在Windows上的Python中演示多核加速的一些示例代码是什么？

Question

我在Windows上使用Python 3并试图构建一个玩具示例，演示使用多个CPU内核如何加快计算速度。玩具示例是Mandelbrot分形的渲染。

至今：

• 我避免了线程化，因为全局解释器锁定在这种情况下禁止多核
• 我放弃了在Windows上无法使用的示例代码，因为它缺乏Linux的分叉功能
• 尝试使用“多处理”包。我声明p = Pool（8）（8是我的内核数），并使用p.starmap（..）委派工作。这应该会产生多个“子进程”，这些窗口将自动将窗口委派给不同的CPU

但是，无论是由于开销还是没有实际的多重处理，我都无法证明任何加速。因此，具有明显加速性的玩具示例指针将非常有帮助:-)

编辑： 谢谢！这将我推向正确的方向，现在我得到了一个可行的示例，该示例演示了具有4个内核的CPU的速度提高了一倍。
我的代码的副本与“讲义”在这里：https :
//pastebin.com/c9HZ2vAV

我决定使用Pool（），但稍后将尝试@ 16num指出的“ Process”替代方法。下面是Pool（）的代码示例：

    p = Pool(cpu_count())

    #Unlike map, starmap only allows 1 input. "partial" provides a workaround
    partial_calculatePixel = partial(calculatePixel, dataarray=data) 
    koord = []
    for j in range(height):
        for k in range(width):
            koord.append((j,k))

    #Runs the calls to calculatePixel in a pool. "hmm" collects the output
    hmm = p.starmap(partial_calculatePixel,koord)

Answer 1

演示多处理速度非常简单：

import multiprocessing
import sys
import time

# multi-platform precision clock
get_timer = time.clock if sys.platform == "win32" else time.time

def cube_function(num):
    time.sleep(0.01)  # let's simulate it takes ~10ms for the CPU core to cube the number
    return num**3

if __name__ == "__main__":  # multiprocessing guard
    # we'll test multiprocessing with pools from one to the number of CPU cores on the system
    # it won't show significant improvements after that and it will soon start going
    # downhill due to the underlying OS thread context switches
    for workers in range(1, multiprocessing.cpu_count() + 1):
        pool = multiprocessing.Pool(processes=workers)
        # lets 'warm up' our pool so it doesn't affect our measurements
        pool.map(cube_function, range(multiprocessing.cpu_count()))
        # now to the business, we'll have 10000 numbers to quart via our expensive function
        print("Cubing 10000 numbers over {} processes:".format(workers))
        timer = get_timer()  # time measuring starts now
        results = pool.map(cube_function, range(10000))  # map our range to the cube_function
        timer = get_timer() - timer  # get our delta time as soon as it finishes
        print("\tTotal: {:.2f} seconds".format(timer))
        print("\tAvg. per process: {:.2f} seconds".format(timer / workers))
        pool.close()  # lets clear out our pool for the next run
        time.sleep(1)  # waiting for a second to make sure everything is cleaned up

当然，在这里我们只是模拟10ms /数字的计算，您可以cube_function用任何CPU负担的方法代替实际演示。结果符合预期：

Cubing 10000 numbers over 1 processes:
        Total: 100.01 seconds
        Avg. per process: 100.01 seconds
Cubing 10000 numbers over 2 processes:
        Total: 50.02 seconds
        Avg. per process: 25.01 seconds
Cubing 10000 numbers over 3 processes:
        Total: 33.36 seconds
        Avg. per process: 11.12 seconds
Cubing 10000 numbers over 4 processes:
        Total: 25.00 seconds
        Avg. per process: 6.25 seconds
Cubing 10000 numbers over 5 processes:
        Total: 20.00 seconds
        Avg. per process: 4.00 seconds
Cubing 10000 numbers over 6 processes:
        Total: 16.68 seconds
        Avg. per process: 2.78 seconds
Cubing 10000 numbers over 7 processes:
        Total: 14.32 seconds
        Avg. per process: 2.05 seconds
Cubing 10000 numbers over 8 processes:
        Total: 12.52 seconds
        Avg. per process: 1.57 seconds

现在，为什么不100％线性？嗯，首先，它需要一些时间来图/数据分配给子流程，并把它找回来，有一些成本的上下文切换，还有一些用我的CPU不时其他任务，time.sleep()不是完全精确（也不可能在非RT
OS上使用）…但是结果大致上可用于并行处理。