python - 如何在 Python 中的线程之间共享数组索引？

Question

我有以下代码:

def task1():
    for url in splitarr[0]:
        print(url) #these are supposed to be scrape_induvidual_page() . print is just for debugging
def task2():
    for url in splitarr[1]:
        print(url)
def task3():
    for url in splitarr[2]:
        print(url)
def task4():
    for url in splitarr[3]:
        print(url)
def task5():
    for url in splitarr[4]:
        print(url)
def task6():
    for url in splitarr[5]:
        print(url)
def task7():
    for url in splitarr[6]:
        print(url)     
def task8():
    for url in splitarr[7]:
        print(url)   

splitarr=np.array_split(urllist, 8)
t1 = threading.Thread(target=task1, name='t1') 
t2 = threading.Thread(target=task2, name='t2')   
t3 = threading.Thread(target=task3, name='t3')
t4 = threading.Thread(target=task4, name='t4') 
t5 = threading.Thread(target=task5, name='t5')
t6 = threading.Thread(target=task6, name='t6')
t7 = threading.Thread(target=task7, name='t7')
t8 = threading.Thread(target=task8, name='t8')

t1.start() 
t2.start()
t3.start() 
t4.start() 
t5.start()
t6.start() 
t7.start()
t8.start() 

t1.join()
t2.join()
t3.join()
t4.join()
t5.join()
t6.join()
t7.join() 
t8.join() 

它确实具有所需的输出，没有重复或任何东西

https://kickasstorrents.to/big-buck-bunny-1080p-h264-aac-5-1-tntvillage-t115783.html
https://kickasstorrents.to/big-buck-bunny-4k-uhd-hfr-60fps-eng-flac-webdl-2160p-x264-zmachine-t1041079.html
https://kickasstorrents.to/big-buck-bunny-4k-uhd-hfr-60-fps-flac-webrip-2160p-x265-zmachine-t1041689.html
https://kickasstorrents.to/big-buck-bunny-2008-720p-bluray-x264-don-no-rars-t11623.html
https://kickasstorrents.to/tkillaahh-big-buck-bunny-dvd-720p-2lions-team-t87503.html
https://kickasstorrents.to/big-buck-bunny-2008-720p-bluray-nhd-x264-nhanc3-t127050.html
https://kickasstorrents.to/big-buck-bunny-2008-brrip-720p-x264-mitzep-t172753.html

但是，我觉得代码有点多余，所有重复的 def taskx():
所以我尝试通过使用单个任务来压缩代码:

x=0
def task1():
    global x
    for url in splitarr[x]:
        print(url)
        x=x+1
t1 = threading.Thread(target=task1, name='t1') 
t2 = threading.Thread(target=task1, name='t2')   
t3 = threading.Thread(target=task1, name='t3')
t4 = threading.Thread(target=task1, name='t4') 
t5 = threading.Thread(target=task1, name='t5')
t6 = threading.Thread(target=task1, name='t6')
t7 = threading.Thread(target=task1, name='t7')
t8 = threading.Thread(target=task1, name='t8')

t1.start() 
t2.start()
t3.start() 
t4.start() 
t5.start()
t6.start() 
t7.start()
t8.start() 

t1.join()
t2.join()
t3.join()
t4.join()
t5.join()
t6.join()
t7.join() 
t8.join() 

但是，这会产生带有重复项的不良输出:

https://kickasstorrents.to/big-buck-bunny-1080p-h264-aac-5-1-tntvillage-t115783.html
https://kickasstorrents.to/big-buck-bunny-1080p-h264-aac-5-1-tntvillage-t115783.html
https://kickasstorrents.to/big-buck-bunny-4k-uhd-hfr-60-fps-flac-webrip-2160p-x265-zmachine-t1041689.html
https://kickasstorrents.to/big-buck-bunny-2008-720p-bluray-x264-don-no-rars-t11623.html
https://kickasstorrents.to/big-buck-bunny-2008-720p-bluray-x264-don-no-rars-t11623.html
https://kickasstorrents.to/tkillaahh-big-buck-bunny-dvd-720p-2lions-team-t87503.html
https://kickasstorrents.to/big-buck-bunny-2008-brrip-720p-x264-mitzep-t172753.html
https://kickasstorrents.to/big-buck-bunny-2008-brrip-720p-x264-mitzep-t172753.html

如何在具有多个线程的程序中正确地使 x 递增？

Answer 1

for url in splitarr[x]:为 splitarr[x] 中的序列创建一个迭代器.稍后增加 x 并不重要 - 迭代器已经构建。由于您在那里有打印，所有线程很可能都会捕获x当它仍然为零并迭代相同的序列时。
一种解决方案是通过 args 将递增值传递给 task1。 threading.Thread 中的参数.但是线程池更容易。

from multiprocessing.pool import ThreadPool

# generate test array
splitarr = []
for i in range(8):
    splitarr.append([f"url_{i}_{j}" for j in range(4)])

def task(splitarr_column):
    for url in splitarr_column:
        print(url)

with ThreadPool(len(splitarr)) as pool:
    result = pool.map(task, splitarr)

在本例中，len(splitarr)用于在 splitarr 中为每个序列创建一个线程.然后将这些序列中的每一个映射到 task功能。由于我们创建了正确数量的线程来处理所有序列，它们都同时运行。 map 完成后，with子句退出并且池关闭，终止线程。