这是我的答案。纯蟒蛇。使用时间看起来很快。在一个有100000行的日志文件中拖尾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
代码如下:
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:]