Regular expression to match start of filename and filename extension

What is the regex to match filenames that start with 'Run' and have a filename extension of '.py'?

It should match any of the following:

RunFoo.py
RunBar.py
Run42.py

It should not match:

myRunFoo.py
RunBar.py1
Run42.txt

(I am a regex novice and am more familiar with sql wildcards so the equivalent of what I am searching for would be "..LIKE 'Run%.py'...")

From stackoverflow

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

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

  Zing- : looks like a Perl solution for a question tagged python... but I am not a python expert :p and as jobscry pointed out your solution is case-sensitive.

  Zing- : *err has Rob Howard pointed out that is

  Brian : Shouldn't you use .*, rather than \w - punctuation and whitespace etc should probably still be considered part of the filename. eg "Run.foo.py"

• /^Run.*\.py$/
  

  Or, in python specifically:

  import re
  re.match(r"^Run.*\.py$", stringtocheck)
  

  This will match "Runfoobar.py", but not "runfoobar.PY". To make it case insensitive, instead use:

  re.match(r"^Run.*\.py$", stringtocheck, re.I)
  

• mabye:

  ^Run.*\.py$
  

  just a quick try

  Brian : You need .*, rather than .? (which will only match a single character)

  jobscry : doh, thanks Brian

• 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.

  Zing- : 1. you don't have to specify start of line for python regular expression match? 2. * is zero or more match (i.e. so Run.py would be acceptable)

  Zing- : Also, how would you make it case-insensitive?

  Brian : re.match already specifies the start of the string (as opposed to re.search, which doesn't). "Run.py" *should* match, given the definition (It starts with Run, and has a .py extension). For case insensitivity, see the note at the end.

• 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.)

  Ates Goral : We'll get you those rep points :)

  Rob Howard : Cripes, that was quick. Thanks. :-)

• 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)

• 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
  

• 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"]) )