Python删除方括号和它们之间的无关信息

Question

我正在尝试处理一个文件，我需要删除文件中的无关信息；值得注意的是，我正在尝试删除括号 []包括括号内和括号之间的文字 [] []块，说这些块之间的所有内容包括它们本身，但打印其外的所有内容。

下面是我的带有数据示例的文本文件:

$ cat smb
Hi this is my config file.
Please dont delete it

[homes]
  browseable                     = No
  comment                        = Your Home
  create mode                    = 0640
  csc policy                     = disable
  directory mask                 = 0750
  public                         = No
  writeable                      = Yes

[proj]
  browseable                     = Yes
  comment                        = Project directories
  csc policy                     = disable
  path                           = /proj
  public                         = No
  writeable                      = Yes

[]

This last second line.
End of the line.

期望输出:

Hi this is my config file.
Please dont delete it
This last second line.
End of the line.

根据我的理解和重新搜索，我尝试了什么:

$ cat test.py
with open("smb", "r") as file:
  for line in file:
    start = line.find( '[' )
    end = line.find( ']' )
    if start != -1 and end != -1:
      result = line[start+1:end]
      print(result)

输出:

$ ./test.py
   homes
   proj

Answer 1

用一个正则表达式

import re

with open("smb", "r") as f: 
    txt = f.read()
    txt = re.sub(r'(\n\[)(.*?)(\[]\n)', '', txt, flags=re.DOTALL)

print(txt)

正则解释:
(\n\[)找到一个序列，其中有一个换行符后跟一个 [
(\[]\n)找到一个序列，其中有 [] 后跟一个换行符
(.*?)删除 (\n\[) 中间的所有内容和 (\[]\n)re.DOTALL用于防止不必要的回溯

!!! Pandas 更新!!!

可以用pandas进行相同逻辑的相同解决方案

import re
import pandas as pd

# read each line in the file (one raw -> one line)
txt = pd.read_csv('smb',  sep = '\n', header=None)
# join all the line in the file separating them with '\n'
txt = '\n'.join(txt[0].to_list())
# apply the regex to clean the text (the same as above)
txt = re.sub(r'(\n\[)(.*?)(\[]\n)', '\n', txt, flags=re.DOTALL)

print(txt)