[BangPypers] list problem

[Shekhar Tiwatne]

On Thursday 22 July 2010 09:36 PM, Anand Balachandran Pillai wrote:
> On Thu, Jul 22, 2010 at 7:00 PM, steve<steve at lonetwin.net>  wrote:
>
>    
>> Hi,
>>
>>
>> On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
>>
>>      
>>> On Thu, Jul 22, 2010 at 3:21 PM, Vikram<kpguy at rediffmail.com>   wrote:
>>>
>>>    Suppose you have the following list:
>>>        
>>>>   >>>   x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>>>>
>>>>   My problem is that i wish to obtain the following two dictionaries:
>>>>   xdictstart = {'cat':10, 'dog':1}
>>>>   xdictend = {'cat':30, 'dog':5}
>>>>
>>>>          
>>>
>>>    Any nice way to do the above? Thanks.
>>>        
>>>>
>>>>   Yes. Try this.
>>>>          
>>>    x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>>>        
>>>>>>   x.sort()
>>>>>>   xdictstart = dict(reversed(x))
>>>>>>   xdictend = dict(x)
>>>>>>   xdictstart,xdictend
>>>>>>
>>>>>>              
>>>>> ({'dog': 1, 'cat': 10}, {'dog': 5, 'cat': 30})
>>>>>            
>>>        
>> Are you sure that'll work ? Is creating a dict from a sequence guaranteed
>> to be executed serially withing the sequence ?
>>      
>
> Yes, it is. Since the list is sorted, the entries are entered into the
> dictionary
> serially (list is ordered), so hence in normal order we find the larger
> value
> overwrites the others and in reversed order, the smaller value.
>
>   I cannot think of a solution which is shorter than this.
>    

Out of curiosity I tried benchmarking few of these solutions alongwith 
the one I wrote using itertools.groupby.
My benchmarking did include large data sets and worst case.
Results are the shortest solution (by Anand) was way faster than others. 
After that it was Steve's and mine and Navin was almost the same of the 
last spot.

Which one is most readable? Well I leave that for now.

Shekhar