获取文件最后 n 行，类似于 tail-IT科技

摘要：问题描述：我正在为 Web 应用程序编写日志文件查看器，我想对日志文件的行进行分页。文件中的项目按行显示，最新项目位于底部。因此我需要一种tail()可以n从底部读取行并支持偏移的方法。这就是我想到的：def tail(f, n, offset=0): """Reads ...

问题描述：

我正在为 Web 应用程序编写日志文件查看器，我想对日志文件的行进行分页。文件中的项目按行显示，最新项目位于底部。

因此我需要一种tail()可以n从底部读取行并支持偏移的方法。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

这是一种合理的方法吗？推荐使用偏移量来跟踪日志文件的方法是什么？

解决方案 1：

这可能比你的更快。不对行长做任何假设。每次返回一个块，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('
')
        lines_to_go -= lines_found
        block_end_byte -= BLOCK_SIZE
        block_number -= 1
    all_read_text = ''.join(reversed(blocks))
    return '
'.join(all_read_text.splitlines()[-total_lines_wanted:])

我不喜欢对行长做出棘手的假设，因为实际情况是，你永远不可能知道这样的事情。

通常，这将在循环的第一遍或第二遍中找到最后 20 行。如果您的 74 个字符确实准确，则将块大小设为 2048，然后几乎立即找到 20 行。

此外，我并不会花太多脑力来尝试巧妙地与物理操作系统块对齐。使用这些高级 I/O 包，我怀疑您不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，那么您可能会看到速度提升。

更新

对于 Python 3.2 及更高版本，按照文本文件（在模式字符串中没有“b”的打开文件）中对字节进行的处理，只允许相对于文件开头进行搜索（例外是使用 seek(0, 2) 搜索到文件末尾）。：

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'
')
        lines_to_go -= lines_found
        block_end_byte -= BLOCK_SIZE
        block_number -= 1
    all_read_text = b''.join(reversed(blocks))
    return b'
'.join(all_read_text.splitlines()[-total_lines_wanted:])

解决方案 2：

假设在 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 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]

解决方案 3：

这是我的答案。纯 Python。使用 timeit 似乎非常快。跟踪包含 100,000 行的日志文件中的 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

以下是代码：

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:]

解决方案 4：

如果可以接受读取整个文件，那么使用双端队列。

from collections import deque
deque(f, maxlen=n)

在 2.6 之前，deques 没有 maxlen 选项，但是它很容易实现。

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

如果要求从末尾读取文件，则使用疾驰（又称指数）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:]

解决方案 5：

上面的 S.Lott 的回答对我来说几乎有用，但最终只给了我部分行。事实证明，它会破坏块边界上的数据，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('
')
        size -= linesFound
        bytes -= BUFSIZ
        block -= 1
    return ''.join(data).splitlines()[-window:]

解决方案 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], \n                   len(lines) > to_read or pos > 0
        avg_line_length *= 1.3

解决方案 7：

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 -= 1
                    if n == -1:
                        break
            return fm[i + 1 if i else 0:].splitlines()
        finally:
            fm.close()

解决方案 8：

将@papercrane 解决方案更新为 python3。使用以下命令打开文件open(filename, 'rb')：

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('
')
        remaining_bytes -= BUFSIZ
        block -= 1

    return ''.join(data).splitlines()[-window:]

解决方案 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)

解决方案 10：

按照评论者的要求，针对我对类似问题的回答发布一个答案，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这是执行此操作的最佳方法。为了改进现有mmap答案，此版本可在 Windows 和 Linux 之间移植，并且运行速度应该更快（尽管如果不对 32 位 Python 上的 GB 范围内的文件进行一些修改，它将无法工作，请参阅其他答案以获取有关处理此问题的提示，以及如何进行修改以在 Python 2 上工作）。

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'
', 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 
 newlines need to be normalized to 
, and input
        # is ASCII compatible, can normalize newlines with:
        # return mm[startofline:endoflines].replace(os.linesep.encode('ascii'), b'
').splitlines(True)

这假设尾随的行数足够小，您可以安全地将它们全部一次读入内存；您也可以将其设为生成器函数，并通过将最后一行替换为：手动一次读取一行：

        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str出行（Py2）和bytes行（Py3）；如果您想要unicode（Py2）或str（Py3），则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, '
') for line in lines)
        # Python 3.2 and earlier:
        for line in lines:
            yield line
        # 3.3+:
        yield from lines

注意：我在一台无法使用 Python 进行测试的机器上输入了这些内容。如果我打错了什么，请告诉我；这与我的其他答案非常相似，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。

解决方案 11：

一个更干净的 python3 兼容版本，它不会插入但会附加和反转：

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'
')
        end -= nread
    return b'
'.join(b''.join(reversed(data)).splitlines()[-window:])

像这样使用：

with open(path, 'rb') as f:
    last_lines = tail(f, 3).decode('utf-8')

解决方案 12：

根据 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('
')
    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。

解决方案 13：

pypi 上有一些现有的 tail 实现，你可以使用 pip 安装：

工具软件
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。

解决方案 14：

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)

解决方案 15：

简单的：

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

解决方案 16：

为了提高处理非常大的文件的效率（在日志文件情况下，您可能希望使用 tail），您通常希望避免读取整个文件（即使您确实这样做了，而不是一次将整个文件读入内存）但是，您确实需要以某种方式计算出行而不是字符的偏移量。一种可能性是使用 seek() 逐个字符向后读取，但这非常慢。相反，最好以更大的块进行处理。

我之前写过一个实用函数，可以在这里反向读取文件。

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（避免需要逆转两次）

解决方案 17：

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

def readline_backwards(self, f):
    backline = ''
    last = ''
    while not last == '
':
        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 == '
':
        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

解决方案 18：

根据 Eyecue 的回答 (2010 年 6 月 10 日 21:28)：此类向文件对象添加 head() 和 tail() 方法。

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('
')
        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)

解决方案 19：

如果文件不是以 \n 结尾或者无法确保读取完整的第一行，则这些解决方案中的几个都会出现问题。

def tail(file, n=1, bs=1024):
    f = open(file)
    f.seek(-1,2)
    l = 1-f.read(1).count('
') # If file doesn't end in 
, 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('
')
    f.seek(B, 0)
    l = min(l,n) # discard first (incomplete) line if l > n
    lines = f.readlines()[-l:]
    f.close()
    return lines

解决方案 20：

我必须从文件的最后一行读取特定值，然后偶然发现了这个线程。我没有用 Python 重新发明轮子，而是最终得到了一个很小的 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"))

解决方案 21：

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。这个示例很通用：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)

解决方案 22：

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

解决方案 23：

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('
') < 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()

解决方案 24：

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)

解决方案 25：

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N后面的行，而且非常高效。

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))

解决方案 26：

另一个解决方案

如果你的 txt 文件如下所示：老鼠蛇猫蜥蜴狼狗

您可以通过简单地使用 python 中的数组索引来反转此文件'''

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

解决方案 27：

我发现上面的 Popen 是最好的解决方案。它快速而简单，而且有效。对于 Unix 机器上的 Python 2.6，我使用了以下方法

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。要逐行遍历 soutput，请执行以下操作：

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

解决方案 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

解决方案 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]))

解决方案 30：

好吧！我遇到了类似的问题，尽管我只需要LAST LINE ONLY，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] != '
'.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，它花费很少的时间来找到最后一行（甚至不到半秒）