Python 删除配置文件中[]方括号内与之间的内容的方法

示例代码smb文件：

i 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.

删除替换后：

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

1､使用正则表达式实现

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

说明：

(\n\[)：找到一个序列，其中有一个换行符，后跟一个[

(\[]\n)：查找有[]后接换行符的序列

(.*?)：删除(\n\[)和中间的所有内容(\[]\n)

re.DOTALL：正则表达式中点(.)所有字符了，包括换行符(\n)。

2､使用pandas和正则re实现

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)

3､使用pandas实现

df = pd.read_csv('smb', sep='----', header=None)
# mark rows starts with `[`
s = df[0].str.startswith('[')
# drop the lines between `[`
df = df.drop(np.arange(s.idxmax(),s[::-1].idxmax()+1))
# write to file if needed
df.to_csv('clean.txt', header=None, index=None)

或者

df = pd.read_csv(your_file,sep='\t',header=None)
idx = df[df[0].str.contains('\[')].index
df1 = df.loc[~df.index.isin(range(idx[0],idx[-1] + 1))]