Python - string replace between parenthesis with wildcards

Question

I am trying to remove some text from a string. What I want to remove could be any of the examples listed below. Basically any combination of uppercase and lowercase, any combination of integers at the end, and any combination of letters at the end. There could also be a space between or not.

(Disk 1)
(Disk 5)
(Disc2)
(Disk 10)
(Part A)
(Pt B)
(disk a)
(CD 7)
(cD X)

I have a method already to get the beginning "(type"

multi_disk_search = [ '(disk', '(disc', '(part', '(pt', '(prt' ]
if any(mds in fileName.lower() for mds in multi_disk_search): #https://stackoverflow.com/a/3389611
  for mds in multi_disk_search:
    if mds in fileName.lower():
      print(mds)
      break

That returns (disc for example.

I cannot just split by the parenthesis because there could be other tags in other parenthesis. Also there is no specific order to the tags. The one I am searching for is typically last; however many times it is not.

I think the solution will require regex, but I'm really lost when it comes to that.

I tried this, but it returns something that doesn't make any sense to me.

regex = re.compile(r"\s*\%s\s*" % (mds), flags=re.I) #https://stackoverflow.com/a/20782251/11214013
regex.split(fileName)
newName = regex
print(newName)

Which returns re.compile('\\s*\\(disc\\s*', re.IGNORECASE)

What are some ways to solve this?

Casimir et Hippolyte · Answer 1 · 2021-01-31T19:29:47.090

Perhaps something like this:

rx = re.compile(r'''
    \(
     (?: dis[ck] | p(?:a?r)?t )
     [ ]?
     (?: [a-z]+ | [0-9]+ )
     \)''', re.I | re.X)

This pattern uses only basic syntax of regex pattern except eventually the X flag, the Verbose mode (with this one any blank character is ignored in the pattern except when it is escaped or inside a character class). Feel free to read the python manual about the re module. Adding support for CD is let as an exercise.

Matt Miguel · Answer 2 · 2021-01-31T22:33:40.930

0

>>> import re
>>> def remove_parens(s,multi_disk_search):
...     mds = '|'.join([re.escape(x) for x in multi_disk_search])
...     return re.sub(f'\((?:{mds})[0-9A-Za-z ]*\)','',s,0,re.I)
...

>>> multi_disk_search = ['disk','cd','disc','part','pt']
>>> remove_parens('this is a (disc a) string with (123xyz) parens removed',multi_disk_search)
'this is a  string with (123xyz) parens removed'

edited Jan 31 '21 at 22:33

answered Jan 31 '21 at 19:15

Matt Miguel

1,325
3
6

I need to keep the (123xyx) parenthesis, and it could appear before the (disc a) parenthesis. – ReenigneArcher Jan 31 '21 at 20:49
@Archer Because it starts with a number? I modified it to only remove parentheses if the first character inside is a letter. The order doesn't matter for re.sub – Matt Miguel Jan 31 '21 at 21:43
Sorry, that won't work. I need to specifically only remove the items in my multi_disk_search list. I'll read up regex. Thanks – ReenigneArcher Jan 31 '21 at 22:12
Is the list you provided a full list or just a subset of examples? – Matt Miguel Jan 31 '21 at 22:24
1

@Archer changed it to have an explicit list of prefixes to remove – Matt Miguel Jan 31 '21 at 22:35

Python - string replace between parenthesis with wildcards

2 Answers2