并发.futures问题：为什么只有1个worker？

我正在尝试使用concurrent.futures.ProcessPoolExecutor并行化串行任务。串行任务涉及从数字范围中查找给定数字的出现次数。我的代码如下所示。
在执行过程中，我从任务管理器/系统监视器/顶部注意到，尽管给定了 max_workers ，但只有一个 cpu/线程在持续运行processPoolExecutor值大于1。为什么会这样呢？如何使用并行化我的代码concurrent.futures?我的代码是用 python 3.5 执行的。

import concurrent.futures as cf
from time import time

def _findmatch(nmax, number):    
    print('def _findmatch(nmax, number):')
    start = time()
    match=[]
    nlist = range(nmax)
    for n in nlist:
        if number in str(n):match.append(n)
    end = time() - start
    print("found {} in {}sec".format(len(match),end))
    return match

def _concurrent(nmax, number, workers):
    with cf.ProcessPoolExecutor(max_workers=workers) as executor:
        start = time()
        future = executor.submit(_findmatch, nmax, number)
        futures = future.result()
        found = len(futures)
        end = time() - start
        print('with statement of def _concurrent(nmax, number):')
        print("found {} in {}sec".format(found, end))
    return futures

if __name__ == '__main__':
    match=[]
    nmax = int(1E8)
    number = str(5) # Find this number
    workers = 3
    start = time()
    a = _concurrent(nmax, number, workers)
    end = time() - start
    print('main')
    print("found {} in {}sec".format(len(a),end))

您的代码的问题在于，它仅提交一项任务，然后由其中一名工作人员执行该任务，而其余工作人员则不执行任何操作。您需要提交多个可以由工作人员并行执行的任务。

下面的示例将搜索区域分为三个不同的任务，每个任务由不同的工作人员执行。期货返还方式submit https://docs.python.org/3/library/concurrent.futures.html#concurrent.futures.Executor.submit被添加到列表中，一旦全部提交wait https://docs.python.org/3/library/concurrent.futures.html#concurrent.futures.wait用于等待它们全部完成。如果你打电话result https://docs.python.org/3/library/concurrent.futures.html#concurrent.futures.Future.result提交任务后，它将立即阻塞，直到未来完成。

请注意，下面的代码不是生成数字列表，而是只计算其中包含数字 5 的数字，以减少内存使用量：

import concurrent.futures as cf
from time import time

def _findmatch(nmin, nmax, number):
    print('def _findmatch', nmin, nmax, number)
    start = time()
    count = 0
    for n in range(nmin, nmax):
        if number in str(n):
            count += 1
    end = time() - start
    print("found {} in {}sec".format(count,end))
    return count

def _concurrent(nmax, number, workers):
    with cf.ProcessPoolExecutor(max_workers=workers) as executor:
        start = time()
        chunk = nmax // workers
        futures = []

        for i in range(workers):
            cstart = chunk * i
            cstop = chunk * (i + 1) if i != workers - 1 else nmax

            futures.append(executor.submit(_findmatch, cstart, cstop, number))

        cf.wait(futures)
        res = sum(f.result() for f in futures)
        end = time() - start
        print('with statement of def _concurrent(nmax, number):')
        print("found {} in {}sec".format(res, end))
    return res

if __name__ == '__main__':
    match=[]
    nmax = int(1E8)
    number = str(5) # Find this number
    workers = 3
    start = time()
    a = _concurrent(nmax, number, workers)
    end = time() - start
    print('main')
    print("found {} in {}sec".format(a,end))

Output:

def _findmatch 0 33333333 5
def _findmatch 33333333 66666666 5
def _findmatch 66666666 100000000 5
found 17190813 in 20.09431290626526sec
found 17190813 in 20.443560361862183sec
found 22571653 in 20.47660517692566sec
with statement of def _concurrent(nmax, number):
found 56953279 in 20.6196870803833sec
main
found 56953279 in 20.648695707321167sec