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

[Ganesh Pal]

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 at 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
>