Re: re.search - Pattern matching review ( Apologies re sending)

From	Ganesh Pal <ganesh1pal@gmail.com>
Newsgroups	comp.lang.python
Subject	Re: re.search - Pattern matching review ( Apologies re sending)
Date	2016-05-29 13:06 +0530
Message-ID	<mailman.19.1464507412.1839.python-list@python.org> (permalink)
References	<CACT3xuXhA9U1Cwsw_OUW-Zfu96o-Bd9-OHFas80=M9etB0=bsw@mail.gmail.com> <DUB123-W24AEE854E04581EC90DDE283430@phx.gbl> <CACT3xuU-4oEuwjt8bUA=gkU6c=FdUTYeMrqeDGBpt51p_JjOcg@mail.gmail.com> <CACT3xuW7LaYDOW1z_NzMNmjPxu2cMzh587r358Kxz737PzMUFQ@mail.gmail.com>

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The matched.groups() will group the pattern based with ","

(Pdb) matched.groups()
*('1', '0', '1375772672', '8192')*

but I wanted to retain the  output as  *'1,0,1375772672:8192' ,*

(Pdb) matched.groups()
('1', '0', '1375772672', '8192')
(Pdb) matched.group()
'Block Address for 1,0,1376034816:8192 (block *1,0,1375772672:8192*'


Regards,
Ganesh










On Sun, May 29, 2016 at 11:53 AM, Ganesh Pal <ganesh1pal@gmail.com> wrote:

>
>
>
>> Perhaps:
>> map(int,  re.search(search_pat, stdout).groups())
>>
>>
>>
> Thanks Albert map saved me many lines of code  but map returns a list I
> will have to convert  the list to string again
> Below is how Iam planning to teh conversion
> >>> block = map(int,  re.search(search_pat, stdout).groups())
> >>> print block
> ['1,2:122']
> >>> s1 = ','.join(str(n) for n in block)
> >>> print s1
> 1,2:122
> >>> str(s1)
> '1,2:122'
>
> Regards,
> Ganesh
>

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Re: re.search - Pattern matching review ( Apologies re sending) Ganesh Pal <ganesh1pal@gmail.com> - 2016-05-29 13:06 +0530

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