[BangPypers] How to sort the IP(s)

[venkata subramanian m]

Anand Balachandran Pillai wrote:
> Do you really need any kind of additional processing ?
>
>   
But, string sorting might not produce proper sorting right?

 >>> l=['192.168.11.1','172.18.13.2','192.168.2.2','172.19.2.1']
 >>> l.sort()
 >>> l
['172.18.13.2', '172.19.2.1', '192.168.11.1', '192.168.2.2']


But,
['172.18.13.2', '172.19.2.1', '192.168.2.2', '192.168.11.1']
is the expected order, I guess.

Regards,
Venkat



> The basic sort algorithm is smart enough to do this by itself,
>   
>>>> l=['192.168.1.1','172.18.13.2','192.168.3.2','172.19.2.1']
>>>> l.sort()
>>>> l
>>>>         
> ['172.18.13.2', '172.19.2.1', '192.168.1.1', '192.168.3.2']
>
> or use sorted(...) if you don't want to modify in place...
>
> Here is an even closer example to demo this...
>   
>>>> l=['192.168.12.21','192.168.12.15','192.168.11.10','192.168.10.5','192.168.15.1','192.167.10.1']
>>>> sorted(l)
>>>>         
> ['192.167.10.1', '192.168.10.5', '192.168.11.10', '192.168.12.15',
> '192.168.12.21', '192.168.15.1']
>
>
> --Anand
>
> On Thu, May 8, 2008 at 2:41 PM, Anand Chitipothu <anandology at gmail.com> wrote:
>   
>> On Thu, May 8, 2008 at 11:53 PM, Kushal Das <kushaldas at gmail.com> wrote:
>>
>>     
>>> Hi,
>>>       
>>  >  What is the best way to sort IP numbers
>>  >  numbers like
>>  >  192.168.20.1
>>  >  192.168.1.1
>>  >  172.18.13.2
>>
>>  ips = ['192.168.20.1', '192.168.1.1', '172.18.13.2']
>>  sorted(ips, key=lambda ip: [int(x) for x in ip.split('.')])
>>
>>  # ips.sort(key=lambda ip: [int(x) for x in ip.split('.')]) if you want
>>  to sort in-place.
>>
>>
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>>     
>
>
>
>   

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