Python regex replace string

Question

I'm trying to achieve the following:

string = 'C:/some path to mp3/song (7) title and so on (1).mp3'

should become:

C:/some path to mp3/song (7) title and so on.mp3

To match it i'm using the following regex:

pattern = '.*(\s\([0-9]+\))\.mp3'

And the match group contains: (u' (1)',)
however, when i'm trying to substitute the match like so:

processed = re.sub(pattern, '', string)

processed contains an empty string. How can i get re.sub() to only replace the match found above?

Answer 1

You were matching the entire string and replacing it, use a lookahead and only match the whitespace and (1) before the final extension.

Expanded RegEx:

\s*     (?# 0+ characters of leading whitespace)
\(      (?# match ( literally)
[0-9]+  (?# match 1+ digits)
\)      (?# match ) literally)
(?=     (?# start lookahead)
  \.    (?# match . literally)
  mp3   (?# match the mp3 extension)
  $     (?# match the end of the string)
)       (?# end lookeahd)

Demo: Regex101

Implementation:

pattern = '\s*\([0-9]+\)(?=\.mp3$)'
processed = re.sub(pattern, '', string)

Notes:

• mp3 can be replaced by [^.]+ to match any extension or (mp3|mp4) to match multiple extensions.
• use \s+ instead of \s* to require at least some whitespace before (1), thanks @SethMMorton.