Regular expression to match as less characters as possible

You don't need regex, just use str.rsplit, with maxsplit=1, and take the last item:

>>> s.rsplit('/',1)[-1]
'x.py'

when you want to extract filename from path, you should use os.path.split. The os.path.split() method in Python is used to Split the path name into a pair head and tail independent of OS. Here, tail is the last path name component and head is everything leading up to that.

import os

path = 'a/b/c/x.py'
res = os.path.split(path)
print(res[1])

You can also use normpath and os.sep for this solution:

import os

path = 'a/b/c/x.py'
path = os.path.normpath(path)
res = path.split(os.sep)
print(res[-1])

You can use rsplit as @ThePyGuy said in this case to avoid more splitting by changing the line to:

res = path.rsplit(os.sep,1)

import re
s = 'a/b/c/x.py'
res = re.search('\w*\.py', s).group() # It will match alphanumeric
# res = re.search(r'[\w&.-]+$', s).group() 
# The above regex will match alphanumeric and the given special characters

EDIT

To match everything after the last / you can use following regex

res = re.search('[^/]+$', s).group()

I prefer a splitting approach here:

s = 'a/b/c/x.py'
last = s.split('/')[-1]
print(last)  # x.py

If you need to ensure that the element is is the last in a path, you can prepend (?<=\/), a positive lookbehind:

>>> s = 'a/b/c/x.py'
>>> el = re.search(r"(?<=\/)(\w+\.py)", s).group(1)
>>> el
'x.py'

Otherwise, if you need to match also filename.py, you need to remove it:

>>> s2 = 'file.py'
>>> el = re.search(r"(\w+\.py)", s2).group(1)
>>> el
'file.py'