list 2 dict?

[Octavian Rasnita]

From: "Rob Williscroft" <rtw at rtw.me.uk>

> Octavian Rasnita wrote in news:0DB6C288B2274DBBA5463E7771349EFB at teddy in
> gmane.comp.python.general: 
> 
>> Hi,
>> 
>> If I want to create a dictionary from a list, is there a better way
>> than the long line below? 
>> 
>> l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b']
>> 
>> d = dict(zip([l[x] for x in range(len(l)) if x %2 == 0], [l[x] for x
>> in range(len(l)) if x %2 == 1])) 
>> 
>> print(d)
>> 
>> {8: 'b', 1: 2, 3: 4, 5: 6, 7: 'a'}
> 
>>>> dict( zip( l[ :: 2 ], l[ 1 :: 2 ] ) )
> {8: 'b', 1: 2, 3: 4, 5: 6, 7: 'a'}
> 
> If you don't know about slice notation, the synatax I'm using above is:
> 
>    list[ start : stop : step ]
> 
> where I have ommited the "stop" item, which defaults to the length of the
> list.
> 
> 
> <http://docs.python.org/library/stdtypes.html#sequence-types-str-unicode-
> list-tuple-bytearray-buffer-xrange>
> 
> That will make 3 lists before it makes the dict thought, so if the
> list is large:
> 
> >>> dict( ( l[ i ], l[ i + 1 ] ) for i in xrange( 0, len( l ), 2 ) )
> 
> may be better.


Thank you all.

I have also discovered that I can zip 2 lists made with range(0, len(l), 2) and range(1, len(l), 2) but I remembered about that the slice notation accepts that third argument and as Stefan suggested, looks to be a shorter way.

I have first thought to the solution you suggested, but I have forgotten to create a tuple from the pair of elements so it didn't work.
I wasn't thinking to performance, but yes, it may be important for large lists.

It seems that in some cases there are more ways to do it in Python than in Perl. :-)

Octavian