得到最后n线的一个文件，蟒蛇，类似的尾巴

StackOverflow https://stackoverflow.com/questions/136168

题

我正在写记录文件观察者对于网络的应用程序和对于我要分通过的行日志的文件。该项目在该文件是基于行与最新的项目。

所以我需要一个 tail() 方法，可以阅读 n 线从底部并支持抵消。我想出了这样的：

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

这是一个合理的做法？什么是建议的方式尾登录文件的有偏?

有帮助吗？

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

其他提示

这可能比你的更快。不对线长做出假设。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:])

我不喜欢关于线长的棘手假设 - 实际上 - 你永远不会知道这样的事情。

通常，这将在第一次或第二次通过循环时找到最后20行。如果你的74个角色实际上是准确的，你可以使块大小为2048，你几乎可以立即拖出20行。

此外，我不会燃烧大量的脑热量，试图与物理OS块进行精确对齐。使用这些高级I / O包，我怀疑你会看到尝试在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。

假设你可以在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，您可以这样做：

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]

如果读取整个文件是可以接受的，那么请使用双端队列。

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

如果要求从最后读取文件，则使用驰骋（a.k.a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:]

这是我的答案。纯蟒蛇。使用timeit似乎很快。拖尾100行有100,000行的日志文件：

>>> 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:]

S.Lott上面的答案几乎对我有用，但最终给了我部分线条。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持读取块。当调用'.join（数据）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:]

使用mmap进行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()

更简洁的python3兼容版本，不插入但附加＆amp;逆转：

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')

我发现上面的Popen是最好的解决方案。这很快，很脏，而且很有效 对于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将包含代码的最后n行。逐行遍历soutput：

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

将@papercrane解决方案更新为python3。 使用 open（filename，'rb'）和

打开文件 
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:]

根据我对类似问题的回答，在评论者的要求下发布答案，其中使用相同的技术来改变文件的最后一行，不只是得到它。

对于大小合适的文件， mmap 是最好的方法。为了改进现有的 mmap 答案，这个版本可以在Windows和Linux之间移植，并且应该运行得更快（尽管如果没有对GB范围内的文件的32位Python进行一些修改它将无法工作，请参阅有关处理此问题以及修改为在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)

这假设尾线数足够小，您可以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：

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，这是二进制模式下的读取（使用 mmap 所必需的），因此它给出了 str 行（Py2）和 bytes 行（Py3） ）;如果你想要 unicode （Py2）或 str （Py3），迭代方法可以调整为你解码和/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：我在一台我无法访问Python进行测试的机器上输入了这一切。如果我输了什么东西，请告诉我。这与我的其他答案类似，我认为它应该有效，但调整（例如处理偏移量） ）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。

基于S.Lott的最高投票回答（08年9月25日21:43），但修复了小文件。

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

希望这很有用。

有一些现有的实现的尾巴pypi，你可以安装使用画中画:

  • mtFileUtil
  • multitail
  • log4tailer
  • ...

根据你的情况下，可能有优势，使用这些现有的工具。

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()

简单：

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

为了提高非常大的文件的效率（在你可能想要使用tail的日志文件情况下很常见），你通常要避免读取整个文件（即使你没有在不将整个文件一次性读入内存的情况下这样做） ，你需要以某种方式计算出线条而不是字符的偏移量。一种可能性是使用char（）查询向后读取char（），但这非常慢。相反，它更好地处理更大的块。

我之前写过一个实用程序函数来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]添加更多特定版本（避免需要反转两次）

您可以使用f.seek（0,2）转到文件的末尾，然后逐行读取行，并使用以下替换readline（）：

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)

如果文件未以\ 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

我必须从文件的最后一行读取一个特定值，并偶然发现了这个帖子。我没有重新发明Python中的轮子，而是使用了一个小的shell脚本，保存为 的/ usr / local / bin中/ 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示例，而是一个更简单的示例。这个是通用的：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)

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)

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)

第二个想法，这可能和这里的任何事情一样快。

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]

这简单得多。它确实似乎以一种良好的速度撕裂。

我找到了查找文件的第一行或最后N行的最简单方法

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]

这很简单：

def tail(fname,nl):
with open(fname) as f:
    data=f.readlines() #readlines return a list
    print(''.join(data[-nl:]))

虽然这不是大文件的有效方面，但这段代码很简单：

  1. 它读取文件对象 f 。
  2. 它拆分使用换行符返回的字符串 \ n 。

  3. 它使数组列出最后的索引，使用负号表示最后的索引，使用：来获取子数组。

    def tail(f,n):
    return "\n".join(f.read().split("\n")[-n:])
许可以下： CC-BY-SA 和 归因
不隶属于 StackOverflow
scroll top