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
这可能比你的要快。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，你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。如果你的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，我怀疑你会发现尝试在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。
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