ris*_*sen 15 python queue pool multiprocessing
我有成千上万的文本文件,我想以各种方式解析.我想将输出保存到单个文件而不会出现同步问题.我一直在使用多处理池来节省时间,但我无法弄清楚如何组合池和队列.
以下代码将保存infile名称以及文件中连续"x"的最大数量.但是,我希望所有进程将结果保存到同一个文件,而不是像我的示例中那样保存到不同的文件.任何有关这方面的帮助将不胜感激.
import multiprocessing
with open('infilenamess.txt') as f:
filenames = f.read().splitlines()
def mp_worker(filename):
with open(filename, 'r') as f:
text=f.read()
m=re.findall("x+", text)
count=len(max(m, key=len))
outfile=open(filename+'_results.txt', 'a')
outfile.write(str(filename)+'|'+str(count)+'\n')
outfile.close()
def mp_handler():
p = multiprocessing.Pool(32)
p.map(mp_worker, filenames)
if __name__ == '__main__':
mp_handler()
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tde*_*ney 32
多处理池为您实现队列.只需使用池方法将工作者返回值返回给调用者.imap效果很好:
import multiprocessing
import re
def mp_worker(filename):
with open(filename) as f:
text = f.read()
m = re.findall("x+", text)
count = len(max(m, key=len))
return filename, count
def mp_handler():
p = multiprocessing.Pool(32)
with open('infilenamess.txt') as f:
filenames = [line for line in (l.strip() for l in f) if line]
with open('results.txt', 'w') as f:
for result in p.imap(mp_worker, filenames):
# (filename, count) tuples from worker
f.write('%s: %d\n' % result)
if __name__=='__main__':
mp_handler()
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我接受了可接受的答案,并对其进行了简化,以使我对它的工作原理有所了解。我将其张贴在这里,以防其他人使用。
import multiprocessing
def mp_worker(number):
number += 1
return number
def mp_handler():
p = multiprocessing.Pool(32)
numbers = list(range(1000))
with open('results.txt', 'w') as f:
for result in p.imap(mp_worker, numbers):
f.write('%d\n' % result)
if __name__=='__main__':
mp_handler()
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