regex question on .findall and \b
nobody at nowhere.com
Thu Jul 2 14:41:54 EDT 2009
On Thu, 02 Jul 2009 09:38:56 -0700, Ethan Furman wrote:
> 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 : re.findall('\d+','this is test a3 attempt 79')
> Out: ['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 : re.findall('\b\d+\b','this is test a3 attempt 79')
> Out: 
> 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'
> print '\d'
> print r'\b'
Try to get into the habit of using raw strings for regexps.
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