Fwd: Unit testing type comparison

2008-08-04 Thread Alexei Zankevich 
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
two.run( Test() )

*two.py*
def run( a ):
from one import Test
#reuse existing Test
print type(a), Test, type(a) == Test

Regards,
Alexey

On Sat, Aug 2, 2008 at 5:16 PM, Mike Wyatt <[EMAIL PROTECTED]> wrote:

> I have a problem where type comparisons don't work in a second module when
> unit tests in the first module are run.  In the example below, class Test is
> declared in one.py.  When one.py is executed, it calls a method in two.py
> that imports Test from one.py.  The problem is that the Test object passed
> into the run method is in the "__main__" namespace, while the Test class
> imported from one.py is in the "one" namespace.  Comparing type(a) and Test
> returns False, even though they are both Test objects.  How can I tweak
> these modules so that the type comparison returns True?
>
> I suppose one solution is to redesign my solution to not compare types, but
> I would still appreciate a solution for my personal knowledge.
>
>  one.py 
> class Test(object):
> pass
>
> if __name__ == '__main__':
> import two
> two.run( Test() )
>
>  two.py ***
> *def run( a ):
> from one import Test
> print type(a), Test, type(a) == Test
> *
> *** Output ***
> *  False
>
>
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Re: regex question

2008-08-05 Thread Alexei Zankevich 
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)', '', '1234.324 abc.100.abc abc.abc')
'1234324 abc.100.abc abc.abc'

Alexey

On Tue, Aug 5, 2008 at 2:10 PM, Marc 'BlackJack' Rintsch <[EMAIL 
PROTECTED]>wrote:

> On Tue, 05 Aug 2008 11:39:36 +0100, Fred Mangusta wrote:
>
> > In other words I'd like to replace all the instances of a '.' character
> > with something (say nothing at all) when the '.' is representing a
> > decimal separator. E.g.
> >
> > 500.675  >   500675
> >
> > but also
> >
> > 1.000.456.344 > 1000456344
> >
> > I don't care about the fact the the resulting number is difficult to
> > read: as long as it remains a series of digits it's ok: the important
> > thing is to get rid of the period, because I want to keep it only where
> > it marks the end of a sentence.
> >
> > I was trying to do like this
> >
> > s=re.sub("[(\d+)(\.)(\d+)]","... ",s)
> >
> > but I don't know much about regular expressions, and don't know how to
> > get the two groups of numbers and join them in the sub. Moreover doing
> > like this I only match things like "345.000" and not "1.000.000".
> >
> > What's the correct approach?
>
> In [13]: re.sub(r'(\d)\.(\d)', r'\1\2', '1.000.456.344')
> Out[13]: '1000456344'
>
>
> Ciao,
> Marc 'BlackJack' Rintsch
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Re: regex question

2008-08-05 Thread Alexei Zankevich 
=)
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:
> > > In other words I'd like to replace all the instances of a '.' character
> > > with something (say nothing at all) when the '.' is representing a
> > > decimal separator. E.g.
> >
> > > 500.675  >   500675
> >
> > > but also
> >
> > > 1.000.456.344 > 1000456344
> >
> > > I don't care about the fact the the resulting number is difficult to
> > > read: as long as it remains a series of digits it's ok: the important
> > > thing is to get rid of the period, because I want to keep it only where
> > > it marks the end of a sentence.
> >
> > > I was trying to do like this
> >
> > > s=re.sub("[(\d+)(\.)(\d+)]","... ",s)
> >
> > > but I don't know much about regular expressions, and don't know how to
> > > get the two groups of numbers and join them in the sub. Moreover doing
> > > like this I only match things like "345.000" and not "1.000.000".
> >
> > > What's the correct approach?
> >
> > In [13]: re.sub(r'(\d)\.(\d)', r'\1\2', '1.000.456.344')
> > Out[13]: '1000456344'
> >
> > Ciao,
> > Marc 'BlackJack' Rintsch
>
> Even faster:
>
> '1.000.456.344'.replace('.', '') => '1000456344'
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Re: regular expressions.

2008-08-08 Thread Alexei Zankevich 
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, r"vowel")
> >
> > ['o', 'e']
> >
> > It works as expected here.
> >
> > Peter
>
> When I key this input in IDLE it works but when I try to run the
> module it wont work.
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