192

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。

所以我需要一种tail()可以n从底部读取行并支持偏移的方法。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?

4

34 回答 34

129

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:])
于 2008-09-25T21:43:18.447 回答
99

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]
于 2008-09-25T21:30:14.810 回答
38

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:]
于 2012-12-09T18:21:42.133 回答
35

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:]
于 2008-11-11T05:17:18.083 回答
26

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:]
于 2011-08-13T00:43:38.617 回答
23

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
于 2009-03-28T11:13:44.380 回答
14

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()
于 2011-07-25T09:18:10.030 回答
6

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)
于 2019-12-25T04:06:48.277 回答
5

应评论者的要求发布我对类似问题的回答的答案,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,并且应该运行得更快(尽管如果不对 32 位 Python 与 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,并用于修改以在 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'\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str出行(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;这与我认为应该可以使用的其他答案非常相似,但是调整(例如处理 an )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。offset

于 2015-12-01T20:34:50.440 回答
5

将@papercrane 解决方案更新为 python3。使用 and 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('\n')
        remaining_bytes -= BUFSIZ
        block -= 1

    return ''.join(data).splitlines()[-window:]
于 2017-08-30T12:40:01.197 回答
4

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')
于 2018-01-04T01:45:20.553 回答
2

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。

于 2010-06-10T21:28:20.193 回答
2

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
于 2012-04-16T13:26:32.587 回答
2

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。

于 2012-10-06T18:23:37.210 回答
2

简单的 :

with open("test.txt") as f:
data = f.readlines()
tail = data[-2:]
print(''.join(tail)
于 2019-05-15T03:51:54.393 回答
1

对于非常大的文件(在您可能想要使用 tail 的日志文件情况下很常见),为了提高效率,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)

于 2008-09-25T21:40:32.557 回答
1

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
于 2010-11-09T06:50:09.913 回答
1

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)
于 2011-01-20T19:29:37.357 回答
1

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
于 2011-04-12T16:14:42.537 回答
1

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()
于 2016-06-19T01:54:08.593 回答
1

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)
于 2018-12-18T23:49:41.803 回答
1

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))

于 2019-07-30T17:37:10.253 回答
0

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"))
于 2013-05-12T12:11:01.107 回答
0

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)
于 2013-05-12T12:38:55.377 回答
0
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
于 2014-04-25T10:24:36.673 回答
0
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
于 2014-08-22T15:52:53.540 回答
0
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]))
于 2018-03-03T21:54:03.857 回答
0
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) 
于 2018-06-17T06:40:53.000 回答
0

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

于 2019-10-02T17:59:49.903 回答
0

好!我有一个类似的问题,虽然我只需要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] != '\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)

于 2021-04-23T15:00:20.960 回答
-1

再想一想,这可能和这里的任何东西一样快。

def tail( f, window=20 ):
    lines= ['']*window
    count= 0
    for l in f:
        lines[count%window]= l
        count += 1
    print lines[count%window:], lines[:count%window]

这要简单得多。它似乎确实以良好的速度发展。

于 2008-09-25T22:03:09.833 回答
-1

我找到了一种可能最简单的方法来查找文件的第一行或最后 N 行

文件的最后 N 行(例如:N=10)

file=open("xyz.txt",'r")
liner=file.readlines()
for ran in range((len(liner)-N),len(liner)):
    print liner[ran]

文件的前 N ​​行(例如:N=10)

file=open("xyz.txt",'r")
liner=file.readlines()
for ran in range(0,N+1):
    print liner[ran]
于 2018-06-18T12:03:11.507 回答
-2

很简单:

def tail(fname,nl):
with open(fname) as f:
    data=f.readlines() #readlines return a list
    print(''.join(data[-nl:]))
于 2018-06-20T11:49:34.883 回答
-5

虽然这对于大文件来说并不是很有效,但这段代码非常简单:

  1. 它读取文件对象,f.
  2. 它使用换行符分割返回的字符串,\n.
  3. 它获取数组列表最后一个索引,使用负号代表最后一个索引,并:获取一个子数组。

    def tail(f,n):
        return "\n".join(f.read().split("\n")[-n:])
    
于 2016-05-12T02:57:48.817 回答