我正在为 Web 应用程序编写日志文件查看器,为此我想通过日志文件的行进行分页。文件中的项目是基于行的,最新的项目位于底部。
所以我需要一个可以从底部读取 n
行并支持偏移量的 tail()
方法。这是我想出的帽子:
def tail(f, n, offset=0):
"""Reads a n lines from f with an offset of offset lines."""
avg_line_length = 74
to_read = n + offset
while 1:
try:
f.seek(-(avg_line_length * to_read), 2)
except IOError:
# woops. apparently file is smaller than what we want
# to step back, go to the beginning instead
f.seek(0)
pos = f.tell()
lines = f.read().splitlines()
if len(lines) >= to_read or pos == 0:
return lines[-to_read:offset and -offset or None]
avg_line_length *= 1.3
这是一个合理的方法吗?使用偏移量尾随日志文件的推荐方法是什么?
seek(0,2)
然后 tell()
),并使用该值相对于开头进行搜索。
f
文件对象的 open
命令的参数,因为根据 f=open(..., 'rb')
或 f=open(..., 'rt')
必须以不同方式处理 f
这可能比你的更快。不对行长做任何假设。一次一个块地返回文件,直到找到正确数量的 '\n' 字符。
def tail( f, lines=20 ):
total_lines_wanted = lines
BLOCK_SIZE = 1024
f.seek(0, 2)
block_end_byte = f.tell()
lines_to_go = total_lines_wanted
block_number = -1
blocks = [] # blocks of size BLOCK_SIZE, in reverse order starting
# from the end of the file
while lines_to_go > 0 and block_end_byte > 0:
if (block_end_byte - BLOCK_SIZE > 0):
# read the last block we haven't yet read
f.seek(block_number*BLOCK_SIZE, 2)
blocks.append(f.read(BLOCK_SIZE))
else:
# file too small, start from begining
f.seek(0,0)
# only read what was not read
blocks.append(f.read(block_end_byte))
lines_found = blocks[-1].count('\n')
lines_to_go -= lines_found
block_end_byte -= BLOCK_SIZE
block_number -= 1
all_read_text = ''.join(reversed(blocks))
return '\n'.join(all_read_text.splitlines()[-total_lines_wanted:])
我不喜欢关于行长的棘手假设——实际上——你永远无法知道这样的事情。
通常,这将在第一次或第二次通过循环时定位最后 20 行。如果您的 74 个字符实际上是准确的,那么您将块大小设置为 2048,并且您几乎会立即拖尾 20 行。
此外,我不会消耗大量大脑卡路里来尝试与物理操作系统块对齐。使用这些高级 I/O 包,我怀疑您会看到尝试在 OS 块边界上对齐的任何性能后果。如果您使用较低级别的 I/O,那么您可能会看到加速。
更新
对于 Python 3.2 及更高版本,按照在文本文件中(那些在模式字符串中没有“b”打开的文件)中的字节处理过程,只允许相对于文件开头的搜索(例外是搜索到文件末尾与寻找(0, 2))。:
例如:f = open('C:/.../../apache_logs.txt', 'rb')
def tail(f, lines=20):
total_lines_wanted = lines
BLOCK_SIZE = 1024
f.seek(0, 2)
block_end_byte = f.tell()
lines_to_go = total_lines_wanted
block_number = -1
blocks = []
while lines_to_go > 0 and block_end_byte > 0:
if (block_end_byte - BLOCK_SIZE > 0):
f.seek(block_number*BLOCK_SIZE, 2)
blocks.append(f.read(BLOCK_SIZE))
else:
f.seek(0,0)
blocks.append(f.read(block_end_byte))
lines_found = blocks[-1].count(b'\n')
lines_to_go -= lines_found
block_end_byte -= BLOCK_SIZE
block_number -= 1
all_read_text = b''.join(reversed(blocks))
return b'\n'.join(all_read_text.splitlines()[-total_lines_wanted:])
假设您可以在 Python 2 上使用类似 unix 的系统:
import os
def tail(f, n, offset=0):
stdin,stdout = os.popen2("tail -n "+n+offset+" "+f)
stdin.close()
lines = stdout.readlines(); stdout.close()
return lines[:,-offset]
对于 python 3,你可以这样做:
import subprocess
def tail(f, n, offset=0):
proc = subprocess.Popen(['tail', '-n', n + offset, f], stdout=subprocess.PIPE)
lines = proc.stdout.readlines()
return lines[:, -offset]
offset_total = str(n+offset)
并替换此行 stdin,stdout = os.popen2("tail -n "+offset_total+" "+f)
以避免 TypeErrors (cannot concatenate int+str)
这是我的答案。纯蟒蛇。使用 timeit 似乎很快。拖尾具有 100,000 行的日志文件的 100 行:
>>> timeit.timeit('tail.tail(f, 100, 4098)', 'import tail; f = open("log.txt", "r");', number=10)
0.0014600753784179688
>>> timeit.timeit('tail.tail(f, 100, 4098)', 'import tail; f = open("log.txt", "r");', number=100)
0.00899195671081543
>>> timeit.timeit('tail.tail(f, 100, 4098)', 'import tail; f = open("log.txt", "r");', number=1000)
0.05842900276184082
>>> timeit.timeit('tail.tail(f, 100, 4098)', 'import tail; f = open("log.txt", "r");', number=10000)
0.5394978523254395
>>> timeit.timeit('tail.tail(f, 100, 4098)', 'import tail; f = open("log.txt", "r");', number=100000)
5.377126932144165
这是代码:
import os
def tail(f, lines=1, _buffer=4098):
"""Tail a file and get X lines from the end"""
# place holder for the lines found
lines_found = []
# block counter will be multiplied by buffer
# to get the block size from the end
block_counter = -1
# loop until we find X lines
while len(lines_found) < lines:
try:
f.seek(block_counter * _buffer, os.SEEK_END)
except IOError: # either file is too small, or too many lines requested
f.seek(0)
lines_found = f.readlines()
break
lines_found = f.readlines()
# we found enough lines, get out
# Removed this line because it was redundant the while will catch
# it, I left it for history
# if len(lines_found) > lines:
# break
# decrement the block counter to get the
# next X bytes
block_counter -= 1
return lines_found[-lines:]
if len(lines_found) > lines:
真的有必要吗? loop
条件不会也捕获它吗?
os.SEEK_END
是否只是为了清楚起见?据我发现,它的值是恒定的(= 2)。我想知道是否可以省略import os
。感谢您的出色解决方案!
while len(lines_found) < lines
更改为 while len(lines_found) <= lines
。谢谢!
如果可以接受读取整个文件,则使用双端队列。
from collections import deque
deque(f, maxlen=n)
在 2.6 之前,deques 没有 maxlen 选项,但它很容易实现。
import itertools
def maxque(items, size):
items = iter(items)
q = deque(itertools.islice(items, size))
for item in items:
del q[0]
q.append(item)
return q
如果需要从末尾读取文件,则使用疾驰(又名指数)搜索。
def tail(f, n):
assert n >= 0
pos, lines = n+1, []
while len(lines) <= n:
try:
f.seek(-pos, 2)
except IOError:
f.seek(0)
break
finally:
lines = list(f)
pos *= 2
return lines[-n:]
pos *= 2
似乎完全随意。它的意义是什么?
S.Lott 上面的回答几乎对我有用,但最终给了我部分内容。事实证明,它破坏了块边界上的数据,因为数据以相反的顺序保存读取的块。调用 ''.join(data) 时,块的顺序错误。这解决了这个问题。
def tail(f, window=20):
"""
Returns the last `window` lines of file `f` as a list.
f - a byte file-like object
"""
if window == 0:
return []
BUFSIZ = 1024
f.seek(0, 2)
bytes = f.tell()
size = window + 1
block = -1
data = []
while size > 0 and bytes > 0:
if bytes - BUFSIZ > 0:
# Seek back one whole BUFSIZ
f.seek(block * BUFSIZ, 2)
# read BUFFER
data.insert(0, f.read(BUFSIZ))
else:
# file too small, start from begining
f.seek(0,0)
# only read what was not read
data.insert(0, f.read(bytes))
linesFound = data[0].count('\n')
size -= linesFound
bytes -= BUFSIZ
block -= 1
return ''.join(data).splitlines()[-window:]
我最终使用的代码。我认为这是迄今为止最好的:
def tail(f, n, offset=None):
"""Reads a n lines from f with an offset of offset lines. The return
value is a tuple in the form ``(lines, has_more)`` where `has_more` is
an indicator that is `True` if there are more lines in the file.
"""
avg_line_length = 74
to_read = n + (offset or 0)
while 1:
try:
f.seek(-(avg_line_length * to_read), 2)
except IOError:
# woops. apparently file is smaller than what we want
# to step back, go to the beginning instead
f.seek(0)
pos = f.tell()
lines = f.read().splitlines()
if len(lines) >= to_read or pos == 0:
return lines[-to_read:offset and -offset or None], \
len(lines) > to_read or pos > 0
avg_line_length *= 1.3
使用 mmap 的简单快速解决方案:
import mmap
import os
def tail(filename, n):
"""Returns last n lines from the filename. No exception handling"""
size = os.path.getsize(filename)
with open(filename, "rb") as f:
# for Windows the mmap parameters are different
fm = mmap.mmap(f.fileno(), 0, mmap.MAP_SHARED, mmap.PROT_READ)
try:
for i in xrange(size - 1, -1, -1):
if fm[i] == '\n':
n -= 1
if n == -1:
break
return fm[i + 1 if i else 0:].splitlines()
finally:
fm.close()
.rfind
方法向后扫描换行符,而不是在 Python 级别执行一次字节检查;在 CPython 中,替换 Python具有 C 内置调用的级别代码通常会胜出很多)。对于较小的输入,带有 maxlen
的 deque
更简单,并且可能同样快。
将@papercrane 解决方案更新为 python3。使用 open(filename, 'rb')
打开文件并:
def tail(f, window=20):
"""Returns the last `window` lines of file `f` as a list.
"""
if window == 0:
return []
BUFSIZ = 1024
f.seek(0, 2)
remaining_bytes = f.tell()
size = window + 1
block = -1
data = []
while size > 0 and remaining_bytes > 0:
if remaining_bytes - BUFSIZ > 0:
# Seek back one whole BUFSIZ
f.seek(block * BUFSIZ, 2)
# read BUFFER
bunch = f.read(BUFSIZ)
else:
# file too small, start from beginning
f.seek(0, 0)
# only read what was not read
bunch = f.read(remaining_bytes)
bunch = bunch.decode('utf-8')
data.insert(0, bunch)
size -= bunch.count('\n')
remaining_bytes -= BUFSIZ
block -= 1
return ''.join(data).splitlines()[-window:]
assert "b" in file.mode, "File mode must be bytes!"
以检查文件模式是否实际上是字节。
最简单的方法是使用 deque
:
from collections import deque
def tail(filename, n=10):
with open(filename) as f:
return deque(f, n)
应评论者的要求在 my answer to a similar question 上发布答案,其中使用相同的技术来改变文件的最后一行,而不仅仅是获取它。
对于较大的文件,mmap
是执行此操作的最佳方式。为了改进现有的 mmap
答案,此版本可在 Windows 和 Linux 之间移植,并且应该运行得更快(尽管它不会在 32 位 Python 上使用 GB 范围内的文件进行一些修改,请参阅 other answer for hints on handling this, and for modifying to work on Python 2) .
import io # Gets consistent version of open for both Py2.7 and Py3.x
import itertools
import mmap
def skip_back_lines(mm, numlines, startidx):
'''Factored out to simplify handling of n and offset'''
for _ in itertools.repeat(None, numlines):
startidx = mm.rfind(b'\n', 0, startidx)
if startidx < 0:
break
return startidx
def tail(f, n, offset=0):
# Reopen file in binary mode
with io.open(f.name, 'rb') as binf, mmap.mmap(binf.fileno(), 0, access=mmap.ACCESS_READ) as mm:
# len(mm) - 1 handles files ending w/newline by getting the prior line
startofline = skip_back_lines(mm, offset, len(mm) - 1)
if startofline < 0:
return [] # Offset lines consumed whole file, nothing to return
# If using a generator function (yield-ing, see below),
# this should be a plain return, no empty list
endoflines = startofline + 1 # Slice end to omit offset lines
# Find start of lines to capture (add 1 to move from newline to beginning of following line)
startofline = skip_back_lines(mm, n, startofline) + 1
# Passing True to splitlines makes it return the list of lines without
# removing the trailing newline (if any), so list mimics f.readlines()
return mm[startofline:endoflines].splitlines(True)
# If Windows style \r\n newlines need to be normalized to \n, and input
# is ASCII compatible, can normalize newlines with:
# return mm[startofline:endoflines].replace(os.linesep.encode('ascii'), b'\n').splitlines(True)
这假设拖尾的行数足够小,您可以一次安全地将它们全部读入内存;您还可以将其设为生成器函数,并通过将最后一行替换为以下内容来手动读取一行:
mm.seek(startofline)
# Call mm.readline n times, or until EOF, whichever comes first
# Python 3.2 and earlier:
for line in itertools.islice(iter(mm.readline, b''), n):
yield line
# 3.3+:
yield from itertools.islice(iter(mm.readline, b''), n)
最后,以二进制模式读取(必须使用 mmap
),因此它给出 str
行(Py2)和 bytes
行(Py3);如果您想要 unicode
(Py2) 或 str
(Py3),可以调整迭代方法以为您解码和/或修复换行符:
lines = itertools.islice(iter(mm.readline, b''), n)
if f.encoding: # Decode if the passed file was opened with a specific encoding
lines = (line.decode(f.encoding) for line in lines)
if 'b' not in f.mode: # Fix line breaks if passed file opened in text mode
lines = (line.replace(os.linesep, '\n') for line in lines)
# Python 3.2 and earlier:
for line in lines:
yield line
# 3.3+:
yield from lines
注意:我在无法访问 Python 进行测试的机器上输入了这一切。如果我输入任何错误,请告诉我;这与 my other answer 非常相似,我认为它应该可以工作,但调整(例如处理 offset
)可能会导致细微的错误。如果有任何错误,请在评论中告诉我。
一个更干净的 python3 兼容版本,它不插入但追加和反转:
def tail(f, window=1):
"""
Returns the last `window` lines of file `f` as a list of bytes.
"""
if window == 0:
return b''
BUFSIZE = 1024
f.seek(0, 2)
end = f.tell()
nlines = window + 1
data = []
while nlines > 0 and end > 0:
i = max(0, end - BUFSIZE)
nread = min(end, BUFSIZE)
f.seek(i)
chunk = f.read(nread)
data.append(chunk)
nlines -= chunk.count(b'\n')
end -= nread
return b'\n'.join(b''.join(reversed(data)).splitlines()[-window:])
像这样使用它:
with open(path, 'rb') as f:
last_lines = tail(f, 3).decode('utf-8')
基于 S.Lott 的最高投票答案(2008 年 9 月 25 日 21:43),但对于小文件是固定的。
def tail(the_file, lines_2find=20):
the_file.seek(0, 2) #go to end of file
bytes_in_file = the_file.tell()
lines_found, total_bytes_scanned = 0, 0
while lines_2find+1 > lines_found and bytes_in_file > total_bytes_scanned:
byte_block = min(1024, bytes_in_file-total_bytes_scanned)
the_file.seek(-(byte_block+total_bytes_scanned), 2)
total_bytes_scanned += byte_block
lines_found += the_file.read(1024).count('\n')
the_file.seek(-total_bytes_scanned, 2)
line_list = list(the_file.readlines())
return line_list[-lines_2find:]
#we read at least 21 line breaks from the bottom, block by block for speed
#21 to ensure we don't get a half line
希望这是有用的。
pypi 上有一些现有的 tail 实现,您可以使用 pip 安装它们:
mtFileUtil
多尾
log4tailer
...
根据您的情况,使用这些现有工具之一可能会有优势。
tailhead
、tailer
,但它们不起作用。还尝试了 mtFileUtil
。它最初抛出错误,因为 print
语句没有括号(我在 Python 3.6 上)。我在 reverse.py
中添加了这些并且错误消息消失了,但是当我的脚本调用模块 (mtFileUtil.tail(open(logfile_path), 5)
) 时,它不会打印任何内容。
简单的 :
with open("test.txt") as f:
data = f.readlines()
tail = data[-2:]
print(''.join(tail)
我发现上面的 Popen 是最好的解决方案。它又快又脏,它适用于 Unix 机器上的 python 2.6 我使用了以下
def GetLastNLines(self, n, fileName):
"""
Name: Get LastNLines
Description: Gets last n lines using Unix tail
Output: returns last n lines of a file
Keyword argument:
n -- number of last lines to return
filename -- Name of the file you need to tail into
"""
p = subprocess.Popen(['tail','-n',str(n),self.__fileName], stdout=subprocess.PIPE)
soutput, sinput = p.communicate()
return soutput
soutput 将包含最后 n 行代码。逐行遍历 soutput:
for line in GetLastNLines(50,'myfile.log').split('\n'):
print line
为了提高文件的效率(在您可能想要使用 tail 的日志文件情况下很常见),您通常希望避免读取整个文件(即使您这样做而不是一次将整个文件读入内存)但是,您这样做需要以某种方式计算行而不是字符的偏移量。一种可能性是用 seek() 逐个字符向后读取,但这非常慢。相反,最好在更大的块中处理。
我有一个我不久前编写的实用程序函数,用于向后读取可以在这里使用的文件。
import os, itertools
def rblocks(f, blocksize=4096):
"""Read file as series of blocks from end of file to start.
The data itself is in normal order, only the order of the blocks is reversed.
ie. "hello world" -> ["ld","wor", "lo ", "hel"]
Note that the file must be opened in binary mode.
"""
if 'b' not in f.mode.lower():
raise Exception("File must be opened using binary mode.")
size = os.stat(f.name).st_size
fullblocks, lastblock = divmod(size, blocksize)
# The first(end of file) block will be short, since this leaves
# the rest aligned on a blocksize boundary. This may be more
# efficient than having the last (first in file) block be short
f.seek(-lastblock,2)
yield f.read(lastblock)
for i in range(fullblocks-1,-1, -1):
f.seek(i * blocksize)
yield f.read(blocksize)
def tail(f, nlines):
buf = ''
result = []
for block in rblocks(f):
buf = block + buf
lines = buf.splitlines()
# Return all lines except the first (since may be partial)
if lines:
result.extend(lines[1:]) # First line may not be complete
if(len(result) >= nlines):
return result[-nlines:]
buf = lines[0]
return ([buf]+result)[-nlines:]
f=open('file_to_tail.txt','rb')
for line in tail(f, 20):
print line
[编辑] 添加了更具体的版本(避免需要反转两次)
您可以使用 f.seek(0, 2) 转到文件末尾,然后使用以下 readline() 替换逐行读取:
def readline_backwards(self, f):
backline = ''
last = ''
while not last == '\n':
backline = last + backline
if f.tell() <= 0:
return backline
f.seek(-1, 1)
last = f.read(1)
f.seek(-1, 1)
backline = last
last = ''
while not last == '\n':
backline = last + backline
if f.tell() <= 0:
return backline
f.seek(-1, 1)
last = f.read(1)
f.seek(-1, 1)
f.seek(1, 1)
return backline
基于 Eyecue 的回答(2010 年 6 月 10 日 21:28):此类将 head() 和 tail() 方法添加到文件对象。
class File(file):
def head(self, lines_2find=1):
self.seek(0) #Rewind file
return [self.next() for x in xrange(lines_2find)]
def tail(self, lines_2find=1):
self.seek(0, 2) #go to end of file
bytes_in_file = self.tell()
lines_found, total_bytes_scanned = 0, 0
while (lines_2find+1 > lines_found and
bytes_in_file > total_bytes_scanned):
byte_block = min(1024, bytes_in_file-total_bytes_scanned)
self.seek(-(byte_block+total_bytes_scanned), 2)
total_bytes_scanned += byte_block
lines_found += self.read(1024).count('\n')
self.seek(-total_bytes_scanned, 2)
line_list = list(self.readlines())
return line_list[-lines_2find:]
用法:
f = File('path/to/file', 'r')
f.head(3)
f.tail(3)
如果文件不以 \n 结尾或确保读取完整的第一行,则其中一些解决方案会出现问题。
def tail(file, n=1, bs=1024):
f = open(file)
f.seek(-1,2)
l = 1-f.read(1).count('\n') # If file doesn't end in \n, count it anyway.
B = f.tell()
while n >= l and B > 0:
block = min(bs, B)
B -= block
f.seek(B, 0)
l += f.read(block).count('\n')
f.seek(B, 0)
l = min(l,n) # discard first (incomplete) line if l > n
lines = f.readlines()[-l:]
f.close()
return lines
这是一个非常简单的实现:
with open('/etc/passwd', 'r') as f:
try:
f.seek(0,2)
s = ''
while s.count('\n') < 11:
cur = f.tell()
f.seek((cur - 10))
s = f.read(10) + s
f.seek((cur - 10))
print s
except Exception as e:
f.readlines()
f.seek
之前 try 的用法?为什么不在 with open
之前?另外,为什么在 except
你做一个 f.readlines()
?
有一个非常有用的 module 可以做到这一点:
from file_read_backwards import FileReadBackwards
with FileReadBackwards("/tmp/file", encoding="utf-8") as frb:
# getting lines by lines starting from the last line up
for l in frb:
print(l)
更新A.Coady给出的答案
适用于 python 3。
这使用 Exponential Search 并且只会从后面缓冲 N
行并且非常有效。
import time
import os
import sys
def tail(f, n):
assert n >= 0
pos, lines = n+1, []
# set file pointer to end
f.seek(0, os.SEEK_END)
isFileSmall = False
while len(lines) <= n:
try:
f.seek(f.tell() - pos, os.SEEK_SET)
except ValueError as e:
# lines greater than file seeking size
# seek to start
f.seek(0,os.SEEK_SET)
isFileSmall = True
except IOError:
print("Some problem reading/seeking the file")
sys.exit(-1)
finally:
lines = f.readlines()
if isFileSmall:
break
pos *= 2
print(lines)
return lines[-n:]
with open("stream_logs.txt") as f:
while(True):
time.sleep(0.5)
print(tail(f,2))
我不得不从文件的最后一行读取一个特定的值,然后偶然发现了这个线程。我没有在 Python 中重新发明轮子,而是得到了一个很小的 shell 脚本,保存为 /usr/local/bin/get_last_netp:
#! /bin/bash
tail -n1 /home/leif/projects/transfer/export.log | awk {'print $14'}
在 Python 程序中:
from subprocess import check_output
last_netp = int(check_output("/usr/local/bin/get_last_netp"))
不是第一个使用双端队列的例子,而是一个更简单的例子。这是通用的:它适用于任何可迭代的对象,而不仅仅是文件。
#!/usr/bin/env python
import sys
import collections
def tail(iterable, N):
deq = collections.deque()
for thing in iterable:
if len(deq) >= N:
deq.popleft()
deq.append(thing)
for thing in deq:
yield thing
if __name__ == '__main__':
for line in tail(sys.stdin,10):
sys.stdout.write(line)
This is my version of tailf
import sys, time, os
filename = 'path to file'
try:
with open(filename) as f:
size = os.path.getsize(filename)
if size < 1024:
s = size
else:
s = 999
f.seek(-s, 2)
l = f.read()
print l
while True:
line = f.readline()
if not line:
time.sleep(1)
continue
print line
except IOError:
pass
import time
attemps = 600
wait_sec = 5
fname = "YOUR_PATH"
with open(fname, "r") as f:
where = f.tell()
for i in range(attemps):
line = f.readline()
if not line:
time.sleep(wait_sec)
f.seek(where)
else:
print line, # already has newline
import itertools
fname = 'log.txt'
offset = 5
n = 10
with open(fname) as f:
n_last_lines = list(reversed([x for x in itertools.islice(f, None)][-(offset+1):-(offset+n+1):-1]))
abc = "2018-06-16 04:45:18.68"
filename = "abc.txt"
with open(filename) as myFile:
for num, line in enumerate(myFile, 1):
if abc in line:
lastline = num
print "last occurance of work at file is in "+str(lastline)
另一种解决方案
如果你的 txt 文件看起来像这样:鼠标蛇猫蜥蜴狼狗
你可以通过简单地在 python ''' 中使用数组索引来反转这个文件
contents=[]
def tail(contents,n):
with open('file.txt') as file:
for i in file.readlines():
contents.append(i)
for i in contents[:n:-1]:
print(i)
tail(contents,-5)
结果:狗狼蜥蜴猫
出色地!我有一个类似的问题,虽然我只需要最后一行,所以我想出了自己的解决方案
def get_last_line(filepath):
try:
with open(filepath,'rb') as f:
f.seek(-1,os.SEEK_END)
text = [f.read(1)]
while text[-1] != '\n'.encode('utf-8') or len(text)==1:
f.seek(-2, os.SEEK_CUR)
text.append(f.read(1))
except Exception as e:
pass
return ''.join([t.decode('utf-8') for t in text[::-1]]).strip()
此函数返回文件中的最后一个字符串我有一个 1.27gb 的日志文件,找到最后一行所需的时间非常少(甚至不到半秒)
不定期副业成功案例分享
io.UnsupportedOperation: can't do nonzero end-relative seeks
我可以将偏移量更改为 0,但这违背了函数的目的。