Use the print statement:
import re
vowel = r'[aeiou]'
print re.findall(vowel, r"vowel")
Alexey
On Fri, Aug 8, 2008 at 2:17 PM, Atul. <[EMAIL PROTECTED]> wrote:
>
> > Yes. You didn't paste the traceback into your message.
> >
> > >>> import re
> > >>> vowel = r'[aeiou]'
> > >>> re.findall(vowel,
=)
Indeed. But it will replace all dots including ordinary strings instead of
numbers only.
On Tue, Aug 5, 2008 at 3:23 PM, Jeff <[EMAIL PROTECTED]> wrote:
> On Aug 5, 7:10 am, Marc 'BlackJack' Rintsch <[EMAIL PROTECTED]> wrote:
> > On Tue, 05 Aug 2008 11:39:36 +0100, Fred Mangusta wrote:
> > > I
No, there is a bad way - because of the example doesn't solve arbitrary
amount of ... blocks.
But the python regexp engine supports for lookahead (?=pattern) and
lookbehind (?<=pattern).
In those cases patterns are not included into the replaced sequence of
characters:
>>> re.sub('(?<=\d)\.(?=\d)',
Hello Mike,
The reason of the problem is that the class Test was not pushed into the
sys.modules.
Use one more separate module for that stuff:
*one.py*
class Test(object):
'''just define'''
*three.py*
from one import Test
#push one.pyc to sys.modules
if __name__ == '__main__':
import two
