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 s

Answer 1

If you write a slightly more complex regular expression, you can get an extra feature: extract the bit between "Run" and ".py":

>>> import re
>>> regex = '^Run(?P<name>.*)\.py$'
>>> m = re.match(regex, 'RunFoo.py')
>>> m.group('name')
'Foo'

(the extra bit is the parentheses and everything between them, except for '.*' which is as in Rob Howard's answer)

Answer 2

This probably doesn't fully comply with file-naming standards, but here it goes:

/^Run[\w]*?\.py$/

Answer 3

You don't need a regular expression, you can use glob, which takes wildcards e.g. Run*.py

For example, to get those files in your current directory...

import os, glob
files = glob.glob( "".join([ os.getcwd(), "\\Run*.py"]) )

Answer 4

mabye:

^Run.*\.py$

just a quick try

Answer 5

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 6

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