Thank you to all of you guys.
It's exactly I was looking for.
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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