How to remove from the list of strings, using regex for all the strings in between 'a' and 'b'?

I wrote a function that searches for a given text in the file and returns all the lines where this text appeared (in the new_list). I would like to delete all the text that's between a character '/* ' and ' */' (including these characters), as these are comments, and are not necessary in the returned list. So now it returns something like:

new_list = [ '123456_xxxx /* cccccccccccccc */ /\n", '67890_yyyy /* cccccccccccccc */ /\n", ... ]

and the goal is:

new_list = [ '123456_xxxx", '67890_yyyy", ... ]

I tried using re library and regex (?<=§).*?(?=;)

new_list = re.sub('(?<=/*).*?(?=*/)', '', str(new_list)) but this one gives an error:

error: nothing to repeat at position 13


PS I also tried with replace but it only deletes the characters and not everything between them:

new_list = [s.replace(' /*', ' ') for s in new_list]

import re

list = [ f'123456_xxxx /* cccccccccccccc */ /\n',  f'67890_yyyy /* cccccccccccccc */ /\n']
newList = []

pattern = f'\/\*.*\*\/'
for str in list:
    newList.append( re.sub(pattern, '', str))
    
print(newList)

['123456_xxxx  /\n', '67890_yyyy  /\n']

if you don't want and \n
change the pattern:

pattern = f' \/\*.*\*\/ \/\\n'

['123456_xxxx', '67890_yyyy']

Just use \/\*.*?\*\/ and remember to supply the re.S flag to search across multiple lines.

Both the / and the * are special and need to be prefaced with a backslash () to be taken literally.

You could use re.search with the following regular expression.

^(?:(?!\/\*).)+(?<!\s)

Python demo<-\(ツ)/->Regex demo

(?:(?!\/\*).) matches a single character in the string (.), starting at the beginning of the string (^). (?!\/\*) is a negative lookahead that asserts that the following two characters in the string are not /*. In other words any character is matched until and if it is / and the next character is *. This is called the tempered greedy token technique.

I've added the negative lookbehind (?<!\s) to ensure that the match does not end in a whitespace.

Using index of strings

Code

old_list = ['123456_xxxx /* cccccccccccccc */ /\n',  '67890_yyyy /* cccccccccccccc */ /\n']

new_list = []

for i in old_list:
    new_list.append(i[:10])

print(new_list)

Output

['123456_xxx', '67890_yyyy']