Python多处理仅使用一个核心

Question

我正在尝试使用标准python文档中的代码片段来学习如何使用多处理模块.代码将粘贴在此消息的末尾.我在Ubuntu 11.04上使用Python 2.7.1在四核机器上(根据系统监视器,由于超线程而给我八个核心)

问题:尽管已启动多个进程,但所有工作负载似乎只安排在一个核心上,接近100%的利用率.有时,所有工作负载都迁移到另一个核心,但工作负载从不在它们之间分配.

任何想法为什么会这样？

最好的祝福,

保罗

#
# Simple example which uses a pool of workers to carry out some tasks.
#
# Notice that the results will probably not come out of the output
# queue in the same in the same order as the corresponding tasks were
# put on the input queue.  If it is important to get the results back
# in the original order then consider using `Pool.map()` or
# `Pool.imap()` (which will save on the amount of code needed anyway).
#
# Copyright (c) 2006-2008, R Oudkerk
# All rights reserved.
#

import time
import random

from multiprocessing import Process, Queue, current_process, freeze_support

#
# Function run by worker processes
#

def worker(input, output):
    for func, args in iter(input.get, 'STOP'):
        result = calculate(func, args)
        output.put(result)

#
# Function used to calculate result
#

def calculate(func, args):
    result = func(*args)
    return '%s says that %s%s = %s' % \
        (current_process().name, func.__name__, args, result)

#
# Functions referenced by tasks
#

def mul(a, b):
    time.sleep(0.5*random.random())
    return a * b

def plus(a, b):
    time.sleep(0.5*random.random())
    return a + b


def test():
    NUMBER_OF_PROCESSES = 4
    TASKS1 = [(mul, (i, 7)) for i in range(500)]
    TASKS2 = [(plus, (i, 8)) for i in range(250)]

    # Create queues
    task_queue = Queue()
    done_queue = Queue()

    # Submit tasks
    for task in TASKS1:
        task_queue.put(task)

    # Start worker processes
    for i in range(NUMBER_OF_PROCESSES):
        Process(target=worker, args=(task_queue, done_queue)).start()

    # Get and print results
    print 'Unordered results:'
    for i in range(len(TASKS1)):
       print '\t', done_queue.get()

    # Add more tasks using `put()`
    for task in TASKS2:
        task_queue.put(task)

    # Get and print some more results
    for i in range(len(TASKS2)):
        print '\t', done_queue.get()

    # Tell child processes to stop
    for i in range(NUMBER_OF_PROCESSES):
        task_queue.put('STOP')

test()

Answer 1

多处理并不意味着您将使用处理器的所有核心，您只是获得多个进程而不是多核进程，这将由操作系统处理并且不确定，评论中发布的问题@Devraj 已经提供了完成您的任务的答案欲望。