regex question on .findall and \b

[Ethan Furman]

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?

~Ethan~

 

[Nobody]

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.