Regular expression to match start of filename and filename extension

问题

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' ....

回答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.

回答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.)

回答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

回答4:

/^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)

回答5:

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"]) )

回答6:

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)

回答7:

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

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

回答8:

mabye:

^Run.*\.py$

just a quick try

来源：https://stackoverflow.com/questions/185378/regular-expression-to-match-start-of-filename-and-filename-extension

标签

python

sql

regex

sql-like