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 1,0,1376034816:81
> 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())
>>> p
> Date: Sat, 28 May 2016 23:48:16 +0530
> Subject: re.search - Pattern matching review ( Apologies re sending)
> From: [email protected]
> To: [email protected]
>
> Dear Python friends,
>
> I am on Python 2.7 and Linux . I am trying to extract the address
> "1,5,147456:8192" from the bel
