获取文件的最后n行，类似于tail

Question

我正在为 Web 应用程序编写一个日志文件查看器，并且我想对日志文件的行进行分页。 文件中的项目基于行，最新项目位于底部。

所以我需要一个 tail() 方法，它可以从底部读取 n 行并支持偏移量。 这就是我想到的：

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？ 使用偏移量跟踪日志文件的推荐方法是什么？

Answer 1

这可能比你的更快。 不对线路长度做出任何假设。 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:])

我不喜欢关于行长度的棘手假设，因为实际上，你永远无法知道这样的事情。

一般来说，这将在第一次或第二次循环中找到最后 20 行。 如果你的 74 个字符的东西实际上是准确的，那么你将块大小设置为 2048，你几乎会立即尾随 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进行巧妙的对齐。 使用这些高级 I/O 包，我怀疑您会看到尝试在操作系统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，请按照字节处理进行操作 在文本文件中（模式字符串中没有 "b" 打开的文件），仅查找相对值允许到文件开头（例外情况是使用seek(0, 2) 查找到文件末尾）。：

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:])

Answer 2

假设 Python 2 上有一个类 UNIX 系统，您可以执行以下操作：

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，您可以执行以下操作：

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]

Answer 3

这是我的答案。 纯蟒蛇。 使用 timeit 看起来相当快。 对包含 100,000 行的日志文件进行尾部 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

代码如下：

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:]

Answer 4

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，则使用快速（也称为指数）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:]

Answer 5

S.Lott 上面的答案几乎对我有用，但最终给了我部分台词。 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。 当调用 ''.join(data) 时，块的顺序是错误的。 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:]

Answer 6

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

Answer 7

使用 mmap 简单快速的解决方案：

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()

Answer 8

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:]

Answer 9

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)

Answer 10

应评论者的要求发布答案我对类似问题的回答，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。

对于较大大小的文件， mmap 为做到这一点的最好方法。 为了改进现有的 mmap 答案，这个版本可以在 Windows 和 Linux 之间移植，并且运行速度应该更快（尽管如果不进行一些修改，它就无法在 32 位 Python 上使用 GB 范围内的文件运行，请参阅有关处理此问题的提示以及修改以在 Python 2 上工作的提示的其他答案）。

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)

这假设尾部的行数足够小，您可以安全地将它们一次性全部读入内存； 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一次手动读取一行：

        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（需要使用 mmap），因此它给出 str 行 (Py2) 和 bytes 行 (Py3)； 如果您想要 unicode (Py2) 或 str (Py3)，可以调整迭代方法来为您解码和/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：我把这一切都输入了在我无法访问 Python 进行测试的机器上。 如果我输入了什么内容，请告诉我； 这与我的其他答案非常相似，我认为它应该有效，但是需要进行一些调整（例如处理偏移量）可能会导致微妙的错误。 如果有任何错误，请在评论中告诉我。

Answer 11

一个更干净的 python3 兼容版本，不插入但附加 & 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')

Answer 12

简单的 ：

with open("test.txt") as f:
data = f.readlines()
tail = data[-2:]
print(''.join(tail)

Answer 13

基于 S.Lott 的最高投票答案（2008 年 9 月 25 日 21:43），但针对小文件进行了修复。

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

希望这有用。

Answer 14

pypi 上有一些现有的 tail 实现，您可以使用 pip 安装：

• mtFileUtil
• multitail
• log4tailer
• ...

根据您的情况，使用这些现有工具之一可能会有优势。

Answer 15

有一个非常有用的 模块 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)

Answer 16

我发现上面的 Popen 是最好的解决方案。 它又快又脏而且有效
对于 Unix 机器上的 python 2.6，我使用以下

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 将包含代码的最后 n 行。 要逐行迭代 soutput，请执行以下操作：

for line in GetLastNLines(50,'myfile.log').split('\n'):
    print line

Answer 17

为了提高非常大文件的效率（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（即使您这样做时没有立即将整个文件读入内存）但是，您这样做需要以某种方式计算出行而不是字符的偏移量。 一种可能是使用seek()逐个字符地向后读取，但这非常慢。 相反，最好以更大的块进行处理。

我不久前编写了一个实用函数来向后读取文件，可以在此处使用。

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）

Answer 18

您可以使用 f.seek(0, 2) 转到文件末尾，然后使用以下 readline() 替换来逐行读取：

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

Answer 19

基于 Eyecue 答案（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)

Answer 20

如果文件不以 \n 结尾或确保读取完整的第一行，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

Answer 21

我必须从文件的最后一行读取特定值，并偶然发现了这个线程。 我没有用 Python 重新发明轮子，而是得到了一个很小的 ​​shell 脚本，另存为
/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"))

Answer 22

这不是第一个使用双端队列的示例，而是一个更简单的示例。 这是通用的：它适用于任何可迭代对象，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)

Answer 23

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

Answer 24

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()

Answer 25

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) 

Answer 26

更新 A.Coady 给出的答案

适用于 python 3。

这使用指数搜索并且只会缓冲前后的N行非常高效。

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))

Answer 27

另一种解决方案

如果您的 txt 文件如下所示，则 ：
老鼠
蛇
猫
蜥蜴
狼
狗

你可以通过简单地使用Python中的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

Answer 28

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

Answer 29

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]))

Answer 30

出色地！ 我遇到了类似的问题，尽管我只需要仅最后一行，
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()

这个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，找到最后一行花费的时间非常少（甚至不到半秒）