使用Python获取文件的最后n行，类似于tail

Question

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？建议使用带偏移量尾部日志文件的推荐方式是什么？

Answer 1

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

Answer 2

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。