问题:使用Python获取文件的最后n行,类似于tail
我正在为Web应用程序编写日志文件查看器,为此,我想在日志文件的各行中进行分页。文件中的项目是基于行的,底部是最新的项目。
因此,我需要一种tail()
可以n
从底部读取行并支持偏移量的方法。我想到的是这样的:
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
这是合理的方法吗?推荐使用偏移量尾部日志文件的推荐方法是什么?
回答 0
这可能比您的要快。不假设行长。一次返回一个文件块,直到找到正确数量的’\ n’字符为止。
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,您永远都不知道那样的事情。
通常,这将在循环的第一遍或第二遍中定位最后20行。如果您的74个字符实际上是准确的,则将块大小设置为2048,并且几乎立即尾随20行。
另外,我不会消耗大量的大脑卡路里来尝试与物理OS块进行精确对齐。使用这些高级I / O程序包,我怀疑您会发现尝试在OS块边界上对齐会对性能产生任何影响。如果使用较低级别的I / O,则可能会看到加速。
更新
对于Python 3.2及更高版本,请按照字节上的说明进行操作,如在文本文件中(那些在模式字符串中打开而没有“ b”的文件),则仅允许相对于文件开头的查找(exceptions是查找到文件末尾)与seek(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:])
回答 1
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,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]
回答 2
这是我的答案。纯Python。使用timeit似乎非常快。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:]
回答 3
如果可以读取整个文件,请使用双端队列。
from collections import deque
deque(f, maxlen=n)
在2.6之前,双端队列没有maxlen选项,但实施起来很容易。
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
如果需要从头开始读取文件,请使用疾驰(也就是指数)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:]
回答 4
S.Lott的上述回答几乎对我有用,但最终给了我一些局限性。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,块顺序错误。这样可以解决此问题。
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:]
回答 5
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
回答 6
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()
回答 7
甚至更清洁的python3兼容版本,不会插入但会附加和反向:
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')
回答 8
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:]
回答 9
根据评论者的要求,将答案发布到我对类似问题的回答上,其中使用相同的技术来改变文件的最后一行,而不仅仅是获取文件。
对于很大的文件,这mmap
是最好的方法。为了改善现有mmap
答案,此版本可在Windows和Linux之间移植,并且运行速度应更快(尽管如果不对文件大小在GB范围内的32位Python进行一些修改,它将无法正常运行,请参见其他答案以获取处理此问题的提示) ,并进行修改以在Python 2上运行)。
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)
假设尾行的数目足够小,您可以安全地一次将它们全部读取到内存中;您还可以将其作为生成器函数,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(必须使用mmap
)进行读取,因此得到str
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所有内容。如果我有错字,请告诉我;这与我认为应该起作用的其他答案足够相似,但是这些调整(例如,处理)可能会导致细微的错误。如果有任何错误,请在评论中让我知道。offset
回答 10
我发现上面的Popen是最好的解决方案。它既快速又肮脏,并且适用于Unix机器上的python 2.6,我使用了以下命令
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
回答 11
基于S.Lott的最高票选答案(08年9月25日在21:43),但针对小文件而固定。
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。
回答 12
pypi上有一些tail的现有实现,您可以使用pip安装:
- mtFileUtil
- 多尾
- log4tailer
- …
根据您的情况,使用这些现有工具之一可能会有好处。
回答 13
简单:
with open("test.txt") as f:
data = f.readlines()
tail = data[-2:]
print(''.join(tail)
回答 14
为了提高大文件的效率(在日志文件中通常需要使用tail的情况下很常见),通常希望避免读取整个文件(即使这样做时也没有立即将整个文件读入内存),但是需要以某种方式而不是字符来计算偏移量。一种可能性是通过char用seek()char向后读取,但这非常慢。相反,最好在较大的块中进行处理。
我有一阵子我写的实用程序函数,可以向后读取文件,可以在这里使用。
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]添加了更特定的版本(避免需要反向两次)
回答 15
您可以使用f.seek(0,2)到文件末尾,然后用readline()的以下替换内容逐行读取:
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
回答 16
基于Eyecue的答案(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)
回答 17
如果文件未以\ n结尾或确保读取完整的第一行,则其中的几种解决方案都存在问题。
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
回答 18
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()
回答 19
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)
回答 20
另一种解决方案
如果您的txt文件如下所示:鼠标蛇猫蜥蜴狼狗
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
回答 21
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)
回答 22
我不得不从文件的最后一行读取特定值,并偶然发现了该线程。我没有重新发明Python的工作原理,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"))
回答 23
不是第一个使用双端队列的示例,而是一个更简单的示例。这是一般性的: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)
回答 24
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
回答 25
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
回答 26
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]))
回答 27
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)
回答 28
适用于python 3。
这使用指数搜索,将仅缓冲N
来自后面的行,并且非常高效。
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))
回答 29
再次考虑,这可能和这里的一切一样快。
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。而且它似乎确实在快速发展。