Create dictionary based of x items per key from two lists

[Mark Lawrence]

On 31/01/2015 02:38, Chris Angelico wrote:
> On Sat, Jan 31, 2015 at 1:27 PM,  <rajanbond at gmail.com> wrote:
>> l1 =  ["a","b","c","d","e","f","g","h","i","j"]
>> l2 = ["aR","bR","cR"]
>>
>> l2 will always be smaller or equal to l1
>>
>> numL1PerL2 = len(l1)/len(l2)
>>
>> I want to create a dictionary that has key from l1 and value from l2 based on numL1PerL2
>>
>> So
>>
>> {
>> a:aR,
>> b:aR,
>> c:aR,
>> d:bR,
>> e:bR,
>> f:bR,
>> g:cR,
>> h:cR,
>> i:cR,
>> j:cR
>> }
>>
>> So last item from l2 is key for remaining items from l1
>
> So the Nth element of l1 will always be paired with the
> (N/numL1PerL2)th element of l2 (with the check at the end)? Seems easy
> enough.
>
> dups = len(l1)/len(l2)
> l2.append(l2[-1])
> result = {x:l2[i/dups] for i,x in enumerate(l1)}
>
> This mutates l2 for convenience, but you could also adjust the index
> to take care of the excess. As a one-liner:
>
> result = {x:l2[min(i/(len(l1)/len(l2)),len(l2)-1)] for i,x in enumerate(l1)}
>
> But the one-liner is not better code :)
>
> ChrisA
>

The one-liner might not be better code, but it must be better speed wise 
precisely because it's on one line, right? :)

-- 
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence