如何使用多處理獲取數字列表中的最大數字

Question

我有一個隨機數列表，我想使用multiprocessing獲得最大的數字。

這是我用來生成列表的代碼：

import random
randomlist = []
for i in range(100000000):
    n = random.randint(1,30000000)
    randomlist.append(n)

要使用串行過程獲得最大數字：

import time

greatest = 0 # global variable

def f(n):
    global greatest
    if n>greatest:
        greatest = n

if __name__ == "__main__":
    global greatest

    t2 = time.time()
    greatest = 0

    for x in randomlist:
        f(x)    
    
    print("serial process took:", time.time()-t2)
    print("greatest = ", greatest)

這是我嘗試使用多處理獲得最大數量的嘗試：

from multiprocessing import Pool
import time

greatest = 0 # the global variable

def f(n):
    global greatest
    if n>greatest:
        greatest = n

if __name__ == "__main__":
    global greatest
    greatest = 0
    t1 = time.time()
    p = Pool() #(processes=3) 
    result = p.map(f,randomlist)
    p.close()
    p.join()
    print("pool took:", time.time()-t1)
    print("greatest = ", greatest)

這里的輸出是0，很明顯沒有全局變量。 如何在不影響性能的情況下解決此問題？

Answer 1

正如@Barmar建議，將您randomlist分割成塊，然后處理從每個塊，最后計算全球最大的局部最大local_maximum_list ：

import multiprocessing as mp
import numpy as np

CHUNKSIZE = 10000

def local_maximum(l):
    m = max(l)
    print(f"Local maximum: {m}")
    return m

if __name__ == '__main__':
    randomlist = np.random.randint(1, 30000000, 100000000)
    chunks = (randomlist[i:i+CHUNKSIZE]
                  for i in range(0, len(randomlist), CHUNKSIZE))

    with mp.Pool(mp.cpu_count()) as pool:
        local_maximum_list = pool.map(local_maximum, chunks)
    print(f"Global maximum: {max(local_maximum_list)}")