On Thu, 02 Jul 2009 09:38:56 -0700, Ethan Furman wrote:
> Greetings!
>
> My closest to successfull attempt:
>
> Python 2.5.4 (r254:67916, Dec 23 2008, 15:10:54) [MSC v.1310 32 bit (Intel)]
> Type "copyright", "credits" or "license" for more information.
>
> IPython 0.9.1 -- An enhanced Interactive Python.
>
> In [161]: re.findall('\d+','this is test a3 attempt 79')
> Out[161]: ['3', '79']
>
> What I really want in just the 79, as a3 is not a decimal number, but
> when I add the \b word boundaries I get:
>
> In [162]: re.findall('\b\d+\b','this is test a3 attempt 79')
> Out[162]: []
>
> What am I missing?
You need to use a raw string (r'...') to prevent \b from being interpreted
as a backspace:
re.findall(r'\b\d+\b','this is test a3 attempt 79')
\d isn't a recognised escape sequence, so it doesn't get interpreted:
> print '\b'
^H
> print '\d'
\d
> print r'\b'
\b
Try to get into the habit of using raw strings for regexps.
--
http://mail.python.org/mailman/listinfo/python-list
