Regular expression to match start of filename and filename extension

Question

What is the regular expression to match strings (in this case, file names) that start with 'Run' and have a filename extension of '.py'?

The regular expression should match any of the following:

RunFoo.py
RunBar.py
Run42.py

It should not match:

myRunFoo.py
RunBar.py1
Run42.txt

The SQL equivalent of what I am looking for is ... LIKE 'Run%.py' ....

Answer 1

For a regular expression, you would use:

re.match(r'Run.*\.py$') 

A quick explanation:

• . means match any character.
• * means match any repetition of the previous character (hence .* means any sequence of chars)
• \ is an escape to escape the explicit dot
• $ indicates "end of the string", so we don't match "Run_foo.py.txt"

However, for this task, you're probably better off using simple string methods. ie.

filename.startswith("Run") and filename.endswith(".py") 

Note: if you want case insensitivity (ie. matching "run.PY" as well as "Run.py", use the re.I option to the regular expression, or convert to a specific case (eg filename.lower()) before using string methods.

Answer 2

Warning:

• jobscry's answer ("^Run.?.py$") is incorrect (will not match "Run123.py", for example).
• orlandu63's answer ("/^Run[\w]*?.py$/") will not match "RunFoo.Bar.py".

(I don't have enough reputation to comment, sorry.)

Answer 3

I don't really understand why you're after a regular expression to solve this 'problem'. You're just after a way to find all .py files that start with 'Run'. So this is a simple solution that will work, without resorting to compiling an running a regular expression:

import os
for filename in os.listdir(dirname):
    root, ext = os.path.splitext(filename)
    if root.startswith('Run') and ext == '.py':
        print filename