在python中读取标头之间的文件

我有一个大文本文件，其中的值由以“#”开头的标题分隔。如果条件与标头中的条件匹配，我想读取文件直到下一个标头“#”并跳过文件的其余部分。

为了测试我正在尝试读取以下名为 test234.txt 的文本文件：

# abcdefgh
1fnrnf
mrkfr
nfoiernfr
nerfnr
# something
njndjen kj
ejkndjke
#vcrvr

我写的代码是：

file_t = open('test234.txt')
cond = True
while cond:
    for line_ in file_t:
        print(line_)
        if file_t.read(1) == "#":
            cond = False
file_t.close()

但是，我得到的输出是：

# abcdefgh

fnrnf

rkfr

foiernfr

erfnr

something

jndjen kj

jkndjke

vcrvr

相反，我希望两个标头之间的输出由“#”分隔，即：

1fnrnf
mrkfr
nfoiernfr
nerfnr      

我怎样才能做到这一点？谢谢！

EDIT: python中使用指定分隔符逐块读取文件 https://stackoverflow.com/questions/38655176/reading-in-file-block-by-block-using-specified-delimiter-in-python谈论读取由标题分隔的组中的文件，但我不想读取所有标题。我只想读取满足给定条件的标头，并且一旦该行到达标有“#”的下一个标头，它就会停止读取文件。

itertools.groupby https://docs.python.org/3/library/itertools.html?highlight=groupby#itertools.groupby可以帮助：

from io import StringIO
from itertools import groupby

text = '''# abcdefgh
1fnrnf
mrkfr
nfoiernfr
nerfnr
# something
njndjen kj
ejkndjke
#vcrvr'''


with StringIO(text) as file:
    lines = (line.strip() for line in file)  # removing trailing '\n'
    for key, group in groupby(lines, key=lambda x: x[0]=='#'):

        if key is True:
            # found a line that starts with '#'
            print('found header: {}'.format(next(group)))

        if key is False:
            # group now contanins all lines that do not start with '#'
            print('\n'.join(group))

请注意，所有这些都是lazy。您只能将两个标题之间的所有项目存储在内存中。

你必须更换with StringIO(text) as file: with; with open('test234.txt', 'r') as file:...

您的测试的输出是：

found header: # abcdefgh
1fnrnf
mrkfr
nfoiernfr
nerfnr
found header: # something
njndjen kj
ejkndjke
found header: #vcrvr

UPDATE正如我误解的那样。这是一个新的尝试：

from io import StringIO
from collections import deque
from itertools import takewhile

from_line = '# abcdefgh'
to_line = '# something'

with StringIO(text) as file:
    lines = (line.strip() for line in file)  # removing trailing '\n'

    # fast-forward up to from_line
    deque(takewhile(lambda x: x != from_line, lines), maxlen=0)

    for line in takewhile(lambda x: x != to_line, lines):
        print(line)

我在哪里使用itertools.takewhile https://docs.python.org/3/library/itertools.html?highlight=takewhile#itertools.takewhile获取行上的迭代器，直到满足条件（直到在您的情况下找到第一个标头）。

the deque部分只是consume pattern https://docs.python.org/3/library/itertools.html?highlight=consume#itertools-recipesitertools 食谱中建议。它只是快进到给定条件不再成立的点。