带计数器的多处理嵌套 for 循环

Question

我正在寻找一种简单的解决方案，它可以帮助我充分利用 PC 的能力来处理我的数据。我认为，将任务划分到不同的核心将有助于减少处理时间，但我不知道该怎么做，我在 stackoverflow 上搜索过类似的问题，但没有任何解决方案可以解决我的问题。我正在处理大约长度的数据：3000，并且由于我使用嵌套 for 循环来查找列表中相似（在 +- 0.5 范围内）元素的数量，它将运行 3000x3000 次，大约需要 2 分钟，我想减少花费的时间。

repeat= []
values = []
for i in completeList:
    count = 0
    for j in completeList:
        if isfloat(i) and isfloat(j):
            if float(i)-0.5 <= float(j) <= float(i)+0.5:
                count = count + 1
    repeat.append(count)
    values.append(i)

任何帮助将不胜感激。

关于， 马尼什

Answer 1

由于您仍然没有发布isfloat 的实际代码或显示completeList 的元素是什么样的，我能做的最好的就是推测它们可能是什么。这有所不同，因为正如我所提到的，执行 isfloat 和 float 以转换 completeList 的元素所需的 CPU 越多，使用多处理获得的收益就越大。

对于CASE 1，我假设completeList 是由字符串组成的，isfloat 需要使用正则表达式来确定字符串是否与我们预期的浮点格式匹配并且@987654328 @ 因此需要从字符串转换。这将是我想象中 CPU 最密集的情况。对于CASE 2，completeList 由浮点数组成，isfloat 只返回True，float 不必做任何真正的转换。

我的桌面有 8 个核心处理器：

案例 1

import multiprocessing as mp
import time
import random
import re
from functools import partial

def isfloat(s):
    return not re.fullmatch(r'\d*\.\d+', s) is None

def single_process(complete_list):
    #repeat = []
    values = []
    for idx_i, v_i in enumerate(complete_list):
        count = 0
        for idx_j, v_j in enumerate(complete_list):
            if idx_i == idx_j:
                continue # don't compare an element with itself
            if isfloat(v_i) and isfloat(v_j):
                f_i = float(v_i)
                if f_i-0.5 <= float(v_j) <= f_i+0.5:
                    count = count + 1
        # repeat will end up being a copy of complete_list
        # why are we doing this?
        #repeat.append(v_i)
        values.append(count) # these are actually counts
    return values


def multi_worker(complete_list, index_range):
    values = []
    for idx_i in index_range:
        v_i = complete_list[idx_i]
        count = 0
        for idx_j, v_j in enumerate(complete_list):
            if idx_i == idx_j:
                continue # don't compare an element with itself
            if isfloat(v_i) and isfloat(v_j):
                f_i = float(v_i)
                if f_i-0.5 <= float(v_j) <= f_i+0.5:
                    count = count + 1
        values.append(count) # these are actually counts
    return values


def multi_process(complete_list):

    def split(a, n):
        k, m = divmod(len(a), n)
        return (a[i * k + min(i, m):(i + 1) * k + min(i + 1, m)] for i in range(n))

    n = len(complete_list)
    POOL_SIZE = mp.cpu_count()
    range_splits = split(range(0, n), POOL_SIZE)
    pool = mp.Pool(POOL_SIZE)
    value_lists = pool.map(partial(multi_worker, complete_list), range_splits)
    values = []
    # join results together:
    for value_list in value_lists:
        values.extend(value_list)
    return values

def main():
    # generate 3000 random numbers:
    random.seed(0)
    complete_list = [str(random.uniform(1.0, 3.0)) for _ in range(3000)]
    t = time.time()
    values = single_process(complete_list)
    print(time.time() - t, values[0:10], values[-10:-1])

    t = time.time()
    values = multi_process(complete_list)
    print(time.time() - t, values[0:10], values[-10:-1])


# required for Windows:
if __name__ == '__main__':
    main()

打印：

27.7540442943573 [1236, 1491, 1464, 1477, 1494, 1472, 1410, 1450, 1502, 1537] [1485, 1513, 1513, 1501, 1283, 1538, 804, 1459, 1457]
7.187546253204346 [1236, 1491, 1464, 1477, 1494, 1472, 1410, 1450, 1502, 1537] [1485, 1513, 1513, 1501, 1283, 1538, 804, 1459, 1457]

案例 2

import multiprocessing as mp
import time
import random
from functools import partial

def isfloat(s):
    return True

def single_process(complete_list):
    values = []
    for idx_i, v_i in enumerate(complete_list):
        count = 0
        for idx_j, v_j in enumerate(complete_list):
            if idx_i == idx_j:
                continue # don't compare an element with itself
            if isfloat(v_i) and isfloat(v_j):
                f_i = float(v_i)
                if f_i-0.5 <= float(v_j) <= f_i+0.5:
                    count = count + 1
        values.append(count) # these are actually counts
    return values


def multi_worker(complete_list, index_range):
    values = []
    for idx_i in index_range:
        v_i = complete_list[idx_i]
        count = 0
        for idx_j, v_j in enumerate(complete_list):
            if idx_i == idx_j:
                continue # don't compare an element with itself
            if isfloat(v_i) and isfloat(v_j):
                f_i = float(v_i)
                if f_i-0.5 <= float(v_j) <= f_i+0.5:
                    count = count + 1
        values.append(count) # these are actually counts
    return values


def multi_process(complete_list):

    def split(a, n):
        k, m = divmod(len(a), n)
        return (a[i * k + min(i, m):(i + 1) * k + min(i + 1, m)] for i in range(n))

    n = len(complete_list)
    POOL_SIZE = mp.cpu_count()
    range_splits = split(range(0, n), POOL_SIZE)
    pool = mp.Pool(POOL_SIZE)
    value_lists = pool.map(partial(multi_worker, complete_list), range_splits)
    values = []
    # join results together:
    for value_list in value_lists:
        values.extend(value_list)
    return values

def main():
    # generate 3000 random numbers:
    random.seed(0)
    complete_list = [random.uniform(1.0, 3.0) for _ in range(3000)]
    t = time.time()
    values = single_process(complete_list)
    print(time.time() - t, values[0:10], values[-10:-1])

    t = time.time()
    values = multi_process(complete_list)
    print(time.time() - t, values[0:10], values[-10:-1])


# required for Windows:
if __name__ == '__main__':
    main()

打印：

4.181002378463745 [1236, 1491, 1464, 1477, 1494, 1472, 1410, 1450, 1502, 1537] [1485, 1513, 1513, 1501, 1283, 1538, 804, 1459, 1457]
1.325998067855835 [1236, 1491, 1464, 1477, 1494, 1472, 1410, 1450, 1502, 1537] [1485, 1513, 1513, 1501, 1283, 1538, 804, 1459, 1457]

结果

对于案例 1，加速比为 3.86，对于案例 2，加速比仅为 3.14。