
            the_new_population[i, ii]  = (lower_bound + decimal *  ((upper_bound-lower_bound)/(math.pow(Base_Value,resolution[:,ii])-1)))
    return the_new_population


# without_parallelizing
# the_new_population = Discretiz(lower_bound, upper_bound, population_in, resolution)


# wit_parallelizing
pool = mp.Pool(mp.cpu_count())
the_new_population = [pool.apply(Discretiz, args=(lower_bound, upper_bound, population_in, resolution))]


print (the_new_population)

标签： python-3.xnumpyparallel-processingpython-multiprocessing

解决方案

和：

population_in = np.random.choice ( np.linspace ( lower_bound , upper_bound , Base_Value ** MinResolution ) , size = ( Popula_size , dimension ) , replace = True )

你制作一个二维数组(Popula_size, dimension)形状。这被传递为DiscPopulation.

resolution = np.random.randint(1, 8, size = (1, dimension))

双迭代函数可以替换为对整个数组进行操作而没有缓慢迭代的函数：

def Discretiz(lower_bound, upper_bound, DiscPopulation, resolution):
    pop_size = DiscPopulation[0]  # no need for the 'int'
    num = DiscPopulation - lower_bound
    divisor = (upper_bound-lower_bound)/(Base_value**resolution-1)
    decimal = num/divisor
    # this divide does (pop,dimension)/(1,dimension); ok by broadcasting)
    decimal = np.round(decimal)  # no need for int
    the_new_population = lower_bound + decimal * divisor
    return the_new_population

我在这里就地写了这个。它在语法上是正确的，但我没有尝试运行它。

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