Thank you to all of you guys.
It's exactly I was looking for.
> You provide scant information for this task. For example, is the
> new list ordered or unordered? Can the list corresponding to the
> 'transports' key contain duplicates?
>
unordered is fine.. I will be storing it in DB from that list.
It should not happen to get 'transports' duplicates in my application.
It's for testing domains where are running SIP servers.
Lada
Bart Ogryczak wrote:
> On 12 jul, 04:49, anethema <[EMAIL PROTECTED]> wrote:
>
>>> li = [ {'index': 0, 'transport': 'udp', 'service_domain':
>>> 'dp0.example.com'},
>>> {'index': 1, 'transport': 'udp', 'service_domain':
>>> 'dp1.example.com'},
>>> {'index': 0, 'transport': 'tcp', 'service_domain':
>>> 'dp0.example.com'},
>>> {'index': 1, 'transport': 'tcp', 'service_domain':
>>> 'dp1.example.com'}]
>>>
>> I like this solution:
>>
>> [{ 'transports' : [d['transport'] for d in li if
>> d['service_domain'] == dom],
>> 'service_domain': dom,
>> } for dom in set(d2['service_domain'] for d2 in li)]
>>
>> merely because it takes one line. Humorously enough, it appears to be
>> twice as efficient,
>>
>
> Correct me if I´m wrong, that is a O(n**2) solution, to O(n) problem.
>
>
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