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分页。文件中的项目是基于行的，底部是最新项目。
所以我需要一个tail()方法，可以从底部读取n行并支持偏移量。我想出的是这样的：
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
这是一种合理的方法吗？使用偏移量尾部日志文件的推荐方法是什么？
答案 0 :(得分：117)
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。
答案 1 :(得分：76)
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]
答案 2 :(得分：29)
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:]
答案 3 :(得分：26)
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:]
答案 4 :(得分：25)
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:]
答案 5 :(得分：19)
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
答案 6 :(得分：13)
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()
答案 7 :(得分：4)
更清晰的python3兼容版本，不会插入但附加＆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')
答案 8 :(得分：3)
将@papercrane解决方案更新为python3。
使用open(filename, 'rb')和
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:]
答案 9 :(得分：3)
在my answer to a similar question上的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，而不仅仅是获取它。
对于大小合适的文件，mmap是执行此操作的最佳方式。为了改进现有的mmap答案，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{{ 3}}）。
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，这是以二进制模式读取的（使用mmap所必需的），因此它提供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{em>认为它应该有效的other answer for hints on handling this, and for modifying to work on Python 2类似，但是调整（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。
答案 10 :(得分：3)
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
答案 11 :(得分：2)
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。
答案 12 :(得分：2)
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()
答案 13 :(得分：2)
pypi上有一些现有的tail实现，您可以使用pip安装：
根据您的具体情况，使用其中一种现有工具可能会有好处。
答案 14 :(得分：1)
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）
答案 15 :(得分：1)
简单：
with open("test.txt") as f:
data = f.readlines()
tail = data[-2:]
print(''.join(tail)
答案 16 :(得分：1)
另一种解决方案
如果您的txt文件如下所示： 鼠 蛇 猫 蜥蜴 狼 狗
您可以通过简单地在python中使用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
答案 17 :(得分：1)
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
答案 18 :(得分：1)
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)
答案 19 :(得分：1)
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
答案 20 :(得分：0)
更新A.Coady给出的答案
使用 python 3 。
这使用Exponential Search，并且只会从后面缓冲N行，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))
答案 21 :(得分：0)
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)
答案 22 :(得分：0)
有一个非常有用的module可以做到：
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)
答案 23 :(得分：0)
好吧！我有一个类似的问题，虽然我只需要最后一行， 所以我想出了我自己的解决方案
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，找到最后一行的时间非常少（甚至不到半秒）
答案 24 :(得分：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)
答案 25 :(得分：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]))
答案 26 :(得分：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
答案 27 :(得分：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
答案 28 :(得分：0)
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)
答案 29 :(得分：0)
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"))
答案 30 :(得分：-1)
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]
答案 31 :(得分：-1)
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。
答案 32 :(得分：-2)
非常简单：
def tail(fname,nl):
with open(fname) as f:
data=f.readlines() #readlines return a list
print(''.join(data[-nl:]))
答案 33 :(得分：-4)
虽然这并不是大文件的有效方面，但这段代码非常简单：
f。\n分割返回的字符串。它使数组列出最后一个索引，使用负号代表最后一个索引，:得到一个子数组。
def tail(f,n):
return "\n".join(f.read().split("\n")[-n:])