Python：使用正则表达式从字符串中删除字符

Question

The whole file consists of lines like below.

\"ansText\" : \"11\",
\"boundsX\" : 0,
\"string\" : \"11\"

For any lines starting with title , I would like to remove the character ; from the string after it. Below is an example of expected output

Input:

\"title\" : \"244442424268391(:)7)$(.:$?3.&!&3$83;767:2\",

Output:

\"title\" : \"244442424268391(:)7)$(.:$?3.&!&3$83767:2\",

I know how to set the regular expression to find the expression title by using:

 str0 = re.sub(r'\"title.*',"\"title\" : ",str0) 

But I am not too certain how can I keep the original string but remove only one character.

Answer 1

You can simply do this use str.replace() and str.startswith() without RegEx like this:

>>> str0 = r'\"title\" : \"244442424268391(:)7)$(.:$?3.&!&3$83;767:2\",'
>>> str0 = str0.replace(';', '') if str0.startswith(r'\"title\" : ') else str0
>>> str0
'\\"title\\" : \\"244442424268391(:)7)$(.:$?3.&!&3$83767:2\\",'

Answer 2

You could use something like so: (\\\\"title\\\\" : \\\\".+?);(.+?\\\\") (example here ) and replace the string with regex groups number 1 and 2. This expression will look for strings containing \\"title\\" and a ; character within it and use this information to create two regular expression groups, this given \\"title\\" : \\"244442424268391(:)7)$(.:$?3.&!&3$83;767:2\\", , the output would be:;

Group 1: \"title\" : \"244442424268391(:)7)$(.:$?3.&!&3$83
Group 2: 767:2\"

When you combine these 2 strings, you will get the result which you are after.

 str0 = re.sub(r'(\\"title\\" : \\".+?);(.+?\\")',r"\1\2", str0)