我可能正在接近这一切,但这仍然是我所处的地方。我有非常大的日志文件,我试图搜索,在某些情况下高达30GB。我正在编写一个脚本来提取信息,并且一直在玩多进程以加快速度。现在我正在测试同时运行两个函数从顶部和底部搜索以获得结果,这似乎有效。我想知道是否有可能将一个功能从另一个功能中停止。例如,如果top函数找到结果,它们都会停止。这样我就可以根据需要进行构建。
from file_read_backwards import FileReadBackwards
from multiprocessing import Process
import sys
z = "log.log"
#!/usr/bin/env python
rocket = 0
def top():
target = "test"
with open(z) as src:
found= None
for line in src:
if len(line) == 0: break #happens at end of file, then stop loop
if target in line:
found= line
break
print(found)
def bottom():
target = "text"
with FileReadBackwards(z) as src:
found= None
for line in src:
if len(line) == 0: break #happens at end of file, then stop loop
if target in line:
found= line
break
print(found)
if __name__=='__main__':
p1 = Process(target = top)
p1.start()
p2 = Process(target = bottom)
p2.start()
答案 0 :(得分:2)
以下是我在评论中提到的方法的概念验证:
import os
import random
import sys
from multiprocessing import Process, Value
def search(proc_no, file_name, seek_to, max_size, find, flag):
stop_at = seek_to + max_size
with open(file_name) as f:
if seek_to:
f.seek(seek_to - 1)
prev_char = f.read(1)
if prev_char != '\n':
# Landed in the middle of a line. Skip back one (or
# maybe more) lines so this line isn't excluded. Start
# by seeking back 256 bytes, then 512 if necessary, etc.
exponent = 8
pos = seek_to
while pos >= seek_to:
pos = f.seek(max(0, pos - (2 ** exponent)))
f.readline()
pos = f.tell()
exponent += 1
while True:
if flag.value:
break
line = f.readline()
if not line:
break # EOF
data = line.strip()
if data == find:
flag.value = proc_no
print(data)
break
if f.tell() > stop_at:
break
if __name__ == '__main__':
# list.txt contains lines with the numbers 1 to 1000001
file_name = 'list.txt'
info = os.stat(file_name)
file_size = info.st_size
if len(sys.argv) == 1:
# Pick a random value from list.txt
num_lines = 1000001
choices = list(range(1, num_lines + 1))
choices.append('XXX')
find = str(random.choice(choices))
else:
find = sys.argv[1]
num_procs = 4
chunk_size, remainder = divmod(file_size, num_procs)
max_size = chunk_size + remainder
flag = Value('i', 0)
procs = []
print(f'Using {num_procs} processes to look for {find} in {file_name}')
for i in range(num_procs):
seek_to = i * chunk_size
proc = Process(target=search, args=(i + 1, file_name, seek_to, max_size, find, flag))
procs.append(proc)
for proc in procs:
proc.start()
for proc in procs:
proc.join()
if flag.value:
print(find, 'found by proc', flag.value)
else:
print(find, 'not found')
在阅读了有关使用多处理和多线程读取文件的各种帖子[1]后,似乎由于潜在的磁盘抖动和序列化读取,这两种方法都不是很好的方法。所以这里有一个不同的,更简单的方法方式更快(至少对于我试用它的一百万行的文件):
import mmap
import sys
def search_file(file_name, text, encoding='utf-8'):
text = text.encode(encoding)
with open(file_name) as f:
with mmap.mmap(f.fileno(), 0, flags=mmap.ACCESS_READ, prot=mmap.PROT_READ) as m:
index = m.find(text)
if index > -1:
# Found a match; now find beginning of line that
# contains match so we can grab the whole line.
while index > 0:
index -= 1
if m[index] == 10:
index += 1
break
else:
index = 0
m.seek(index)
line = m.readline()
return line.decode(encoding)
if __name__ == '__main__':
file_name, search_string = sys.argv[1:]
line = search_file(file_name, search_string)
sys.stdout.write(line if line is not None else f'Not found in {file_name}: {search_string}\n')
我很好奇这将如何使用30GB的日志文件。
[1]包括this one
答案 1 :(得分:0)
使用multiprocessing.Pool
和回调函数的简单示例。
结果返回后终止剩余的池进程。
您可以使用此方法添加任意数量的进程以从文件中的不同偏移进行搜索。
import math
import time
from multiprocessing import Pool
from random import random
def search(pid, wait):
"""Sleep for wait seconds, return PID
"""
time.sleep(wait)
return pid
def done(result):
"""Do something with result and stop other processes
"""
print("Process: %d done." % result)
pool.terminate()
print("Terminate Pool")
pool = Pool(2)
pool.apply_async(search, (1, math.ceil(random() * 3)), callback=done)
pool.apply_async(search, (2, math.ceil(random() * 3)), callback=done)
# do other stuff ...
# Wait for result
pool.close()
pool.join() # block our main thread
答案 2 :(得分:0)
这与Blurp的答案基本相同,但我缩短了它并使其更加通用。你可以看到top应该是一个无限循环,但是bottom会立即停止顶部。
from multiprocessing import Process
valNotFound = True
def top():
i=0
while ValNotFound:
i += 1
def bottom():
ValNotFound = False
p1 = Process(target = top)
p2 = Process(target = bottom)
p1.start()
p2.start()