On Sat, Jan 31, 2015 at 1:27 PM, <rajanb...@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
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