Dưới đây là cách tiếp cận Python 3.8+, sử dụng hai bộ đệm chuỗi, với kết hợp phụ giống như GREP (hoặc chỉ đơn giản là lặp lại từng dòng nếu phần nền trống được truyền). Tôi hy vọng điều này sẽ hiệu quả hơn về bộ nhớ hơn là tải tất cả các tệp vào bộ nhớ (bạn có thể kiểm soát kích thước bộ đệm, đôi khi là mong muốn), ví dụ: Nếu bạn chỉ muốn tìm một cái gì đó ở cuối tệp. Gist ở đây. 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
Dưới đây là một số thử nghiệm để hiển thị nó hoạt động, với 3 tệp đầu vào thử nghiệm được thực hiện bằng cách đếm tới 100 người nói 'hi 0', 'hi 9', 'hi 18', ...: - ... và cho số 27 một dòng kép mới
- ... và cho kết thúc tệp không có dòng mới
- ... và cho phần cuối của Tệp 2 Newlines
# 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("------------------------------------------------------------")
⇣ ------------------------------------------------------------
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']
------------------------------------------------------------
Readline có trả lại EOF trong Python không?
Phương thức readline () không kích hoạt điều kiện cuối tệp. Thay vào đó, khi dữ liệu cạn kiệt, nó sẽ trả về một chuỗi trống. Lưu ý rằng chúng ta cũng phải loại bỏ ký tự dòng mới và chuyển đổi chuỗi thành số nguyên.. Instead, when data is exhausted, it returns an empty string. Notice that we must also remove the new-line character and convert the string to an integer.
Làm thế nào để Python đại diện cho phần cuối của một dòng khi đọc các tệp?
Nhân vật dòng mới trong Python là \ n.Nó được sử dụng để chỉ ra sự kết thúc của một dòng văn bản.Bạn có thể in các chuỗi mà không cần thêm một dòng mới có kết thúc =, đó là ký tự sẽ được sử dụng để tách các dòng.\n . It is used to indicate the end of a line of text. You can print strings without adding a new line with end = , which is the character that will be used to separate the lines. |
