如何使用python以相反的顺序读取文件？我想从最后一行读取一个文件到第一行。
问问题
180131 次
22 回答
174
作为生成器编写的正确、有效的答案。
import os
def reverse_readline(filename, buf_size=8192):
"""A generator that returns the lines of a file in reverse order"""
with open(filename) as fh:
segment = None
offset = 0
fh.seek(0, os.SEEK_END)
file_size = remaining_size = fh.tell()
while remaining_size > 0:
offset = min(file_size, offset + buf_size)
fh.seek(file_size - offset)
buffer = fh.read(min(remaining_size, buf_size))
remaining_size -= buf_size
lines = buffer.split('\n')
# The first line of the buffer is probably not a complete line so
# we'll save it and append it to the last line of the next buffer
# we read
if segment is not None:
# If the previous chunk starts right from the beginning of line
# do not concat the segment to the last line of new chunk.
# Instead, yield the segment first
if buffer[-1] != '\n':
lines[-1] += segment
else:
yield segment
segment = lines[0]
for index in range(len(lines) - 1, 0, -1):
if lines[index]:
yield lines[index]
# Don't yield None if the file was empty
if segment is not None:
yield segment
于 2014-05-14T05:09:51.237 回答
87
for line in reversed(open("filename").readlines()):
print line.rstrip()
在 Python 3 中：
for line in reversed(list(open("filename"))):
print(line.rstrip())
于 2010-02-20T10:10:52.363 回答
30
您也可以使用 python 模块file_read_backwards。
安装后，通过pip install file_read_backwards(v1.2.1)，您可以通过以下方式以内存有效的方式向后（逐行）读取整个文件：
#!/usr/bin/env python2.7
from file_read_backwards import FileReadBackwards
with FileReadBackwards("/path/to/file", encoding="utf-8") as frb:
for l in frb:
print l
它支持“utf-8”、“latin-1”和“ascii”编码。
也可支持 python3。更多文档可以在http://file-read-backwards.readthedocs.io/en/latest/readme.html找到
于 2017-01-01T19:05:11.820 回答
23
像这样的东西怎么样：
import os
def readlines_reverse(filename):
with open(filename) as qfile:
qfile.seek(0, os.SEEK_END)
position = qfile.tell()
line = ''
while position >= 0:
qfile.seek(position)
next_char = qfile.read(1)
if next_char == "\n":
yield line[::-1]
line = ''
else:
line += next_char
position -= 1
yield line[::-1]
if __name__ == '__main__':
for qline in readlines_reverse(raw_input()):
print qline
由于文件是以相反的顺序逐个字符读取的，因此即使在非常大的文件上也可以工作，只要各个行适合内存即可。
于 2014-11-05T00:42:49.443 回答
15
接受的答案不适用于内存无法容纳的大文件的情况（这种情况并不罕见）。
正如其他人所指出的那样，@srohde 的答案看起来不错，但它还有下一个问题：
- 打开文件看起来是多余的，当我们可以传递文件对象并将其留给用户决定应该以哪种编码读取时，
- 即使我们重构以接受文件对象，它也不适用于所有编码：我们可以选择具有
utf-8编码和非 ascii 内容的文件，例如
й
通过buf_size等于1并且将有
UnicodeDecodeError: 'utf8' codec can't decode byte 0xb9 in position 0: invalid start byte
当然，文本可能更大，但buf_size可能会被拾取，因此会导致像上面这样的混淆错误，
- 我们不能指定自定义行分隔符，
- 我们不能选择保留行分隔符。
因此，考虑到所有这些问题，我编写了单独的函数：
- 一种适用于字节流的，
- 第二个处理文本流并将其底层字节流委托给第一个并解码结果行。
首先让我们定义下一个实用函数：
ceil_division用于使用天花板进行划分（与标准//的地板划分相比，可以在此线程中找到更多信息）
def ceil_division(left_number, right_number):
"""
Divides given numbers with ceiling.
"""
return -(-left_number // right_number)
split用于从右端通过给定分隔符拆分字符串，并能够保留它：
def split(string, separator, keep_separator):
"""
Splits given string by given separator.
"""
parts = string.split(separator)
if keep_separator:
*parts, last_part = parts
parts = [part + separator for part in parts]
if last_part:
return parts + [last_part]
return parts
read_batch_from_end从二进制流的右端读取批处理
def read_batch_from_end(byte_stream, size, end_position):
"""
Reads batch from the end of given byte stream.
"""
if end_position > size:
offset = end_position - size
else:
offset = 0
size = end_position
byte_stream.seek(offset)
return byte_stream.read(size)
之后，我们可以定义以相反顺序读取字节流的函数，例如
import functools
import itertools
import os
from operator import methodcaller, sub
def reverse_binary_stream(byte_stream, batch_size=None,
lines_separator=None,
keep_lines_separator=True):
if lines_separator is None:
lines_separator = (b'\r', b'\n', b'\r\n')
lines_splitter = methodcaller(str.splitlines.__name__,
keep_lines_separator)
else:
lines_splitter = functools.partial(split,
separator=lines_separator,
keep_separator=keep_lines_separator)
stream_size = byte_stream.seek(0, os.SEEK_END)
if batch_size is None:
batch_size = stream_size or 1
batches_count = ceil_division(stream_size, batch_size)
remaining_bytes_indicator = itertools.islice(
itertools.accumulate(itertools.chain([stream_size],
itertools.repeat(batch_size)),
sub),
batches_count)
try:
remaining_bytes_count = next(remaining_bytes_indicator)
except StopIteration:
return
def read_batch(position):
result = read_batch_from_end(byte_stream,
size=batch_size,
end_position=position)
while result.startswith(lines_separator):
try:
position = next(remaining_bytes_indicator)
except StopIteration:
break
result = (read_batch_from_end(byte_stream,
size=batch_size,
end_position=position)
+ result)
return result
batch = read_batch(remaining_bytes_count)
segment, *lines = lines_splitter(batch)
yield from lines[::-1]
for remaining_bytes_count in remaining_bytes_indicator:
batch = read_batch(remaining_bytes_count)
lines = lines_splitter(batch)
if batch.endswith(lines_separator):
yield segment
else:
lines[-1] += segment
segment, *lines = lines
yield from lines[::-1]
yield segment
最后一个用于反转文本文件的函数可以定义为：
import codecs
def reverse_file(file, batch_size=None,
lines_separator=None,
keep_lines_separator=True):
encoding = file.encoding
if lines_separator is not None:
lines_separator = lines_separator.encode(encoding)
yield from map(functools.partial(codecs.decode,
encoding=encoding),
reverse_binary_stream(
file.buffer,
batch_size=batch_size,
lines_separator=lines_separator,
keep_lines_separator=keep_lines_separator))
测试
准备工作
我使用fsutil命令生成了 4 个文件：
- empty.txt没有内容，大小 0MB
- tiny.txt，大小为 1MB
- 大小为 10MB 的small.txt
- 大小为 50MB 的large.txt
我还重构了@srohde 解决方案以使用文件对象而不是文件路径。
测试脚本
from timeit import Timer
repeats_count = 7
number = 1
create_setup = ('from collections import deque\n'
'from __main__ import reverse_file, reverse_readline\n'
'file = open("{}")').format
srohde_solution = ('with file:\n'
' deque(reverse_readline(file,\n'
' buf_size=8192),'
' maxlen=0)')
azat_ibrakov_solution = ('with file:\n'
' deque(reverse_file(file,\n'
' lines_separator="\\n",\n'
' keep_lines_separator=False,\n'
' batch_size=8192), maxlen=0)')
print('reversing empty file by "srohde"',
min(Timer(srohde_solution,
create_setup('empty.txt')).repeat(repeats_count, number)))
print('reversing empty file by "Azat Ibrakov"',
min(Timer(azat_ibrakov_solution,
create_setup('empty.txt')).repeat(repeats_count, number)))
print('reversing tiny file (1MB) by "srohde"',
min(Timer(srohde_solution,
create_setup('tiny.txt')).repeat(repeats_count, number)))
print('reversing tiny file (1MB) by "Azat Ibrakov"',
min(Timer(azat_ibrakov_solution,
create_setup('tiny.txt')).repeat(repeats_count, number)))
print('reversing small file (10MB) by "srohde"',
min(Timer(srohde_solution,
create_setup('small.txt')).repeat(repeats_count, number)))
print('reversing small file (10MB) by "Azat Ibrakov"',
min(Timer(azat_ibrakov_solution,
create_setup('small.txt')).repeat(repeats_count, number)))
print('reversing large file (50MB) by "srohde"',
min(Timer(srohde_solution,
create_setup('large.txt')).repeat(repeats_count, number)))
print('reversing large file (50MB) by "Azat Ibrakov"',
min(Timer(azat_ibrakov_solution,
create_setup('large.txt')).repeat(repeats_count, number)))
注意：我已经使用collections.deque类来排放发电机。
输出
对于 Windows 10 上的 PyPy 3.5：
reversing empty file by "srohde" 8.31e-05
reversing empty file by "Azat Ibrakov" 0.00016090000000000028
reversing tiny file (1MB) by "srohde" 0.160081
reversing tiny file (1MB) by "Azat Ibrakov" 0.09594989999999998
reversing small file (10MB) by "srohde" 8.8891863
reversing small file (10MB) by "Azat Ibrakov" 5.323388100000001
reversing large file (50MB) by "srohde" 186.5338368
reversing large file (50MB) by "Azat Ibrakov" 99.07450229999998
对于 Windows 10 上的 CPython 3.5：
reversing empty file by "srohde" 3.600000000000001e-05
reversing empty file by "Azat Ibrakov" 4.519999999999958e-05
reversing tiny file (1MB) by "srohde" 0.01965560000000001
reversing tiny file (1MB) by "Azat Ibrakov" 0.019207699999999994
reversing small file (10MB) by "srohde" 3.1341862999999996
reversing small file (10MB) by "Azat Ibrakov" 3.0872588000000007
reversing large file (50MB) by "srohde" 82.01206720000002
reversing large file (50MB) by "Azat Ibrakov" 82.16775059999998
因此，我们可以看到它的性能类似于原始解决方案，但更通用且没有上面列出的缺点。
广告
我已将此添加到具有许多经过良好测试的功能/迭代实用程序的包0.3.0版本（需要Python 3.5 +）中。lz
可以像这样使用
import io
from lz.reversal import reverse
...
with open('path/to/file') as file:
for line in reverse(file, batch_size=io.DEFAULT_BUFFER_SIZE):
print(line)
于 2018-11-23T14:07:59.947 回答
8
import re
def filerev(somefile, buffer=0x20000):
somefile.seek(0, os.SEEK_END)
size = somefile.tell()
lines = ['']
rem = size % buffer
pos = max(0, (size // buffer - 1) * buffer)
while pos >= 0:
somefile.seek(pos, os.SEEK_SET)
data = somefile.read(rem + buffer) + lines[0]
rem = 0
lines = re.findall('[^\n]*\n?', data)
ix = len(lines) - 2
while ix > 0:
yield lines[ix]
ix -= 1
pos -= buffer
else:
yield lines[0]
with open(sys.argv[1], 'r') as f:
for line in filerev(f):
sys.stdout.write(line)
于 2010-02-20T10:36:27.773 回答
7
for line in reversed(open("file").readlines()):
print line.rstrip()
如果你在linux上，你可以使用tac命令。
$ tac file
于 2010-02-20T10:13:43.277 回答
2
在这里你可以找到我的实现，你可以通过改变“缓冲区”变量来限制内存的使用，有一个错误，程序在开始打印一个空行。
如果没有超过缓冲区字节的新行，内存使用量可能会增加，“泄漏”变量将增加，直到看到新行（“\n”）。
这也适用于比我的总内存大的 16 GB 文件。
import os,sys
buffer = 1024*1024 # 1MB
f = open(sys.argv[1])
f.seek(0, os.SEEK_END)
filesize = f.tell()
division, remainder = divmod(filesize, buffer)
line_leak=''
for chunk_counter in range(1,division + 2):
if division - chunk_counter < 0:
f.seek(0, os.SEEK_SET)
chunk = f.read(remainder)
elif division - chunk_counter >= 0:
f.seek(-(buffer*chunk_counter), os.SEEK_END)
chunk = f.read(buffer)
chunk_lines_reversed = list(reversed(chunk.split('\n')))
if line_leak: # add line_leak from previous chunk to beginning
chunk_lines_reversed[0] += line_leak
# after reversed, save the leakedline for next chunk iteration
line_leak = chunk_lines_reversed.pop()
if chunk_lines_reversed:
print "\n".join(chunk_lines_reversed)
# print the last leaked line
if division - chunk_counter < 0:
print line_leak
于 2012-04-17T14:04:34.760 回答
2
感谢@srohde 的回答。它有一个使用“is”运算符检查换行符的小错误，我无法评论 1 信誉的答案。此外，我想管理在外部打开的文件，因为这使我能够嵌入我的 luigi 任务。
我需要改变的形式是：
with open(filename) as fp:
for line in fp:
#print line, # contains new line
print '>{}<'.format(line)
我很想改成：
with open(filename) as fp:
for line in reversed_fp_iter(fp, 4):
#print line, # contains new line
print '>{}<'.format(line)
这是一个修改后的答案，需要一个文件句柄并保留换行符：
def reversed_fp_iter(fp, buf_size=8192):
"""a generator that returns the lines of a file in reverse order
ref: https://stackoverflow.com/a/23646049/8776239
"""
segment = None # holds possible incomplete segment at the beginning of the buffer
offset = 0
fp.seek(0, os.SEEK_END)
file_size = remaining_size = fp.tell()
while remaining_size > 0:
offset = min(file_size, offset + buf_size)
fp.seek(file_size - offset)
buffer = fp.read(min(remaining_size, buf_size))
remaining_size -= buf_size
lines = buffer.splitlines(True)
# the first line of the buffer is probably not a complete line so
# we'll save it and append it to the last line of the next buffer
# we read
if segment is not None:
# if the previous chunk starts right from the beginning of line
# do not concat the segment to the last line of new chunk
# instead, yield the segment first
if buffer[-1] == '\n':
#print 'buffer ends with newline'
yield segment
else:
lines[-1] += segment
#print 'enlarged last line to >{}<, len {}'.format(lines[-1], len(lines))
segment = lines[0]
for index in range(len(lines) - 1, 0, -1):
if len(lines[index]):
yield lines[index]
# Don't yield None if the file was empty
if segment is not None:
yield segment
于 2017-10-14T15:18:16.987 回答
1
创建第二个反转文件的简单函数（仅限linux）：
import os
def tac(file1, file2):
print(os.system('tac %s > %s' % (file1,file2)))
如何使用
tac('ordered.csv', 'reversed.csv')
f = open('reversed.csv')
于 2015-04-22T21:57:47.473 回答
1
使用 open("filename") 作为 f：
print(f.read()[::-1])
于 2019-08-13T18:54:26.487 回答
1
如果您担心文件大小/内存使用情况，内存映射文件并向后扫描换行符是一种解决方案：
于 2018-02-08T10:29:46.683 回答
1
这是一种 Python 3.8+ 方法，使用两个字符串缓冲区，具有类似 grep 的子字符串匹配（或者如果传递了空子字符串，则只是简单地迭代每一行）。我希望这比将所有文件加载到内存中更节省内存（您可以控制缓冲区大小，这有时是可取的），例如，如果您只想在文件末尾找到一些东西。要点在这里。
from __future__ import annotations
from io import StringIO, SEEK_END
from pathlib import Path
from typing import Iterator, TextIO
def grep_backwards(
fh: TextIO,
match_substr: str,
line_ending: str = "\n",
strip_eol: bool = False,
step: int = 10,
) -> Iterator[str]:
"""
Helper for scanning a file line by line from the end, imitating the behaviour of
the Unix command line tools ``grep`` (when passed ``match_substr``) or ``tac`` (when
``match_substr`` is the empty string ``""``, i.e. matching all lines).
Args:
fh : The file handle to read from
match_substr : Substring to match at. If given as the empty string, gives a
reverse line iterator rather than a reverse matching line iterator.
line_ending : The line ending to split lines on (default: "\n" newline)
strip_eol : Whether to strip (default: ``True``) or keep (``False``) line
endings off the end of the strings returned by the iterator.
step : Number of characters to load into chunk buffer (i.e. chunk size)
"""
# Store the end of file (EOF) position as we are advancing backwards from there
file_end_pos = fh.seek(0, SEEK_END) # cursor has moved to EOF
# Keep a reversed string line buffer as we are writing right-to-left
revlinebuf = StringIO()
# Keep a [left-to-right] string buffer as we read left-to-right, one chunk at a time
chunk_buf = StringIO()
# Initialise 'last chunk start' at position after the EOF (unreachable by ``read``)
last_chunk_start = file_end_pos + 1
line_offset = 0 # relative to SEEK_END
has_EOF_newline = False # may change upon finding first newline
# In the worst case, seek all the way back to the start (position 0)
while last_chunk_start > 0:
# Ensure that read(size=step) will read at least 1 character
# e.g. when step=4, last_chunk_start=3, reduce step to 3 --> chunk=[0,1,2]
if step > last_chunk_start:
step = last_chunk_start
chunk_start = last_chunk_start - step
fh.seek(chunk_start)
# Read in the chunk for the current step (possibly after pre-existing chunks)
chunk_buf.write(fh.read(step))
while chunk := chunk_buf.getvalue():
# Keep reading intra-chunk lines RTL, leaving any leftovers in revlinebuf
lhs, EOL_match, rhs = chunk.rpartition(line_ending)
if EOL_match:
if line_offset == 0:
has_EOF_newline = rhs == ""
# Reverse the right-hand-side of the rightmost line_ending and
# insert it after anything already in the reversed line buffer
if rhs:
# Only bother writing rhs to line buffer if there's anything in it
revlinebuf.write(rhs[::-1])
# Un-reverse the line buffer --> full line after the line_ending match
completed_line = revlinebuf.getvalue()[::-1] # (may be empty string)
# Clear the reversed line buffer
revlinebuf.seek(0)
revlinebuf.truncate()
# `grep` if line matches (or behaves like `tac` if match_substr == "")
if line_offset == 0:
if not has_EOF_newline and match_substr in completed_line:
# The 0'th line from the end (by definition) cannot get an EOL
yield completed_line
elif match_substr in (completed_line + line_ending):
if not strip_eol:
completed_line += line_ending
yield completed_line
line_offset += 1
else:
# If line_ending not found in chunk then add entire [remaining] chunk,
# in reverse, onto the reversed line buffer, before chunk_buf is cleared
revlinebuf.write(chunk_buf.getvalue()[::-1])
# The LHS of the rightmost line_ending (if any) may contain another line
# ending so truncate the chunk to that and re-iterate (else clear chunk_buf)
chunk_buf.seek(len(lhs))
chunk_buf.truncate()
last_chunk_start = chunk_start
if completed_line := revlinebuf.getvalue()[::-1]:
# Iteration has reached the line at start of file, left over in the line buffer
if line_offset == 0 and not has_EOF_newline and match_substr in completed_line:
# The 0'th line from the end (by definition) cannot get an EOL
yield completed_line
elif match_substr in (
completed_line + (line_ending if line_offset > 1 or has_EOF_newline else "")
):
if line_offset == 1:
if has_EOF_newline and not strip_eol:
completed_line += line_ending
elif not strip_eol:
completed_line += line_ending
yield completed_line
else:
raise StopIteration
这里有一些测试表明它是有效的，通过计数多达 100 个说 'Hi 0'、'Hi 9'、'Hi 18'、... 生成 3 个测试输入文件：
- ...并给 27 号一个双换行符
- ...并且文件末尾没有换行符
- ...并给出文件 2 换行符的结尾
# Write lines counting to 100 saying 'Hi 0', 'Hi 9', ... give number 27 a double newline
str_out = "".join([f"Hi {i}\n" if i != 27 else f"Hi {i}\n\n" for i in range(0, 100, 9)])
example_file = Path("example.txt")
no_eof_nl_file = Path("no_eof_nl.txt") # no end of file newline
double_eof_nl_file = Path("double_eof_nl.txt") # double end of file newline
with open(example_file, "w") as f_out:
f_out.write(str_out)
with open(no_eof_nl_file, "w") as f_out:
f_out.write(str_out.rstrip("\n"))
with open(double_eof_nl_file, "w") as f_out:
f_out.write(str_out + "\n")
file_list = [example_file, no_eof_nl_file, double_eof_nl_file]
labels = [
"EOF_NL ",
"NO_EOF_NL ",
"DBL_EOF_NL",
]
print("------------------------------------------------------------")
print()
print(f"match_substr = ''")
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
lines_rev_from_iterator = list(grep_backwards(fh=fh, match_substr=""))
with open(each_file, "r") as fh:
lines_rev_from_readline = list(reversed(fh.readlines()))
print(label, f"{lines_rev_from_iterator == lines_rev_from_readline=}")
print()
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
reverse_iterator = grep_backwards(fh=fh, match_substr="")
first_match = next(reverse_iterator)
print(label, f"{first_match=}")
print()
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
all_matches = list(grep_backwards(fh=fh, match_substr=""))
print(label, f"{all_matches=}")
print()
print()
print("------------------------------------------------------------")
print()
print(f"match_substr = 'Hi 9'")
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
reverse_iterator = grep_backwards(fh=fh, match_substr="Hi 9")
first_match = next(reverse_iterator)
print(label, f"{first_match=}")
print()
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
all_matches = list(grep_backwards(fh=fh, match_substr="Hi 9"))
print(label, f"{all_matches=}")
print()
print("------------------------------------------------------------")
print()
print(f"match_substr = '\\n'")
for len_flag in (True, False):
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
lines_rev_from_iterator = list(grep_backwards(fh=fh, match_substr="\n"))
if len_flag:
print(label, f"{len(lines_rev_from_iterator)=}")
else:
print(label, f"{lines_rev_from_iterator=}")
print()
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
reverse_iterator = grep_backwards(fh=fh, match_substr="\n")
first_match = next(reverse_iterator)
print(label, f"{first_match=}")
print()
for label, each_file in zip(labels, file_list):
with open(each_file, "r") as fh:
all_matches = list(grep_backwards(fh=fh, match_substr="\n"))
print(label, f"{all_matches=}")
print()
print("------------------------------------------------------------")
⇣</p>
------------------------------------------------------------
match_substr = ''
EOF_NL lines_rev_from_iterator == lines_rev_from_readline=True
NO_EOF_NL lines_rev_from_iterator == lines_rev_from_readline=True
DBL_EOF_NL lines_rev_from_iterator == lines_rev_from_readline=True
EOF_NL first_match='Hi 99\n'
NO_EOF_NL first_match='Hi 99'
DBL_EOF_NL first_match='\n'
EOF_NL all_matches=['Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
NO_EOF_NL all_matches=['Hi 99', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
DBL_EOF_NL all_matches=['\n', 'Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
------------------------------------------------------------
match_substr = 'Hi 9'
EOF_NL first_match='Hi 99\n'
NO_EOF_NL first_match='Hi 99'
DBL_EOF_NL first_match='Hi 99\n'
EOF_NL all_matches=['Hi 99\n', 'Hi 90\n', 'Hi 9\n']
NO_EOF_NL all_matches=['Hi 99', 'Hi 90\n', 'Hi 9\n']
DBL_EOF_NL all_matches=['Hi 99\n', 'Hi 90\n', 'Hi 9\n']
------------------------------------------------------------
match_substr = '\n'
EOF_NL len(lines_rev_from_iterator)=13
NO_EOF_NL len(lines_rev_from_iterator)=12
DBL_EOF_NL len(lines_rev_from_iterator)=14
EOF_NL lines_rev_from_iterator=['Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
NO_EOF_NL lines_rev_from_iterator=['Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
DBL_EOF_NL lines_rev_from_iterator=['\n', 'Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
EOF_NL first_match='Hi 99\n'
NO_EOF_NL first_match='Hi 90\n'
DBL_EOF_NL first_match='\n'
EOF_NL all_matches=['Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
NO_EOF_NL all_matches=['Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
DBL_EOF_NL all_matches=['\n', 'Hi 99\n', 'Hi 90\n', 'Hi 81\n', 'Hi 72\n', 'Hi 63\n', 'Hi 54\n', 'Hi 45\n', 'Hi 36\n', '\n', 'Hi 27\n', 'Hi 18\n', 'Hi 9\n', 'Hi 0\n']
------------------------------------------------------------
于 2021-08-08T01:15:52.783 回答
0
大多数答案需要在做任何事情之前阅读整个文件。这个样本从最后读取越来越大的样本。
我在写这个答案时只看到了 Murat Yükselen 的答案。这几乎是一样的，我认为这是一件好事。下面的示例还处理 \r 并在每一步增加其缓冲区大小。我也有一些单元测试来支持这段代码。
def readlines_reversed(f):
""" Iterate over the lines in a file in reverse. The file must be
open in 'rb' mode. Yields the lines unencoded (as bytes), including the
newline character. Produces the same result as readlines, but reversed.
If this is used to reverse the line in a file twice, the result is
exactly the same.
"""
head = b""
f.seek(0, 2)
t = f.tell()
buffersize, maxbuffersize = 64, 4096
while True:
if t <= 0:
break
# Read next block
buffersize = min(buffersize * 2, maxbuffersize)
tprev = t
t = max(0, t - buffersize)
f.seek(t)
lines = f.read(tprev - t).splitlines(True)
# Align to line breaks
if not lines[-1].endswith((b"\n", b"\r")):
lines[-1] += head # current tail is previous head
elif head == b"\n" and lines[-1].endswith(b"\r"):
lines[-1] += head # Keep \r\n together
elif head:
lines.append(head)
head = lines.pop(0) # can be '\n' (ok)
# Iterate over current block in reverse
for line in reversed(lines):
yield line
if head:
yield head
于 2018-09-10T21:11:41.580 回答
0
import sys
f = open(sys.argv[1] , 'r')
for line in f.readlines()[::-1]:
print line
于 2018-12-03T22:13:22.627 回答
0
with在处理文件时始终使用它，因为它会为您处理一切：
with open('filename', 'r') as f:
for line in reversed(f.readlines()):
print line
或者在 Python 3 中：
with open('filename', 'r') as f:
for line in reversed(list(f.readlines())):
print(line)
于 2016-08-02T15:19:59.193 回答
0
def reverse_lines(filename):
y=open(filename).readlines()
return y[::-1]
于 2015-11-13T05:05:54.393 回答
0
您需要首先以读取格式打开文件，将其保存到变量中，然后以写入格式打开第二个文件，您将在其中使用 [::-1] 切片写入或附加变量，完全反转文件。您还可以使用 readlines() 将其变成行列表，您可以对其进行操作
def copy_and_reverse(filename, newfile):
with open(filename) as file:
text = file.read()
with open(newfile, "w") as file2:
file2.write(text[::-1])
于 2018-06-07T12:47:47.890 回答
0
逐行读取文件，然后以相反的顺序将其添加到列表中。
这是一个代码示例：
reverse = []
with open("file.txt", "r") as file:
for line in file:
line = line.strip()
reverse[0:0] = line
于 2018-11-29T15:19:59.280 回答
0
def previous_line(self, opened_file):
opened_file.seek(0, os.SEEK_END)
position = opened_file.tell()
buffer = bytearray()
while position >= 0:
opened_file.seek(position)
position -= 1
new_byte = opened_file.read(1)
if new_byte == self.NEW_LINE:
parsed_string = buffer.decode()
yield parsed_string
buffer = bytearray()
elif new_byte == self.EMPTY_BYTE:
continue
else:
new_byte_array = bytearray(new_byte)
new_byte_array.extend(buffer)
buffer = new_byte_array
yield None
使用：
opened_file = open(filepath, "rb")
iterator = self.previous_line(opened_file)
line = next(iterator) #one step
close(opened_file)
于 2019-12-10T02:40:21.740 回答
-1
我认为之前没有提到过，但是使用dequefromcollections并且reverse对我有用：
from collections import deque
fs = open("test.txt","rU")
fr = deque(fs)
fr.reverse() # reverse in-place, returns None
for li in fr:
print li
fs.close()
于 2020-09-06T20:29:42.800 回答
-3
前段时间我不得不这样做并使用下面的代码。它通过管道连接到外壳。恐怕我已经没有完整的剧本了。如果您使用的是 unixish 操作系统，则可以使用“tac”，但是在 Mac OSX tac 命令不起作用时，请使用 tail -r。下面的代码片段测试您所在的平台，并相应地调整命令
# We need a command to reverse the line order of the file. On Linux this
# is 'tac', on OSX it is 'tail -r'
# 'tac' is not supported on osx, 'tail -r' is not supported on linux.
if sys.platform == "darwin":
command += "|tail -r"
elif sys.platform == "linux2":
command += "|tac"
else:
raise EnvironmentError('Platform %s not supported' % sys.platform)
于 2010-02-20T11:26:23.410 回答